Electrochemistry: Class 12 Chemistry Chapter 2 Hinglish mein

Chemistry Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

125 min
1 simulation
38 practice questions
41 MCQs

Is chapter mein aap kya seekhenge

Galvanic aur electrolytic cells ko samjhao, cell notation likho, aur anode, cathode aur electron flow ki direction pehchano
Standard electrode potentials se cell emf nikaalo aur predict karo ki koi redox reaction spontaneous hai ya nahi
Nernst equation lagakar kisi bhi concentration par cell aur electrode potentials nikaalo
Cell ke standard potential ko uski reaction ke standard Gibbs energy change aur equilibrium constant se jodo
Resistivity, conductivity, molar conductivity aur cell constant define karo, aur samjhao ki conductivity kaise measure hoti hai aur concentration ke saath kaise badalti hai
Kohlrausch's law batao aur usse limiting molar conductivities aur weak electrolytes ke dissociation constants nikaalo
Faraday's laws lagao aur electrolysis ke products predict karo
Samjhao ki batteries aur fuel cells kaise kaam karte hain, aur corrosion ko ek electrochemical process ki tarah samjhao aur batao ki use kaise roka jaata hai

NCERT Chemistry Class 12, Chapter 2, pp. 31–60 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Tumhari phone battery screen jalaane ke liye ek chemical reaction ko aage chalati hai, aur raat bhar charger usi reaction ko ulta chalata hai. Chemical energy aur electrical energy ke beech yeh do-taraf ka traffic hi electrochemistry (विद्युत रसायन) hai.

Yeh tumhe poore chapter mein milega: tumhari jeb ke lithium-ion cell mein, un brine cells mein jo water treatment ke liye chlorine banate hain, railing par lage rust mein, aur tumhari apni nerves mein bhi, jahan cell membranes ke aar-paar jaate ions signals le jaate hain. Hum ideas ko NCERT ke order mein banayenge: cells aur electrode potentials, Nernst equation, solutions kaise conduct karte hain, electrolysis, batteries, fuel cells aur corrosion.

2.1 Electrochemical Cells

Class 11 mein tum Daniell cell se mil chuke ho: zinc sulphate solution mein ek zinc rod aur copper sulphate solution mein ek copper rod, jo bahar ek wire se aur andar ek salt bridge se jude hain. Zinc jo electrons chhodta hai, woh wire se hokar copper ions tak jaate hain, isliye redox reaction

\begin{aligned} Z n (s) + C u^{2 +} (a q) \\ \overset{}{\to} Z n^{2 +} (a q) \\ + C u (s) \end{aligned}

(2.1)

beaker ko sirf garam karne ki jagah electrical work karti hai. Jab dono ions 1 mol L⁻¹ par hon, to cell 1.10 V deta hai.Yahan concentration asal mein har ion ki activity ki jagah use ho raha hai; activity ek tarah ka 'effective concentration' hai jo sirf dilute solutions mein asli concentration se kaafi milta hai. Aisa device, jismein apne aap chalne wali reaction electrons ko circuit mein ghumati hai, galvanic (voltaic) cell (गैल्वेनी सेल) kehlata hai.

Ab ek adjustable power supply aise jodo ki woh cell ke against push kare, aur uska voltage $E_{ext}$ zero se dheere-dheere badhao (Fig. 2.1). 1.10 V se neeche cell hi jeetta hai: electrons zinc ko chhodte hain, zinc ghulta hai aur copper jamta hai, bas dheere. Theek 1.10 V par dono push cancel ho jaate hain, current ruk jaata hai aur reaction bhi. 1.10 V se upar supply jeetti hai. Current ulti taraf behta hai, zinc ions wapas zinc metal ban jaate hain aur copper ghulne lagta hai. Wahi hardware ab ek electrolytic cell hai: electrical energy ek non-spontaneous reaction ko zabardasti chala rahi hai.

3/3Supply 1.10 V se zyada: current ulta ho jaata hai aur cell electrolytic ban jaata hai.

Fig. 2.1Opposing voltage badhane par Daniell cell pehle dheema hota hai, phir rukta hai, phir ulta chalne lagta hai.NCERT Fig. 2.2 · p. 32

Ek loop circuit left mein 1.10 V mark kiye Daniell cell ko right mein ek adjustable supply se jodta hai; ek dial supply voltage dikhata hai. Pehle step mein dial 0.80 V padhta hai, ek arrow zinc side se nikalte electrons dikhata hai, aur notes kehte hain ki zinc ghulta hai aur copper jamta hai. Doosre step mein dial 1.10 V padhta hai, arrow gayab hai aur ek note kehta hai na current, na reaction. Teesre step mein dial 1.40 V padhta hai, electron arrow ulti taraf point karta hai aur notes kehte hain ki zinc jamta hai aur copper ghulta hai, yaani device ab electrolytic cell hai.

Definition

Galvanic (voltaic) cell

Aisa cell jismein ek spontaneous redox reaction electrical energy paida karti hai.

ExampleBulb jalata torch cell.

Definition

Electrolytic cell

Aisa cell jismein electrical energy ka external source ek non-spontaneous reaction chalata hai.

ExampleRecharge hoti battery, ya silver-plate hota chammach.

Compare

Electrochemical cell ke do prakaar

Galvanic cell

chemical energy → electrical energy
reaction spontaneous hai ( $Δ_{r} G < 0$ )
anode negative terminal hai
dry cell, fuel cell, use ho rahi battery

Electrolytic cell

electrical energy → chemical energy
external supply ek non-spontaneous reaction chalati hai
anode positive terminal se juda hota hai
electroplating, copper ki refining, charge ho rahi battery

2.2 Galvanic Cells

Galvanic cell ek redox reaction se uski Gibbs energy heat ki jagah electrical work ke roop mein nikalwata hai. Copper sulphate solution mein zinc dust daalo to wahi reaction chalti hai, lekin beaker bas garam hota hai, kyunki electrons zinc atoms se seedhe unhe chhoote copper ions par kood jaate hain. Dono partners ko alag rakho aur electrons ko ek external wire se guzarne par majboor karo, to woh wire ek LED jala sakti hai ya chhoti motor ghuma sakti hai.

Har cell reaction ko do half-reactions mein toda ja sakta hai, har electrode par ek. Daniell cell ke liye yeh hain

reduction, copper electrode par

\begin{aligned} C u^{2 +} (a q) + 2 e^{-} \\ \overset{}{\to} C u (s) \end{aligned}

(2.2)

oxidation, zinc electrode par

\begin{aligned} Z n (s) \\ \overset{}{\to} Z n^{2 +} (a q) + 2 e^{-} \end{aligned}

(2.3)

Inhe jodo to electrons cancel ho jaate hain aur reaction ((2.1)) wapas mil jaati hai. Har half-reaction apne compartment mein chalti hai, jise half-cell kehte hain; ise redox couple bhi kehte hain aur oxidised form/reduced form likhte hain: $C u^{2 +} / C u$ , $Z n^{2 +} / Z n$ . Koi bhi do half-cells jodo aur ek naya cell ban jaata hai, isliye possible combinations hazaaron mein hain. Wiring hamesha ek jaisi hoti hai (Fig. 2.2): meter (aur switch) wali wire dono electrodes ko jodti hai, aur ek salt bridge (लवण सेतु) dono solutions ko jodta hai. Bridge ki zaroorat sirf tab nahi hoti jab dono electrodes ek hi electrolyte share kar sakein, jaise Section 2.6.2 ki lead storage battery mein.

3/3 $Z n^{2 +}$ left solution mein aata hai aur $C u^{2 +}$ right wale se nikalta hai; bridge anions ko left aur cations ko right bhejta hai.

Fig. 2.2Ek Daniell cell. Electrons wire se guzarte hain; ions solutions aur salt bridge mein charge le jaate hain.NCERT Fig. 2.1 · p. 32

Do beakers aamne-saamne rakhe hain, left mein zinc sulphate mein zinc aur right mein copper sulphate mein copper; ek ulta U-shaped salt bridge dono mein dooba hai aur rods ko jodti wire par ek voltmeter hai. Pehla step parts ke labels dikhata hai. Doosra step zinc ko anode mark karta hai jahan oxidation hota hai, negative, aur copper ko cathode jahan reduction hota hai, positive, aur electrons wire se zinc se copper ki taraf jaate hain. Teesra step ions dikhata hai: zinc ions zinc rod se nikal rahe hain, copper ions copper rod par jam rahe hain, aur salt bridge nitrate ions ko zinc side ki taraf aur potassium ions ko copper side ki taraf bhej raha hai.

Salt bridge ek tube hai jismein koi unreactive electrolyte, jaise $K N O_{3}$ ya $K C l$ , agar jelly mein jamaya hota hai. Jaise-jaise zinc ions left beaker mein aate hain aur copper ions right wale se nikalte hain, left solution mein jaldi hi positive excess aur right mein negative excess ho jaata, aur flow ruk jaata. Bridge se nikalte ions aisa hone nahi dete. Anions zinc side ki taraf aur cations copper side ki taraf jaate hain, dono solutions neutral rehte hain aur circuit poora hota hai.

Voltage aakhir aata hi kyun hai? Zinc rod ko zinc sulphate solution mein dubao. Kuch zinc atoms $Z n^{2 +}$ banakar paani mein chale jaate hain, har ek metal mein do electrons chhod deta hai, jabki kuch $Z n^{2 +}$ ions ulta karte hain. Zinc ke liye pehla process thoda bhari padta hai, isliye rod par electrons ka halka excess ho jaata hai aur uske paas liquid mein positive ions ki ek parat jama ho jaati hai: ek charged double layer, ek microscopic capacitor jaisi. Copper sulphate mein copper ke liye balance doosri taraf jhukta hai, aur rod thoda positive ho jaata hai. Is double layer ke across potential difference hi half-cell ka electrode potential hai.

Electrode potentials hamesha reduction ki tarah likhi half-reaction ke liye tabulate kiye jaate hain (IUPAC convention), isiliye purani books inhe 'reduction potentials' kehti hain. Jab har species apne standard state mein ho, yaani dissolved species 1 mol L⁻¹ par, gases 1 bar par aur solids aur liquids pure, to measured value standard electrode potential, $E^{\circ}$ , kehlati hai.

Definition

Standard electrode potential

Reduction ki tarah likhe half-cell ka potential, jab uski saari species apne standard states mein hon, standard hydrogen electrode ke against measure kiya gaya.

Example $E_{C u^{2 +} / C u}^{\circ} = + 0.34 V$

Kaunsa electrode kaunsa hai? Electrons ko follow karo. Oxidation unhe chhodta hai, isliye jis electrode par yeh hota hai, yaani anode, wahan electrons ka surplus jama hota hai aur woh galvanic cell ka negative terminal hai. Reduction cathode par electrons use karta hai, isliye woh positive hai. Circuit band hote hi electrons anode se wire ke through nikalte hain aur cathode par pahunchte hain; conventional current, apne purane sign convention ke kaaran, ulti taraf dikhaya jaata hai.

Har baar 'zinc sulphate solution mein zinc rod, jo salt bridge se copper sulphate solution mein copper rod se juda hai' likhna jaldi hi bore karta hai, isliye chemists cell ko ek line mein likhte hain:

anode half-cell pehle, left mein, aur cathode half-cell aakhir mein, right mein likho;
ek single vertical line, |, do phases ke beech ki boundary mark karti hai, jaise metal aur solution;
double vertical line, ||, salt bridge mark karti hai;
concentrations ya pressures har species ke baad brackets mein jaate hain.

Is tarah padho to cell ka potential right-hand electrode ka reduction potential minus left-hand wale ka hai: cathode minus anode. Jo cell usi tarah likha ho jaise woh sach mein chalta hai, uske liye answer positive aata hai:

\begin{array}{c} E_{cell} = E_{right} - E_{left} \\ = E_{cathode} - E_{anode} \end{array}

(2.7)

Example ke liye ek cell lo jismein lead silver ions ko reduce karta hai, $P b (s) + 2 A g^{+} (a q)$ $\overset{}{\to} P b^{2 +} (a q) + 2 A g (s)$ . Lead oxidise hota hai, isliye woh anode hai aur left mein jaata hai: $P b (s) ∣ P b^{2 +} (a q) ∣ ∣ A g^{+} (a q) ∣ A g (s)$ . Har ion 1 mol L⁻¹ par ho to $E_{cell}^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{P b^{2 +} / P b}^{\circ} = 0.80 V - (- 0.13 V) = 0.93 V$ . $A g^{+}$ ke aage ka 2 uske electrode potential ko double nahi karta. Potential per unit charge energy hai, isliye yeh is baat par depend nahi karta ki equation kitne electrons ke liye likhi gayi hai.

Terminals par laga voltmeter asal mein kya dikhata hai? Unke beech ka potential difference cell potential hai, lekin meter poora $E_{cathode} - E_{anode}$ sirf tab padhta hai jab cell lagbhag koi current nahi de raha. Current behte hi voltage ka kuch hissa cell ke apne internal resistance se charge chalane mein lag jaata hai, aur reading gir jaati hai (Fig. 2.3). Zero-current wali value cell ka emf (विद्युत वाहक बल) hai, aur tables aur calculations mein $E_{cell}$ ka matlab yahi hota hai.

SimulationEmf aur terminal voltage

Ek simple circuit jismein 4 ohm internal resistance wala 1.10 volt ka cell, ek switch, ek ammeter aur lamp label wala 10 ohm resistor hai. Voltmeter ko cell ya lamp ke across lagaya ja sakta hai, aur readouts current aur terminal voltage dikhate hain. Play dabane par electrons negative terminal se circuit mein ghoomte dikhte hain. Sliders cell voltage aur resistance badalte hain.

Fig. 2.34 Ω internal resistance wala 1.10 V ka cell ek 10 Ω lamp chalata hai. Voltmeter ko cell ke across set karo, phir switch kholo aur band karo.

Try karo

Voltmeter ko cell ke across lagao aur switch kholo. Woh 1.10 V padhta hai: current nahi hai, to tum emf measure kar rahe ho.
Switch band karo. Reading lagbhag 0.79 V tak gir jaati hai, kyunki cell ke andar 4 Ω se behta 0.079 A current 0.31 V kha jaata hai.
Lamp ka resistance 100 Ω tak badhao. Current ghatta hai aur reading wapas 1.10 V ki taraf chadhti hai. Emf ko (lagbhag) zero current par hi kyun measure karna chahiye?

Ghar par try karo

Ek lemon cell

Tumhe chahiye

2 nimbu (lemons)
2 galvanised nails aur 1 plain iron nail
copper ki 2 strips (ya moti copper wire)
ek digital multimeter
crocodile clips wale connecting wires

Pehle guess karo

Copper–zinc lemon cell mein kaunsa metal negative terminal hoga?

copperzinckoi nahi: lemon juice hi terminal hai

Ek galvanised (zinc-coated) nail aur saaf copper ki ek strip nimbu mein lagbhag 2 cm ki doori par ghusao, dhyaan rahe ki dono chhoo na jayein.
Digital multimeter ko uski 2 V DC range par set karo. Red lead copper se aur black lead nail se chhuao, aur reading note karo.
Do lemon cells ko series mein jodo (ek ka copper doosre ki nail se) aur free ends ke across measure karo.
Ek nimbu mein galvanised nail ki jagah plain iron nail lagao aur phir se measure karo.

Table 2.1Lemon cell readings (measure karne se pehle har ek ke aage apna prediction likho)

Set-up	Reading (V)
Cu + Zn, ek nimbu
series mein do nimbu
Cu + Fe, ek nimbu

Tumhari entries apne aap save hoti hain.

Copper aur zinc lagbhag 0.9 V dete hain: tumhari ummeed ke 1.10 V se thoda kam, kyunki lemon juice kisi bhi ion ka 1 mol L⁻¹ solution nahi hai aur cell ka internal resistance zyada hai. Series mein do nimbu lagbhag double dete hain. Copper aur plain iron kam dete hain, lagbhag 0.4–0.6 V, kyunki iron aur copper ke electrode potentials zinc aur copper ki tulna mein ek doosre ke zyada kareeb hain. Meter positive reading tabhi dikhata hai jab uski red lead zyada positive metal par ho, jo har case mein copper hai.

Quick check

3 mein se 1

Cell

N i (s) ∣ N i^{2 +} (a q) ∣ ∣ A g^{+} (a q) ∣ A g (s)

mein external wire ke electrons behte hain…

Ag se Ni ki tarafsalt bridge se hokar Ni se Ag ki tarafNi se Ag ki tarafpehle ek taraf, phir doosri taraf

Answer dekho

Answer (C) Ni se Ag ki taraf

Left wala nickel anode hai: woh oxidise hota hai aur electrons chhodta hai, jo wire se hokar silver cathode tak jaate hain. Salt bridge ions le jaata hai, electrons kabhi nahi.

2.2.1 Electrode Potential ka Measurement

Voltmeter ke do terminals hote hain, isliye woh kabhi akele ek half-cell ka potential measure nahi kar sakta; sirf do ke beech ka difference measure ho sakta hai. Iska hal wahi hai jo hum heights ke liye use karte hain. Pahadon ki height sea level se naapi jaati hai, jo ek agreed zero hai. Electrode potentials standard hydrogen electrode (SHE) ke against measure kiye jaate hain, jise har temperature par theek 0 V potential diya gaya hai:

\begin{aligned} H^{+} (a q) + e^{-} \\ \overset{}{\to} \frac{1}{2} H_{2} (g) \\ E^{\circ} = 0 V \end{aligned}

SHE ko ek aisi surface chahiye jahan hydrogen gas aur hydrogen ions jaldi se electrons ka len-den kar sakein. Yeh surface ek platinum foil hai jis par finely divided platinum ('platinum black') ki coating karke use khurdura banaya gaya hai, jisse bahut bada area milta hai aur woh catalyst ki tarah kaam karti hai. Foil $[H^{+}] = 1 m o l L^{- 1}$ wale acid mein rakhi hoti hai aur uske upar 1 bar par pure hydrogen behti hai. Short mein: $P t (s) ∣ H_{2} (g, 1 b a r) ∣ H^{+} (a q, 1 M)$ (Fig. 2.4).

2/3Copper cathode hai: meter $+ 0.34$ V padhta hai, isliye $E_{C u^{2 +} / C u}^{\circ} = + 0.34$ V.

Fig. 2.4Kisi half-cell ko SHE ke saath jodne se uska standard electrode potential seedha mil jaata hai, sign ke saath.NCERT Fig. 2.3 · p. 34

Left mein 1 M acid ke beaker mein platinised platinum foil ke chaaron taraf ek glass hood hai, aur andar 1 bar par hydrogen bubble ho rahi hai. Ek wire voltmeter se hokar right mein doosre half-cell tak jaati hai. Pehla step hydrogen electrode ko akele uske labels ke saath dikhata hai. Doosre step mein right half-cell 1 M copper ions mein copper hai; meter plus 0.34 V padhta hai aur electrons hydrogen electrode se copper ki taraf behte hain. Teesre step mein right half-cell 1 M zinc ions mein zinc hai; meter minus 0.76 V padhta hai kyunki ab electrons zinc se hydrogen electrode ki taraf behte hain.

Kisi bhi aur couple ka standard potential nikaalne ke liye ek cell banao jismein SHE left mein ho aur unknown half-cell, standard conditions par, right mein. Tab $E_{cell}^{\circ} = E_{R}^{\circ} - E_{L}^{\circ} = E_{R}^{\circ} - 0 = E_{R}^{\circ}$ : reading hi electrode potential hai. Copper ke saath meter $+ 0.34$ V dikhata hai aur copper cathode hai, isliye $E_{C u^{2 +} / C u}^{\circ} = + 0.34 V$ . Zinc ke saath cell asal mein ulta chalta hai: zinc oxidise hota hai aur platinum par hydrogen gas nikalti hai. SHE ko left mein likhne par emf isliye negative, $- 0.76$ V, aata hai, aur $E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V$ .

Lab mein try karo: dilute hydrochloric acid ki do tubes mein zinc granules aur copper turnings daalo. Zinc mein bubbles uthte hain; copper waisa hi pada rehta hai. Signs ne yahi predict kiya tha. Zinc ka negative $E^{\circ}$ batata hai ki woh hydrogen se zyada aasaani se electrons chhodta hai, isliye $H^{+}$ unhe le leta hai aur hydrogen bubbles nikalti hain. Copper ka positive $E^{\circ}$ batata hai ki $C u^{2 +}$ ko $H^{+}$ se zyada electrons chahiye, isliye $H^{+}$ copper ko oxidise nahi kar sakta. (Copper nitric acid mein ghulta hai, lekin wahan nitrate ion oxidising agent hai.)

Jab dono values pata hon, to Daniell cell ke liye naye measurement ki zaroorat nahi: $E_{cell}^{\circ} = 0.34 V - (- 0.76 V) = 1.10 V$ .

Kuch couples ka apna koi metal nahi hota, kyunki dono partners ions, gases ya dissolved molecules hote hain. Woh ek inert electrode udhaar lete hain, aksar platinum (kabhi gold ya graphite), sirf milne ki jagah ke roop mein: metal khud nahi badalta, woh bas dissolved partners aur wire ke beech electrons pass karta hai. SHE ke alawa examples hain chlorine electrode, $P t (s) ∣ C l_{2} (g) ∣ C l^{-} (a q)$ , aur iron(III)/iron(II) electrode, $P t (s) ∣ F e^{3 +} (a q), F e^{2 +} (a q)$ , jismein dono ions ek hi solution mein hain.

Is tarah measure karke saikdon couples ke standard potentials tabulate kiye gaye hain. Table 2.2 mein kuch chune hue diye hain, sabse zyada $E^{\circ}$ se sabse kam tak.

Table 2.2Kuch chune hue standard electrode potentials (298 K). Dissolved ions aqueous hain; (s) solid aur (g) gas mark karta hai.NCERT Table 2.1 · p. 37

Reduction half-reaction	$E^{\circ}$ / V
$F_{2} (g) + 2 e^{-} \overset{}{\to} 2 F^{-}$	+2.87
$S_{2} O_{8}^{2 -} + 2 e^{-} \overset{}{\to} 2 S O_{4}^{2 -}$	+2.01
$H_{2} O_{2} + 2 H^{+} + 2 e^{-} \overset{}{\to} 2 H_{2} O$	+1.78
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-}$ $\overset{}{\to} M n^{2 +} + 4 H_{2} O$	+1.51
$C l_{2} (g) + 2 e^{-} \overset{}{\to} 2 C l^{-}$	+1.36
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-}$ $\overset{}{\to} 2 C r^{3 +} + 7 H_{2} O$	+1.33
$O_{2} (g) + 4 H^{+} + 4 e^{-} \overset{}{\to} 2 H_{2} O$	+1.23
$B r_{2} + 2 e^{-} \overset{}{\to} 2 B r^{-}$	+1.09
$N O_{3}^{-} + 4 H^{+} + 3 e^{-}$ $\overset{}{\to} N O (g) + 2 H_{2} O$	+0.96
$A g^{+} + e^{-} \overset{}{\to} A g (s)$	+0.80
$F e^{3 +} + e^{-} \overset{}{\to} F e^{2 +}$	+0.77
$I_{2} (s) + 2 e^{-} \overset{}{\to} 2 I^{-}$	+0.54
$O_{2} (g) + 2 H_{2} O + 4 e^{-} \overset{}{\to} 4 O H^{-}$	+0.40
$C u^{2 +} + 2 e^{-} \overset{}{\to} C u (s)$	+0.34
$2 H^{+} + 2 e^{-} \overset{}{\to} H_{2} (g)$	0.00
$P b^{2 +} + 2 e^{-} \overset{}{\to} P b (s)$	−0.13
$S n^{2 +} + 2 e^{-} \overset{}{\to} S n (s)$	−0.14
$N i^{2 +} + 2 e^{-} \overset{}{\to} N i (s)$	−0.25
$F e^{2 +} + 2 e^{-} \overset{}{\to} F e (s)$	−0.44
$Z n^{2 +} + 2 e^{-} \overset{}{\to} Z n (s)$	−0.76
$2 H_{2} O + 2 e^{-} \overset{}{\to} H_{2} (g) + 2 O H^{-}$	−0.83
$A l^{3 +} + 3 e^{-} \overset{}{\to} A l (s)$	−1.66
$M g^{2 +} + 2 e^{-} \overset{}{\to} M g (s)$	−2.36
$N a^{+} + e^{-} \overset{}{\to} N a (s)$	−2.71
$K^{+} + e^{-} \overset{}{\to} K (s)$	−2.93
$L i^{+} + e^{-} \overset{}{\to} L i (s)$	−3.04

Har row ko electrons ke liye ek tug-of-war samjho, jiska score $E^{\circ}$ hai. Row jitni upar, uski left-hand species utni zor se kheenchti hai, isliye sabse acche oxidising agents top left mein hain ( $F_{2}$ , acidified $M n O_{4}^{-}$ , $C l_{2}$ ). Row jitni neeche, uski right-hand species utni aasaani se electrons chhodti hai, isliye sabse acche reducing agents bottom right mein hain (Li, K, Na). Har strong player ka partner kamzor hota hai: fluoride ion mushkil se electrons chhodta hai, aur $L i^{+}$ mushkil se unhe leta hai. Negative $E^{\circ}$ wale couples $H_{2}$ se behtar reducing agents hain; positive wale kamzor.

Isse left ke kisi oxidising agent aur right ke kisi reducing agent ke beech reaction ka quick test mil jaata hai. Standard conditions mein reaction spontaneous hoti hai jab oxidising agent table mein reducing agent se upar ho, kyunki tab $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ positive hota hai (Fig. 2.5).

2/4 $C u^{2 +}$ Zn se upar hai, isliye copper ions zinc ko oxidise karte hain: $E^{\circ} = + 1.10$ V.

Fig. 2.5Oxidising agent ladder par apne neeche ke kisi bhi reducing agent ke saath react karta hai.

Das steps (rungs) upar +2.87 V par fluorine se neeche −3.04 V par lithium tak jaate hain; oxidised forms left mein aur reduced forms right mein hain. Side ke arrows batate hain ki oxidising strength left mein upar ki taraf aur reducing strength right mein neeche ki taraf badhti hai. Doosre step mein copper aur zinc rungs highlight hain aur ek curved arrow right mein neeche wale zinc ko left mein upar wale copper ions se jodta hai, saath mein tick aur E° = +1.10 V. Teesre step mein copper metal aur hydrogen ions highlight hain; copper ions hydrogen ions se upar hain, isliye arrow uphill jaata aur ek cross dikhata hai ki reaction nahi hogi. Chauthe step mein chlorine aur iodide ek tick wale arrow se jude hain, E° = +0.82 V.

Measured potentials lab ka ek tool bhi hain. pH meter ek electrochemical cell hai; titration ko end point par potential ki jump dekhkar follow kiya ja sakta hai; aur cell potentials se equilibrium constants aur solubility products milte hain jo kisi aur tarah measure karna mushkil hai (Section 2.3.1).

Example

Kaunse metals acid se hydrogen nikaalte hain?

Table 2.2 use karke tay karo ki nickel, silver aur aluminium mein se kaunse 1 mol L⁻¹ hydrochloric acid mein hydrogen chhodte hue ghulne chahiye, aur jo ghulte hain unke liye $E_{cell}^{\circ}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 3 questions

Intext questions 2.1–2.3: answers ke saath

Short answerEasyNCERT Intext Question 2.1
Tumhare paas ek nickel rod, 1 M nickel sulphate solution aur lab ka normal saamaan hai. Samjhao ki $N i^{2 +} / N i$ ka $E^{\circ}$ kaise nikaaloge, kaunsa electrode anode niklega, aur $E^{\circ}$ ka sign kya hoga.
Solution dikhao
AnswerIse SHE ke saath jodo; nickel anode nikalta hai, isliye $E^{\circ}$ negative ( $- 0.25$ ~V) hai.
298 K par $P t (s) ∣ H_{2} (g, 1 b a r)$ $∣ H^{+} (a q, 1 M)$ $∣ ∣ N i^{2 +} (a q, 1 M) ∣ N i (s)$ set up karo aur emf ko high-resistance voltmeter se measure karo, taaki lagbhag koi current na bahe.
Kyunki SHE ke liye $E_{L}^{\circ} = 0$ hai, likhe gaye cell ka emf $E_{N i^{2 +} / N i}^{\circ}$ ke barabar hai. Jodne par nickel ghulta hai aur hydrogen nikalti hai, isliye nickel asal mein anode hai, likhe gaye cell ka emf negative hai, aur $E_{N i^{2 +} / N i}^{\circ} = - 0.25 V$ .
Quick answerEasyNCERT Intext Question 2.2
Ek student silver nitrate solution ko ek hafte tak copper ke bowl mein chhod deti hai. Use kya milega?
Kuch nahi: copper silver se kam reactive haiBowl mein gaddhe pad jaate hain aur us par silver ki parat jam jaati hai, aur solution blue ho jaata haiKuch nahi: koi bhi metal nitrate solutions ke saath react nahi kartaBrown gas, kyunki silver ions nitrate ko oxidise karte hain
Answer dekho
Answer (B) Bowl mein gaddhe pad jaate hain aur us par silver ki parat jam jaati hai, aur solution blue ho jaata hai
$C u + 2 A g^{+} \overset{}{\to} C u^{2 +} + 2 A g$ ke liye $E_{cell}^{\circ} = 0.80 - 0.34 = + 0.46 V$ , isliye copper silver ions ko reduce karta hai: silver crystals bowl par jam jaate hain, aur solution mein aata $C u^{2 +}$ use blue kar deta hai. Dono mein copper zyada reactive hai, isliye solution ko copper mein store nahi kiya ja sakta.
2.3
Quick answerMediumNCERT Intext Question 2.3
Table 2.2 use karke har woh species chuno jo standard conditions mein iodide ions ko iodine mein oxidise kar sakti hai. ( $E_{I_{2} / I^{-}}^{\circ} = + 0.54 V$ .)
$F e^{3 +}$ $N i^{2 +}$ $C l_{2}$ $Z n^{2 +}$ acidified $M n O_{4}^{-}$
Saare sahi options chuno
Answer dekho
Answer
(A) $F e^{3 +}$
(C) $C l_{2}$
(E) acidified $M n O_{4}^{-}$
$F e^{3 +}$ (+0.77 V), $C l_{2}$ (+1.36 V) aur acidified $M n O_{4}^{-}$ (+1.51 V) sab +0.54 V se upar hain. $N i^{2 +}$ (−0.25 V) aur $Z n^{2 +}$ (−0.76 V) neeche hain, isliye woh nahi kar sakte.

2.3 Nernst Equation

Ek ghante se on torch cell naye cell se kam voltage deta hai, jabki uske chemicals wahi hain. Badle hain concentrations: standard potentials theek 1 mol L⁻¹ wale solutions ke liye hain, aur chalta hua cell jaldi hi usse door chala jaata hai. Walther Nernst ne 1889 mein iska correction nikaala. Apne ions ke contact mein rakhe metal M ke liye, $M^{n +} (a q) + n e^{-} ⟶ M (s)$ , temperature $T$ par electrode potential (SHE ke against) hai

Electrode ke liye Nernst equation

\begin{array}{c} E_{M^{n +} / M} = E_{M^{n +} / M}^{\circ} \\ - \frac{R T}{n F} \ln \frac{1}{[M^{n +}]} \end{array}

(2.8)

jahan

$R$: gas constant, $8.314 J K^{- 1} m o l^{- 1}$
$T$: kelvin mein temperature
$n$: half-reaction mein electrons
$F$: Faraday constant, $96 485 C m o l^{- 1}$
$[M^{n +}]$: ion ka molar concentration

Poore form mein $\ln ([reduced form] / [oxidised form])$ hota hai, lekin pure solid metal ko 1 gina jaata hai, isliye sirf $[M^{n +}]$ bachta hai. Ion concentration kam karne se electrode potential zyada negative ho jaata hai: jamne ke liye kam ions hon, to metal ka ghulne ka tendency bhari pad jaata hai.

DerivationDaniell cell ke liye Nernst equation

Har electrode par (2.8) lagao, dono ke liye $n = 2$ :

cathode

\begin{array}{c} E_{C u^{2 +} / C u} \\ = E_{C u^{2 +} / C u}^{\circ} \\ - \frac{R T}{2 F} \ln \frac{1}{[C u^{2 +}]} \end{array}

(2.9)

anode

\begin{array}{c} E_{Z n^{2 +} / Z n} \\ = E_{Z n^{2 +} / Z n}^{\circ} \\ - \frac{R T}{2 F} \ln \frac{1}{[Z n^{2 +}]} \end{array}

(2.10)

Cell potential cathode minus anode hai. Dono logarithms ek mein mil jaate hain, kyunki $\ln (1 / a) - \ln (1 / b) = \ln (b / a)$ , aur standard potentials milkar $E_{cell}^{\circ}$ bante hain:

\begin{array}{c} E_{cell} = E_{cell}^{\circ} \\ - \frac{R T}{2 F} \ln \frac{[Z n^{2 +}]}{[C u^{2 +}]} \end{array}

(2.11)

298 K par base-10 logarithms mein badalne aur $R$ aur $F$ daalne se $2.303 R T / F = 0.0591 V$ milta hai, isliye

\begin{array}{c} E_{cell} = E_{cell}^{\circ} \\ - \frac{0.0591}{2} \log \frac{[Z n^{2 +}]}{[C u^{2 +}]} \end{array}

(2.12)

Result ko physically padho. Zyada copper ions (reactant) cell potential badhate hain; zyada zinc ions (product) use ghatate hain. Yeh volts mein naapa gaya Le Chatelier's principle hai.

Yahi reasoning kisi bhi cell reaction $a A + b B - > c C + d D$ par chalti hai jo $n$ electrons transfer karti hai. Isse general Nernst equation milti hai, jahan $Q$ reaction quotient hai: pure solids aur liquids usmein nahi aate, aur gases bar mein partial pressures ki tarah aati hain.

\begin{array}{c} E_{cell} = E_{cell}^{\circ} - \frac{R T}{n F} \ln Q \\ Q = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}} \end{array}

(2.13)

Poore cell ke liye ek hi $n$ use karo: balanced overall equation mein transfer hone wale electrons ki sankhya. Example ke liye, $F e (s) + 2 F e^{3 +} (a q) \overset{}{\to} 3 F e^{2 +} (a q)$ mein $n = 2$ aur $Q = [F e^{2 +}]^{3} / [F e^{3 +}]^{2}$ hai. Har concentration ko uske coefficient ki power di jaati hai, aur iron metal aata hi nahi. 298 K par factor $2.303 R T / n F$ ki value $0.0591 / n$ volts hai, isliye do-electron reaction mein $Q$ ke har dus guna change par cell potential lagbhag 30 mV badalta hai (Fig. 2.6).

3/4Line zero se wahan milti hai jahan $Q = K$ : cell equilibrium par hai aur koi work nahi kar sakta.

Fig. 2.6Nernst equation

\log Q

ke against

E_{cell}

ki ek seedhi line hai, jiski slope 298 K par

- 0.0591 / n

volts hai.

Volts mein cell potential upar ki taraf aur log Q neeche ki axis par plot kiya gaya hai, ek aise cell ke liye jiska standard potential 0.12 V hai. Pehle step mein n = 2 ki seedhi line minus 0.0296 V per unit log Q ki slope se girti hai, aur log Q = 0 par ek dot E = E° mark karta hai. Doosre step mein dot left mein log Q = minus 3 par jaata hai, jahan E zyada hai, 0.21 V. Teesre step mein dot wahan hai jahan line zero cross karti hai, log Q lagbhag 4.1 par, label Q = K: cell equilibrium par hai. Chauthe step mein n = 1 ki doosri, double steep line judti hai, jo lagbhag log Q = 2.0 par zero cross karti hai.

Example 2.1NCERT Example 2.1 · p. 39

Aluminium–copper cell ke liye Nernst equation

Ek cell reaction $2 A l (s) + 3 C u^{2 +} (0.010 M)$ $\overset{}{\to} 2 A l^{3 +} (0.0050 M) + 3 C u (s)$ chalata hai. Cell ko standard notation mein likho aur 298 K par uska potential nikaalo. $E_{A l^{3 +} / A l}^{\circ} = - 1.66 V$ aur $E_{C u^{2 +} / C u}^{\circ} = + 0.34 V$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Hydrogen electrode ek badhiya pH meter banata hai. 1 bar hydrogen ke saath $H^{+} (a q) + e^{-} \overset{}{\to} \frac{1}{2} H_{2} (g)$ ke liye $E = 0 - 0.0591 \log (1 / [H^{+}])$ , yaani 298 K par $E = - 0.0591 pH$ volts. pH mein har ek unit ki badhat electrode ko 59 mV zyada negative kar deti hai. Lab ke pH meters iski jagah glass electrode use karte hain, jo $H^{+}$ par usi tarah respond karta hai lekin use hydrogen gas nahi chahiye.

InteractiveCell builder

Ek panel jismein do half-cells beakers ki tarah bane hain, salt bridge aur voltmeter se jude hue. Tum har side ke liye das ki list se metal couple chunte ho, jismein hydrogen electrode bhi hai, aur har ion concentration ko 0.0001 se 1 mole per litre tak ke logarithmic slider par set karte ho. Panel tay karta hai ki kaunsa electrode anode hai, cell ko balanced reaction ke saath standard notation mein likhta hai, aur standard cell potential, reaction quotient, Nernst equation se cell potential, Gibbs energy change aur equilibrium constant dikhata hai. Arrows wire mein electron direction dikhate hain. Discharge button reaction ko chalne deta hai: ion concentrations step by step badalte hain, voltmeter girta hai aur ek chhota graph cell potential trace karta hai, dono solutions ke barabar volumes ke liye, jab tak woh equilibrium par zero tak nahi pahunchta; phir ek message batata hai ki cathode ion kitna bacha hai.

Fig. 2.7Do half-cells aur unke ion concentrations chuno aur cell notation,

E_{cell}^{\circ}

, Nernst-corrected

E_{cell}

Δ_{r} G

aur

K

dekho. Cell ko equilibrium tak run down hote dekhne ke liye Discharge dabao.

Quick check

3 mein se 1

Kaunsa change Daniell cell ka potential badhata hai?

copper sulphate solution ko dilute karnaaur zinc sulphate daalnamoti zinc rod use karnazinc sulphate solution ko dilute karna

Answer dekho

Answer (D) zinc sulphate solution ko dilute karna

E_{cell} = E_{cell}^{\circ} - 0.0296 \log ([Z n^{2 +}] / [C u^{2 +}])

[Z n^{2 +}]

(ek product) kam karne se

Q

chhota aur

E_{cell}

bada hota hai. Moti rod se kuch nahi badalta, kyunki solid

Q

mein aata hi nahi.

2.3.1 Nernst Equation se Equilibrium Constant

Cell aakhir flat kyun ho jaata hai? Lamp se guzarne wala har electron pair ek aur $C u^{2 +}$ ion ko copper aur ek aur zinc atom ko $Z n^{2 +}$ bana deta hai. To $Q = [Z n^{2 +}] / [C u^{2 +}]$ badhta rehta hai, aur Nernst equation kehti hai ki $E_{cell}$ ghatta rehta hai. Cell tab khatam hota hai jab $E_{cell} = 0$ , aur yahi chemical equilibrium ki condition hai: $Q$ equilibrium constant $K_{c}$ ban chuka hai. Nernst equation mein $E_{cell} = 0$ aur $Q = K_{c}$ daalne se milta hai

E_{cell}^{\circ} = \frac{2.303 R T}{n F} \log K_{c}

(2.14)

Daniell cell ke liye $\log K_{c} = 2 \times 1.10 / 0.0591 = 37.2$ , isliye $K_{c} \approx 1.7 \times 10^{37}$ . Equilibrium par $[C u^{2 +}]$ , $[Z n^{2 +}]$ ka lagbhag $10^{- 37}$ guna hai: practically reaction poori ho jaati hai. Koi bhi chemical analysis itna chhota concentration kabhi measure nahi kar sakta, phir bhi ek voltmeter reading equilibrium constant de deti hai.

Example 2.2NCERT Example 2.2 · p. 40

Cell potentials se equilibrium constants

Table 2.2 ki values use karke 298 K par $K_{c}$ nikaalo: (a) $N i (s) + C u^{2 +} (a q)$ $\overset{}{⇌} N i^{2 +} (a q) + C u (s)$ aur (b) $S n (s) + P b^{2 +} (a q)$ $\overset{}{⇌} S n^{2 +} (a q) + P b (s)$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.3.2 Electrochemical Cell aur Reaction ki Gibbs Energy

Ek mole reaction kitna electrical work de sakti hai? Maan lo balanced equation $n$ moles electrons ko move karti hai. Woh $n F$ charge le jaate hain, jahan $F$ Faraday constant hai (ek mole electrons ka charge), aur woh $E_{cell}$ potential difference se girte hain, isliye work $n F E_{cell}$ hai. Yeh sabse zyada possible amount hai: yeh tabhi milta hai jab cell ko infinitely dheere drain kiya jaye, taaki uske andar koi voltage waste na ho (reversible discharge). Class 11 mein tumne dekha ki koi reaction maximum non-expansion work $- Δ_{r} G$ kar sakti hai. Dono ko barabar rakhne par,

Δ_{r} G = - n F E_{cell}

(2.15)

Spontaneous reaction ka $Δ_{r} G < 0$ hota hai, isliye uska cell potential positive hota hai. Equation ke dono sides alag tarah ki quantities hain. $E_{cell}$ intensive hai: button jitna Daniell cell aur bucket jitna Daniell cell, dono 1.10 V dete hain. $Δ_{r} G$ extensive hai: yeh equation se judi hai jaise tum use likhte ho, aur equation ko scale karne par $n$ aur $Δ_{r} G$ saath-saath scale hote hain jabki $E_{cell}$ waisa hi rehta hai (Example 2.3 yeh numbers ke saath dikhata hai).

Jab har species apne standard state mein ho, to yahi relation reaction ki standard Gibbs energy deta hai:

Δ_{r} G^{\circ} = - n F E_{cell}^{\circ}

(2.16)

Kyunki $Δ_{r} G^{\circ} = - R T \ln K$ (Class 11 thermodynamics), ek measured standard cell potential se $Δ_{r} G^{\circ}$ aur $K$ dono mil jaate hain. Teenon quantities ke signs hamesha mel khaate hain (Table 2.3).

Table 2.3

E_{cell}^{\circ}

Δ_{r} G^{\circ}

aur

K

ke signs kaise saath chalte hain

$E_{cell}^{\circ}$	$Δ_{r} G^{\circ}$	$K$	Matlab
positive	negative	1 se zyada	forward reaction favoured hoti hai
zero	zero	1 ke barabar	koi bhi direction favoured nahi hoti
negative	positive	1 se kam	reverse reaction favoured hoti hai

Example 2.3NCERT Example 2.3 · p. 40

Hydrogen–oxygen reaction ki Gibbs energy

Hydrogen–oxygen fuel cell (Section 2.7) reaction $2 H_{2} (g) + O_{2} (g) \overset{}{\to} 2 H_{2} O (l)$ chalata hai, jahan $E_{cell}^{\circ} = 1.23 V$ . Is equation ke liye $Δ_{r} G^{\circ}$ nikaalo, aur phir per mole bane paani ke liye.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek cell reaction ka

E_{cell}^{\circ} = - 0.20 V

hai. Standard conditions mein kya sach hai?

Δ_{r} G^{\circ} < 0

aur

K > 1

Δ_{r} G^{\circ} > 0

aur

K < 1

Δ_{r} G^{\circ} > 0

aur

K > 1

Δ_{r} G^{\circ} = 0

aur

K = 1

Answer dekho

Answer (B) $Δ_{r} G^{\circ} > 0$ aur $K < 1$

Jab

E^{\circ}

negative ho, to

Δ_{r} G^{\circ} = - n F E^{\circ}

positive hota hai, aur tab

\ln K = - Δ_{r} G^{\circ} / R T

negative hota hai, isliye

K < 1

Δ_{r} G^{\circ}

aur

K - 1

ke signs hamesha ulte hote hain.

Practice · 3 questions

Intext questions 2.4–2.6: answers ke saath

Quick answerEasyNCERT Intext Question 2.4
Ek hydrogen electrode (1 bar par hydrogen) pH 3.0 wale soft drink mein dubaya jaata hai. 298 K par uska potential kya hai?
V
Solution dikhao
1 bar par $H^{+} + e^{-} \overset{}{\to} \frac{1}{2} H_{2}$ ke liye $E = 0 - 0.0591 \log (1 / [H^{+}]) = - 0.0591 pH$ . pH 3.0 ke saath $E = - 0.177 V$ .
NumericalMediumNCERT Intext Question 2.5
$F e (s) + 2 A g^{+} (0.0050 M)$ $\overset{}{\to} F e^{2 +} (0.10 M) + 2 A g (s)$ chalane wala cell 298 K par kitna voltage deta hai? ( $E_{cell}^{\circ} = 1.24 V$ .)
V
Solution dikhao
$Q = [F e^{2 +}] / [A g^{+}]^{2} = 0.10 / (0.0050)^{2} = 4.0 \times 10^{3}$ .
$n = 2$ ke saath: $E_{cell} = 1.24 - 0.0296 \log (4.0 \times 10^{3}) = 1.24 - 0.106 = 1.13 V$ .
2.6
NumericalMediumNCERT Intext Question 2.6
Circuit boards se copper ko iron(III) chloride solution se etch karke hataya jaata hai: $C u (s) + 2 F e^{3 +} (a q)$ $\overset{}{\to} C u^{2 +} (a q) + 2 F e^{2 +} (a q)$ , jiske liye 298 K par $E_{cell}^{\circ} = 0.43 V$ .
1. kJ mol⁻¹ mein $Δ_{r} G^{\circ}$ nikaalo.
  kJ mol⁻¹
  Solution dikhao
  $Δ_{r} G^{\circ} = - n F E^{\circ} = - 2 \times 96 485 \times 0.43 = - 8.30 \times 10^{4} J m o l^{- 1} = - 83.0 k J m o l^{- 1}$ .
2. Equilibrium constant $K_{c}$ nikaalo.
  Solution dikhao
  $\log K_{c} = 2 \times 0.43 / 0.0591 = 14.55$ , isliye $K_{c} \approx 3.6 \times 10^{14}$ . ( $\ln K = - Δ_{r} G^{\circ} / R T$ se bhi yahi answer aata hai.)

2.4 Electrolytic Solutions ka Conductance

Cell tabhi kaam karta hai jab charge uske poore circuit mein ghoom sake: wires mein electrons ki tarah aur solutions mein ions ki tarah. Solutions kitna accha conduct karte hain, aur kyun? Hum physics ki jaani-pehchaani quantities se shuru karte hain.

Ek wire se shuru karo. Uski length double karo to uska resistance $R$ double ho jaata hai, kyunki charge ko do guna door tak push karna padta hai; uska cross-sectional area double karo to $R$ aadha ho jaata hai, kyunki ab do guna side-by-side raaste hain. (Physics mein tumne Wheatstone bridge se resistances measure kiye the; SI base units mein ohm, Ω, $k g m^{2} s^{- 3} A^{- 2}$ hai.) Dono effects ek line mein aa jaate hain:

R = ρ \frac{l}{A}

(2.17)

jahan

$ρ$: resistivity, ohm metres (Ω m) mein

Sample ki shape ki saari baat ab $l / A$ mein hai. Jo bacha, $ρ$ (rho), woh material ka hai: yeh resistivity hai (purani books ise specific resistance kehti hain), ohm metres, Ω m, mein. Iska matlab samajhne ke liye $l = 1$ m aur $A = 1$ m² rakho: $ρ$ ek metre cube ke aamne-saamne ke faces ke beech ka resistance hai. Lab values aksar centimetre cube ke liye di jaati hain, aur $1 Ω m = 100 Ω c m$ .

Solutions ke liye yeh poochna zyada natural hai ki charge kitni aasaani se behta hai, isliye dono quantities ko ulta kar dete hain. $R$ ka reciprocal conductance $G$ hai, siemens mein ( $1 S = 1 Ω^{- 1}$ , jise kabhi mho kehte the), aur $ρ$ ka reciprocal conductivity $κ$ (kappa; purana naam specific conductance) hai:

G = \frac{1}{R} = \frac{A}{ρ l} = κ \frac{A}{l}

(2.18)

Jaise $ρ$ unit cube ka resistance hai, waise hi $κ$ unit cube ke aamne-saamne ke faces ke beech ka conductance hai: metre cube ke liye S m⁻¹ ya centimetre cube ke liye S cm⁻¹, jahan $1 S c m^{- 1} = 100 S m^{- 1}$ . Table 2.4 dikhata hai ki $κ$ ek material se doosre material tak kitna zyada badalta hai, ten ki bees se zyada powers tak; kisi ek material ke liye bhi yeh temperature ke saath aur thoda sa pressure ke saath badalta hai.

Table 2.4298 K ke aas-paas approximate conductivitiesNCERT Table 2.2 · p. 42

Material	$κ$ / (S m⁻¹)	Kind
silver	$6.3 \times 10^{7}$	metal
copper	$6.0 \times 10^{7}$	metal
iron	$1.0 \times 10^{7}$	metal
graphite (layers ke along)	lagbhag $2 \times 10^{5}$	non-metal conductor
seawater (samudri paani)	lagbhag 5	ionic solution
germanium (pure)	lagbhag 2	semiconductor
0.100 M KCl	1.29	ionic solution
nal ka paani	$10^{- 3}$ se $10^{- 1}$	ionic solution
silicon (pure)	lagbhag $4 \times 10^{- 4}$	semiconductor
ultra-pure water	$5.5 \times 10^{- 6}$	ionic (self-ionisation)
glass	$10^{- 15}$ se $10^{- 11}$	insulator
PTFE (Teflon)	$10^{- 16}$ se kam	insulator

Table ko neeche se padho. PTFE aur glass jaise insulators mushkil se conduct karte hain. Silicon aur germanium jaise semiconductors beech mein hain, aur kisi doosre element ki zara si matra unki conductivity ko bahut zyada badha deti hai; isi tunability ki wajah se woh har phone aur computer chalate hain. Upar conductors hain: metals aur alloys, aur saath mein graphite jaise kuch non-metals. Scale ke top se bhi aage superconductors hain, jinka resistance bilkul zero hai. 4.2 K se neeche mercury pehla tha, jo 1911 mein mila, aur kuch copper-oxide ceramics ab 77 K se upar superconduct karte hain, jo liquid nitrogen ka boiling point hai.

Metal mein current sirf electrons le jaate hain: electronic (metallic) conductance (Fig. 2.8). Valence electrons kisi ek atom ke nahi hote, aur voltage lagne par woh positive ions ki lattice mein drift karte hain. Wire ke ek end mein ghusne wala har electron door wale end se doosre ko bahar dhakel deta hai, isliye metal khud kabhi nahi badalta. Woh kitna accha conduct karta hai, yeh is par depend karta hai ki har atom kitne valence electrons deta hai aur atoms kaise arranged hain, isiliye silver iron se aage hai. Metal ko garam karne par woh kam conduct karta hai, kyunki zyada zor se vibrate karte ions drift karte electrons ko zyada baar scatter karte hain.

Bahut pure paani mein ek chutki namak ghol do aur uski conductivity hazaaron guna badh jaati hai. Akela paani bahut kam conduct karta hai, apne $10^{- 7} m o l L^{- 1}$ $H^{+}$ aur utne hi $O H^{-}$ se; namak kahin zyada carriers jod deta hai. To solution mein current ions le jaate hain, cations aur anions ulti directions mein drift karte hue: electrolytic (ionic) conductance. Ek solution doosre se behtar conductor kyun hota hai?

Kitne ions: per litre zyada electrolyte matlab zyada carriers.
Kis tarah ke ions: doubly charged ion singly charged ion se do guna charge le jaata hai.
Woh kitne mobile hain: chhote bare ions hamesha tez nahi hote. $L i^{+}$ apne saath water molecules ka bada shell kheenchta hai aur bade $K^{+}$ se dheere chalta hai.
Solvent: zyada viscous solvent har ion ko dheema kar deta hai.
Temperature: garam karne se viscosity kam hoti hai, isliye metals ke ulat, solutions garam hone par behtar conduct karte hain.

2/3Solution mein cations aur anions ulti directions mein drift karte hain.

Fig. 2.8Metal mein electrons current le jaate hain; solution mein dono taraf chalte ions use le jaate hain, aur electrodes par reactions hoti hain.

Upar wale hisse mein ek copper wire hai jismein electrons left se right drift karte hain jabki lattice ke copper ions apni jagah rehte hain. Neeche wale hisse mein ek salt solution hai, left mein negative electrode aur right mein positive electrode ke beech. Pehle step mein sirf wire active hai aur ek note kehta hai ki wire nahi badalti. Doosre step mein solution ke positive ions left mein negative electrode ki taraf drift karte hain jabki negative ions right mein positive electrode ki taraf. Teesre step mein electrodes par products dikhte hain, left mein ek deposit aur right mein gas bubbles, aur ek note kehta hai ki solution badal jaata hai.

Compare

Conduct karne ke do tareeke

Electronic (metals, graphite)

carriers: electrons
conductor mein koi chemical change nahi
temperature badhne par conductance girta hai

Ionic (solutions, molten salts)

carriers: cations aur anions, ulti directions mein chalte hue
electrodes par reactions composition badal deti hain
temperature badhne par conductance badhta hai

Quick check

3 mein se 1

Sodium chloride solution ko 25 °C se 50 °C tak garam karne par uski conductivity…

girti hai, jaise metal wire meinwaisi hi rehti hai, kyunki ions ki sankhya fixed haibadhti hai, kyunki ions zyada aasaani se chalte hainzero tak gir jaati hai

Answer dekho

Answer (C) badhti hai, kyunki ions zyada aasaani se chalte hain

Solvent kam viscous ho jaata hai, isliye wahi ions tez chalte hain aur zyada current le jaate hain. Metal mein ulta hota hai, kyunki lattice vibrations electrons ko zyada scatter karti hain.

2.4.1 Ionic Solutions ki Conductivity ka Measurement

Solution ka resistance measure karna wire se zyada tricky hai. Battery se try karo to tum solution ko electrolyse kar doge (Section 2.5): electrodes par gas ya metal jama hone lagta hai aur tumhare dekhte-dekhte solution badal jaata hai. Isliye measurement lagbhag ek kilohertz ke alternating current se kiya jaata hai; har half-cycle pichhle wale ka shuru kiya kaam ulta kar deta hai, aur kuch jama nahi hota.

Liquid ki apni koi shape bhi nahi hoti, isliye use ek conductivity cell mein rakha jaata hai: do parallel platinum plates, bade surface ke liye platinum black se coated, ek glass holder mein fixed doori par lagi hui. Plates ke beech ka liquid $l$ length aur $A$ cross-section ka ek column hai, isliye

R = ρ \frac{l}{A} = \frac{1}{κ} \cdot \frac{l}{A}

Saari geometry ek number mein aa jaati hai, cell constant $G^{*} = l / A$ (सेल स्थिरांक), m⁻¹ ya cm⁻¹ mein. Spongy platinised plates par ruler par bharosa karne ki jagah chemists calibrate karte hain: cell ko ek standard KCl solution se bharo, jiska $κ$ bahut dhyaan se measure kiya gaya hai, resistance padho aur multiply karo:

G^{*} = \frac{l}{A} = R κ

Classic set-up (Fig. 2.9) mein cell Wheatstone bridge ki chaar arms mein se ek hai, aur baaki fixed resistances $P$ aur $Q$ aur ek variable resistance $S$ hain. Ek oscillator se audio-frequency AC bridge ko chalata hai, isliye beech mein lagaya headphone tab tak hum karta hai jab tak bridge balance nahi hota (modern versions electronic detector use karte hain). $S$ ko tab tak ghumao jab tak hum gayab na ho jaye; tab bridge balanced hai aur

R_{x} = S \frac{P}{Q}

(2.19)

3/3 $S$ ko tab tak adjust karo jab tak detector chup na ho jaye: $R_{x} = S P / Q$ .

Fig. 2.9Solution ka resistance measure karna: AC Wheatstone bridge mein ek conductivity cell.NCERT Fig. 2.5 · p. 44

Left mein ek conductivity cell hai jismein A area ki do platinised platinum plates l doori par hain, aur unke beech solution ka column shaded hai. Right mein ek diamond-shaped bridge hai jiske upar arms P aur Q hain aur neeche variable resistance S aur cell R x, ek diagonal ke across AC oscillator aur doosre ke across detector hai. Pehle step mein sirf cell highlight hai, l aur A mark kiye hue. Doosre step mein bridge unbalanced hai aur detector buzzing signal dikhata hai. Teesre step mein S adjust ho chuka hai, detector chup hai, aur balance condition R x = S guna P upon Q dikhayi gayi hai.

Peene ka paani check karne wala handheld conductivity meter yeh sab ek box ke andar karta hai aur seedha $κ$ dikhata hai; principle wahi hai.

Ek baar $G^{*}$ pata ho, to wahi cell kisi bhi solution ki conductivity uske resistance se de deta hai:

κ = \frac{G^{*}}{R}

(2.20)

Sirf conductivity se electrolytes ko compare karna theek nahi, kyunki zyada concentrated solution sirf zyada ions hone ki wajah se behtar conduct karta hai. Barabari se compare karne ke liye concentration se divide karo. Molar conductivity $Λ_{m}$ (capital lambda) hai

Λ_{m} = \frac{κ}{c}

(2.21)

$κ$ S m⁻¹ mein aur $c$ mol m⁻³ mein ho to $Λ_{m}$ S m² mol⁻¹ mein aata hai. Concentrations aksar mol L⁻¹ mein di jaati hain, aur $1 m o l L^{- 1} = 1000 m o l m^{- 3}$ , isliye pehle convert karo. Chemists S cm² mol⁻¹ bhi use karte hain, $κ$ S cm⁻¹ mein; kyunki 1 L = 1000 cm³, working formula aur dono units ka link yeh hai

Λ_{m} = \frac{1000 κ}{c}

jahan

$Λ_{m}$: molar conductivity, S cm² mol⁻¹ mein
$κ$: conductivity, S cm⁻¹ mein
$c$: concentration, mol L⁻¹ mein

\begin{array}{c} 1 S m^{2} m o l^{- 1} \\ = 10^{4} S c m^{2} m o l^{- 1} \end{array}

Example 2.4NCERT Example 2.4 · p. 45

Cell ko calibrate karna, phir ek salt solution measure karna

0.0100 mol L⁻¹ KCl se bhare conductivity cell ka resistance, jahan KCl ki conductivity $0.1413 S m^{- 1}$ hai, 312 Ω hai. 0.0500 mol L⁻¹ NaCl solution se bharne par wahi cell 79.4 Ω padhta hai. Cell constant aur NaCl solution ki conductivity aur molar conductivity nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2.5NCERT Example 2.5 · p. 46

Acid column ki resistivity

0.80 cm internal diameter aur 20.0 cm length wali ek glass tube 0.100 mol L⁻¹ hydrochloric acid se bhari hai. Acid column ka resistance $1.02 \times 10^{3} Ω$ hai. Acid ki resistivity, conductivity aur molar conductivity nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Solution ka resistance measure karne ke liye DC ki jagah AC kyun use hota hai?

AC per second zyada charge le jaata haiDC solution ko electrolyse karke uska composition badal detaPlatinum electrodes DC nahi le ja sakteAC ions ko tez chalata hai

Answer dekho

Answer (B) DC solution ko electrolyse karke uska composition badal deta

DC mein electrodes par products jama hote hain aur measurement ke dauraan solution badal jaata hai. AC mein har half-cycle pichhle wale ko ulta kar deta hai. AC ions ko tez nahi karta; woh bas aage-peeche oscillate karte hain.

2.4.2 Concentration ke saath Conductivity aur Molar Conductivity ka Variation

Kisi bhi electrolyte solution mein, strong ho ya weak, paani milao aur $κ$ gir jaata hai. Yaad karo $κ$ kya measure karta hai: solution ke ek unit cube ka conductance. Paani daalo to wahi ions zyada cubes mein bant jaate hain, isliye har cube mein kam carriers hote hain.

Molar conductivity ulta karti hai: dilution par woh badhti hai. Socho ki tum woh saara solution ikattha karte ho jismein theek ek mole electrolyte hai aur use ek unit doori par rakhe do lambe electrodes ke beech phaila dete ho, jaise Fig. 2.10 ki tray. Tab plates ka area us volume $V$ ke barabar hona chahiye, isliye $G = κ A / l$ ban jaata hai

Λ_{m} = κ V, V = \frac{1}{c}

(2.22)

Product badhta kyun hai? Dus guna dilute karne se $V$ theek dus guna hota hai lekin $κ$ dus se kam guna ghatta hai, kyunki har cube mein bache ions ek doosre se door hote hain, ek doosre ko kam kheenchte hain aur isliye tez chalte hain (weak electrolyte mein extra molecules bhi ionise hote hain). Jaise $c$ zero ke kareeb aata hai, $Λ_{m}$ ek limiting value ke kareeb pahunchti hai, limiting molar conductivity $Λ_{m}^{\circ}$ , jise infinite dilution par molar conductivity bhi kehte hain. Strong aur weak electrolytes is tak bilkul alag tareekon se pahunchte hain (Fig. 2.11).

2/3Do guna volume: har cube mein aadhe ions, lekin total conductance thoda zyada.

Fig. 2.10Dilution har unit cube mein kam ions chhodta hai (κ girta hai) lekin wahi mole bade area par phaila deta hai (Λm badhta hai).

Ek unit doori par rakhe do lambe electrodes ek tray mein solution rakhte hain jismein theek ek mole electrolyte hai, upar se dekha gaya; ions plus aur minus dots ki tarah bane hain aur ek unit cube outline kiya gaya hai. Right mein bar gauges kappa, yaani unit cube ka conductance, ko lambda m, yaani poori tray ke conductance, se compare karte hain. Pehle step mein tray chhoti hai aur cube mein aath ions hain. Doosre step mein solution do guna dilute hai: tray do guna lambi hai, cube mein chaar ions hain, kappa bar lagbhag aadha hai aur lambda bar thoda lamba hai. Teesre step mein tray chaar guna lambi hai, cube mein do ions hain, kappa lagbhag ek chauthai hai aur lambda aur bhi lamba hai.

Strong electrolytes. KCl jaisa strong electrolyte har concentration par poori tarah dissociated hota hai, isliye dilution koi naye ions nahi jodta. Phir bhi uski molar conductivity thodi badhti hai, kyunki har ion ke chaaron taraf opposite charge wale ions ka ek dheela cloud hota hai jo use peeche kheenchta hai, aur ions ke phailne par yeh kheench kamzor ho jaati hai. Kam concentrations par experiment dikhata hai ki

Λ_{m} = Λ_{m}^{\circ} - A \sqrt{c}

(2.23)

Yeh chhupi hui seedhi line hai: $\sqrt{c}$ ke against $Λ_{m}$ plot karo, intercept $Λ_{m}^{\circ}$ hai aur slope $- A$ (Example 2.6 theek yahi karta hai). $A$ batata hai ki ions ek doosre ko kitni zor se rokte hain, isliye yeh unke charges ke saath badhta hai: doubly charged ions apne aas-paas ghane clouds jama karte hain. Isliye chemists electrolytes ko charge type se label karte hain, pehle cation. $K N O_{3}$ 1-1 electrolyte hai, $B a C l_{2}$ 2-1 aur $C u S O_{4}$ 2-2, aur kisi solvent mein kisi temperature par same type ke electrolytes ka $A$ lagbhag same hota hai.

4/4Steep curve ko extend nahi kar sakte; $Λ_{m}^{\circ}$ (404) Kohlrausch's law se aata hai.

Fig. 2.11Concentration zero ke kareeb aane par strong aur weak electrolytes bahut alag tarah se behave karte hain.NCERT Fig. 2.6 · p. 47

Siemens square centimetres per mole mein molar conductivity upar ki taraf concentration ke square root ke against plot ki gayi hai. Pehle step mein sodium chloride ke measured points 126 se thoda neeche ek halki seedhi line par lagbhag pade hain. Doosre step mein line ko vertical axis tak peeche extend kiya gaya hai, jahan woh 126.4 par milti hai, label limiting molar conductivity. Teesre step mein formic acid, ek weak acid, ka curve joda gaya hai: zyada concentrations par yeh neeche hai aur concentration zero ke kareeb aane par tezi se upar jaata hai. Chauthe step mein ek dashed continuation aur question mark dikhate hain ki curve ko bharose se extend nahi kiya ja sakta, aur iski jagah axis par Kohlrausch's law se mili value 404 mark ki gayi hai.

Example 2.6NCERT Example 2.6 · p. 48

NaCl ki limiting molar conductivity nikaalna

298 K par dilute sodium chloride solutions ki molar conductivity neeche di gayi hai. Dikhao ki $Λ_{m}$ $\sqrt{c}$ ke saath linearly badalti hai, aur $Λ_{m}^{\circ}$ aur $A$ nikaalo.

Table 2.5Dilute NaCl ki molar conductivity

$c$ / (mol L⁻¹)	$Λ_{m}$ / (S cm² mol⁻¹)
0.000400	124.7
0.000900	123.9
0.00250	122.1
0.00490	120.5

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Kohlrausch's law. 1870s mein limiting molar conductivities measure karte hue Friedrich Kohlrausch ne dekha ki woh sums ki tarah behave karti hain. Table 2.6 ki kisi bhi row ke along dekho: chloride ki jagah nitrate lagane se $Λ_{m}^{\circ}$ hamesha 4.8 kam hoti hai, cation koi bhi ho. Kisi column mein neeche dekho: lithium ki jagah sodium lagane se hamesha 11.4 judta hai, anion koi bhi ho.

Table 2.6298 K par kuch salts ki

Λ_{m}^{\circ}

, S cm² mol⁻¹ mein

	chloride	nitrate	$C l^{-}$ − $N O_{3}^{-}$
lithium	115.0	110.2	4.8
sodium	126.4	121.6	4.8
potassium	149.8	145.0	4.8
Na − Li	11.4	11.4	–
K − Li	34.8	34.8	–

Agar partner kuch bhi ho aur difference same rahe, to iska matlab hai ki har ion apna fixed hissa deta hai, jis par doosre ion ka koi asar nahi. Kohlrausch ne ise ek law mein samet diya:

LawIons ke independent migration ka Kohlrausch's law

Infinite dilution par kisi electrolyte ki molar conductivity uske cations aur anions ke alag-alag contributions ka sum hai, aur har ion ek fixed amount deta hai jo uske partner par depend nahi karta.

\begin{array}{c} Λ_{m}^{\circ} (N a C l) \\ = λ_{N a^{+}}^{\circ} + λ_{C l^{-}}^{\circ} \end{array}

(2.24)

Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}

(2.25)

jahan

$ν_{+}, ν_{-}$: ek formula unit mein cations aur anions ki sankhya
$λ_{+}^{\circ}, λ_{-}^{\circ}$: har ion ki apni limiting molar conductivity

Table 2.7298 K par paani mein ions ki limiting molar conductivities

λ^{\circ}

, S cm² mol⁻¹ mein (doubly charged ions ke liye, per mole ions)NCERT Table 2.4 · p. 49

Cation	$λ^{\circ}$	Anion	$λ^{\circ}$
$H^{+}$	349.6	$O H^{-}$	199.1
$L i^{+}$	38.7	$C l^{-}$	76.3
$N a^{+}$	50.1	$B r^{-}$	78.1
$K^{+}$	73.5	$I^{-}$	76.8
$N H_{4}^{+}$	73.5	$N O_{3}^{-}$	71.5
$A g^{+}$	61.9	$H C O O^{-}$	54.6
$M g^{2 +}$	106.0	$C H_{3} C O O^{-}$	40.9
$C a^{2 +}$	119.0	$S O_{4}^{2 -}$	160.0
$B a^{2 +}$	127.2	$C O_{3}^{2 -}$	138.6

Weak electrolytes. Fig. 2.11 mein formic acid ka curve alag kahani kehta hai. Yahan dilution sirf ion clouds ko dheela nahi karta: woh ions banata hai. 0.02 mol L⁻¹ par gyaarah mein se lagbhag ek hi molecule ionised hota hai; sau guna dilution aadhe se zyada ko ionise kar deta hai. Isliye $c$ girne par $Λ_{m}$ aur bhi tezi se chadhta hai, kyunki degree of dissociation $α$ badhti rehti hai, aur $c = 0$ tak seedhi line ka extrapolation sirf andaaza hoga. Tum measure karte-karte limit tak pahunch bhi nahi sakte: har molecule ionise hone se bahut pehle hi solution us paani se mushkil se behtar conduct karta hai jisse woh bana hai. Iski jagah weak electrolyte ki $Λ_{m}^{\circ}$ ko Kohlrausch's law se ion values jodkar banaya jaata hai (Example 2.8).

Ek baar $Λ_{m}^{\circ}$ pata ho, to conductivity se degree of dissociation mil jaati hai. Infinite dilution par har molecule ionised hota hai; concentration $c$ par sirf fraction $α$ hota hai, aur acche approximation mein

α = \frac{Λ_{m}}{Λ_{m}^{\circ}}

(2.26)

Weak acid HA ke liye dissociation constant (Class 11) phir sirf conductivity se nikal aata hai:

\begin{aligned} K_{a} & = \frac{c α^{2}}{1 - α} \\ = \frac{c Λ_{m}^{2}}{Λ_{m}^{\circ} (Λ_{m}^{\circ} - Λ_{m})} \end{aligned}

(2.27)

Kohlrausch's law use karna. Agle teen examples law ko kaam mein lagate hain: ion values jodna (Example 2.7), teen strong electrolytes se ek weak electrolyte banana (Example 2.8), aur ek conductivity reading ko dissociation constant mein badalna (Example 2.9).

Example 2.7NCERT Example 2.7 · p. 50

Ion values se limiting molar conductivities

Table 2.7 use karke barium chloride aur sodium sulphate ke liye $Λ_{m}^{\circ}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2.8NCERT Example 2.8 · p. 50

Weak base ki limiting molar conductivity

298 K par $Λ_{m}^{\circ}$ $N H_{4} C l$ ke liye 149.8, $N a O H$ ke liye 249.2 aur $N a C l$ ke liye 126.4 hai, sab S cm² mol⁻¹ mein. Aqueous ammonia ko $N H_{4} O H$ maankar uski $Λ_{m}^{\circ}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2.9NCERT Example 2.9 · p. 50

Conductivity se ammonia ka dissociation constant

298 K par ammonia ke 0.0100 mol L⁻¹ solution ki conductivity $1.13 \times 10^{- 4} S c m^{- 1}$ hai. Example 2.8 se $Λ_{m}^{\circ} = 272.6 S c m^{2} m o l^{- 1}$ lekar degree of dissociation aur base dissociation constant $K_{b}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek KCl solution ko 0.10 se 0.010 mol L⁻¹ tak dilute kiya jaata hai.

κ

aur

Λ_{m}

ka kya hota hai?

dono girte hain

κ

badhta hai,

Λ_{m}

girta haidono badhte hain

κ

girta hai,

Λ_{m}

badhta hai

Answer dekho

Answer (D) $κ$ girta hai, $Λ_{m}$ badhta hai

Ab har unit cube mein kam ions hain, isliye

κ

girta hai. Lekin ek mole dus guna volume mein phaila hai aur ions ek doosre ko kam rokte hain, isliye

κ V = Λ_{m}

badhta hai.

Practice · 3 questions

Intext questions 2.7–2.9: answers ke saath

Short answerEasyNCERT Intext Question 2.7
Salt solution mein dooba conductivity meter, jaise-jaise tum paani daalte jaate ho, kam se kam padhta jaata hai. Kyun?
Solution dikhao
AnswerHar unit volume mein charge le jaane ke liye kam ions hote hain.
Conductivity solution ke ek unit cube ka aamne-saamne ke faces ke beech conductance hai. Dilute karne se ions zyada solvent mein phail jaate hain, isliye har unit cube mein kam ions hote hain, aur kam carriers matlab same voltage par kam current. Yeh har ion ki speed mein hone wale chhote faayde par bhari padta hai.
Short answerMediumNCERT Intext Question 2.8
Samjhao ki teen strong electrolytes ki limiting molar conductivities se paani ki $Λ_{m}^{\circ}$ kaise nikaali ja sakti hai, aur Table 2.7 se uski value nikaalo.
S cm² mol⁻¹
Solution dikhao
Answer $Λ_{m}^{\circ} (H_{2} O) = 548.7 S c m^{2} m o l^{- 1}$
Paani khud itna kam conduct karta hai ki seedha measure nahi ho sakta, isliye Kohlrausch's law use karo: $Λ_{m}^{\circ} (H_{2} O) = Λ_{m}^{\circ} (H C l) + Λ_{m}^{\circ} (N a O H) - Λ_{m}^{\circ} (N a C l)$ , jismein sodium aur chloride ke contributions cancel ho jaate hain.
Ion values se yeh $λ_{H^{+}}^{\circ} + λ_{O H^{-}}^{\circ} = 349.6 + 199.1 = 548.7 S c m^{2} m o l^{- 1}$ hai.
2.9
NumericalMediumNCERT Intext Question 2.9
0.0050 mol L⁻¹ acetic acid solution ki $Λ_{m} = 22.9 S c m^{2} m o l^{- 1}$ hai. $λ_{H^{+}}^{\circ} = 349.6$ aur $λ_{C H_{3} C O O^{-}}^{\circ} = 40.9 S c m^{2} m o l^{- 1}$ lekar nikaalo:
1. degree of dissociation;
  Solution dikhao
  $Λ_{m}^{\circ} = 349.6 + 40.9 = 390.5 S c m^{2} m o l^{- 1}$ , isliye $α = 22.9 / 390.5 = 0.0586$ .
2. dissociation constant $K_{a}$ .
  mol L⁻¹
  Solution dikhao
  $K_{a} = c α^{2} / (1 - α) = 0.0050 \times {0.0586}^{2} / 0.9414 = 1.83 \times 10^{- 5} m o l L^{- 1}$ .

2.5 Electrolytic Cells aur Electrolysis

Electrical wiring ke copper ko 99.9 % se zyada pure hona chahiye, aur ise paane ka tareeka electrolysis hai. Acidified copper sulphate solution mein crude copper ka ek mota slab aur pure copper ki ek patli sheet latkao, aur DC supply jodo: slab positive terminal se, sheet negative se. Supply electrons ko sheet mein dhakelti hai, jisse woh cathode ban jaati hai, jahan copper ions reduce hokar jam jaate hain:

cathode, reduction

\begin{aligned} C u^{2 +} (a q) + 2 e^{-} \\ \overset{}{\to} C u (s) \end{aligned}

(2.28)

Woh slab se electrons kheenchti hai, jisse woh anode ban jaata hai, jahan copper atoms oxidise hokar ghul jaate hain:

anode, oxidation

\begin{aligned} C u (s) \\ \overset{}{\to} C u^{2 +} (a q) + 2 e^{-} \end{aligned}

(2.29)

Copper bas slab se sheet tak chala jaata hai. Raste mein impurities alag ho jaati hain: iron aur zinc jaise zyada reactive metals ghul to jaate hain lekin solution mein hi rehte hain, kyunki unke ions ko reduce karna $C u^{2 +}$ se mushkil hai, jabki silver aur gold ghulte hi nahi aur neeche anode mud ki tarah gir jaate hain, jise khud nikaalna faayde ka hai. Supply ke bina yahan kuch nahi hota: yeh ek electrolytic cell hai jo aisa kaam kar raha hai jo koi spontaneous reaction nahi karti.

Sabse reactive metals ke liye electrolysis hi practical raasta hai, kyunki unke ions Table 2.2 mein itne neeche hain ki carbon jaisa koi sasta reducing agent unhe reduce nahi kar sakta. Jaise aluminium, molten cryolite ( $N a_{3} A l F_{6}$ ) mein ghule uske oxide ko electrolyse karke banaya jaata hai, jisse cell lagbhag 950 °C par chal jaata hai, jabki akele oxide ko pighlane ke liye 2000 °C ya usse zyada chahiye. Sodium aur magnesium unke molten chlorides se aate hain.

Electrolysis ke quantitative aspects. Ek ghante ka current kitna copper jamata hai? Michael Faraday ne 1833–34 mein aise sawaalon ke jawab experiment se diye, electron ki khoj se dashakon pehle. Aaj uske do laws electrons ka simple hisaab-kitaab lagte hain.

LawElectrolysis ka Faraday's first law

Electrode par react karne wale substance ki amount electrolyte se guzarne wali electricity ki quantity (charge) ke directly proportional hoti hai.

LawElectrolysis ka Faraday's second law

Alag-alag electrolytes se same charge guzaro, to free hone wale substances ke masses unke equivalent masses ke ratio mein hote hain, jahan equivalent mass = molar mass ÷ per ion zaroori electrons ki sankhya.

Modern constant-current supply mein tum time $t$ ke liye current $I$ set karte ho, aur guzarne wala charge bas hai

Q = I t

jahan

$Q$: charge, coulombs mein
$I$: current, amperes mein
$t$: time, seconds mein

(First law ko kabhi ulta, charge measure karne ke liye, use kiya jaata tha. Constant-current supplies se pehle circuit mein ek chhota electrolytic cell lagaya jaata tha jise coulometer kehte the, aur usmein jame silver ya copper ko tola jaata tha.) Coulombs ko moles mein badalne ke liye humein ek mole electrons ka charge chahiye, yaani Faraday constant (फैराडे स्थिरांक): $F = N_{A} e = (6.022 14 \times 10^{23} m o l^{- 1}) (1.602 177 \times 10^{- 19} C) = 96 485 C m o l^{- 1}$ . Itne charge ko ek faraday (1 F) kehte hain. Rough kaam ke liye 96 500 C mol⁻¹ chal jaata hai; purani books 96 487 C mol⁻¹ likhti hain, jisse teen significant figures par koi farq nahi padta.

Electrode equation tay karti hai ki product ke har mole ko kitne faradays chahiye. Ek mole $N a^{+}$ ko ek mole electrons (1 F) chahiye, ek mole $C u^{2 +}$ ko 2 F aur ek mole $A l^{3 +}$ ko 3 F:

\begin{aligned} N a^{+} + e^{-} & \overset{}{\to} N a \\ C u^{2 +} + 2 e^{-} & \overset{}{\to} C u \\ A l^{3 +} + 3 e^{-} & \overset{}{\to} A l \end{aligned}

Sab milakar, time $t$ tak behta current $I$ molar mass $M$ wale aise element ka mass $m$ jamata hai jiske ion ko $z$ electrons chahiye:

m = \frac{M}{z F} I t

Industrial cells bahut bade currents par chalte hain. Ek modern aluminium smelting cell lagbhag 300 000 A le jaata hai, jo har second lagbhag 3.1 mol electrons hai: per second lagbhag 28 g aluminium ke liye kaafi, yaani ek hi cell se roz lagbhag 2.4 tonne (Fig. 2.12 second law ko bench scale par dikhata hai).

3/3Ek faraday 1 mol Ag, ½ mol Cu aur ⅓ mol Au deta hai: 107.9 g, 31.8 g aur 65.7 g.

Fig. 2.12Same charge metals ki amounts unke equivalent masses ke ratio mein jamata hai (Faraday's second law).

Ek DC supply aur ammeter series mein jude teen cells se current chalate hain: silver nitrate, copper sulphate aur ek gold(III) salt. Pehle step mein circuit khaali cathodes ke saath dikhaya gaya hai. Doosre step mein ek label par 1 F = 96 485 coulombs likha hai aur har cathode par metal jam gaya hai: sabse zyada silver cathode par, copper par aadhe moles aur gold par ek tihaai. Teesre step mein bars amounts compare karte hain: 1 mol silver jiska weight 107.9 g hai, aadha mole copper jiska weight 31.8 g hai, aur ek tihaai mole gold jiska weight 65.7 g hai.

Example 2.10NCERT Example 2.10 · p. 52

Chammachon par silver-plating

Kuch chammachon ko silver nitrate solution mein 45 minute tak 0.80 A ke steady current se silver-plate kiya jaata hai. Kitna mass silver jamta hai? (Ag ka molar mass = 107.9 g mol⁻¹.)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Copper ki electrolytic refining mein impure copper ko banaya jaata hai…

cathode, taaki impurities reduce ho jayeinelectrolytesalt bridgeanode, taaki woh ghul jaye

Answer dekho

Answer (D) anode, taaki woh ghul jaye

Impure slab ko ghulna hai, jo oxidation hai, isliye woh anode hai (positive terminal se juda). Pure copper cathode par jamta hai, aur gold aur silver anode mud ki tarah ikattha hote hain.

2.5.1 Electrolysis ke Products

Copper sulphate ko electrolyse karo to copper milta hai; sodium chloride solution ko electrolyse karo to sodium bilkul nahi milta. Products teen cheezein tay karti hain: kaunse ions maujood hain, kya unse compete karne ke liye paani maujood hai, aur kya electrodes react kar sakte hain. Inert electrode (platinum, gold, graphite) sirf electrons aage deta hai; reactive electrode (copper, silver, nickel) khud oxidise ho sakta hai, jaise refining mein copper slab hua tha.

Har electrode ko ek muqabla samjho. Cathode par woh species jeetti hai jise reduce karna sabse aasaan ho: jis couple ka reduction potential sabse zyada ho. Anode par woh species jeetti hai jise oxidise karna sabse aasaan ho: jis couple ka reduction potential sabse kam ho. Lekin potentials sirf yeh batate hain ki kya ho sakta hai. Kuch electrode reactions itni sust hoti hain ki useful rate par chalne se pehle unhe ek extra push chahiye, ek overpotential (अधिविभव), aur paani se oxygen banana mashhoor roop se sust hai.

Molten sodium chloride mein koi muqabla nahi: $N a^{+}$ aur $C l^{-}$ hi akele candidates hain, isliye melt se sodium aur chlorine milte hain:

\begin{aligned} N a^{+} + e^{-} \overset{}{\to} N a \\ C l^{-} \overset{}{\to} \frac{1}{2} C l_{2} + e^{-} \end{aligned}

Salt ko paani mein gholo to paani dono muqablon mein shaamil ho jaata hai (Fig. 2.13). Cathode ka muqabla:

\begin{aligned} N a^{+} + e^{-} \overset{}{\to} N a (s) \\ E^{\circ} = - 2.71 V \\ H^{+} + e^{-} \overset{}{\to} \frac{1}{2} H_{2} (g) \\ E^{\circ} = 0.00 V \end{aligned}

$H^{+}$ ka reduction 2.71 V se jeetta hai. Thode se $H^{+}$ ions jitni tezi se use hote hain, utni hi tezi se paani se wapas aa jaate hain ( $H_{2} O \overset{}{⇌} H^{+} + O H^{-}$ ), isliye cathode reaction ko khud paani ke saath likhna behtar hai, jisse hydroxide bachta hai:

\begin{aligned} H_{2} O (l) + e^{-} \\ \overset{}{\to} \frac{1}{2} H_{2} (g) \\ + O H^{-} (a q) \end{aligned}

(2.35)

Anode ka muqabla:

\begin{aligned} C l^{-} (a q) \\ \overset{}{\to} \frac{1}{2} C l_{2} (g) + e^{-} \\ E_{C l_{2} / C l^{-}}^{\circ} = 1.36 V \end{aligned}

(2.36)

\begin{aligned} 2 H_{2} O (l) \overset{}{\to} O_{2} (g) \\ + 4 H^{+} (a q) + 4 e^{-} \\ E_{O_{2} / H_{2} O}^{\circ} = 1.23 V \end{aligned}

(2.37)

Yahan potentials ulti taraf ishaara karte hain: kam $E^{\circ}$ wale paani ko jeetna chahiye. Woh oxygen overpotential ki wajah se haarta hai, isliye concentrated brine chlorine deta hai. Dono electrode reactions jodkar, aur halves hataane ke liye double karke, overall change milta hai:

\begin{aligned} 2 N a C l (a q) \\ + 2 H_{2} O (l) \\ \overset{}{\to} 2 N a O H (a q) \\ + H_{2} (g) + C l_{2} (g) \end{aligned}

Ek process, teen bikne wale products: hydrogen, chlorine aur sodium hydroxide. Yeh chlor-alkali process India aur duniya bhar mein sabse badi electrochemical industries mein se ek hai.

3/4O₂ evolution dheema hai (overpotential), isliye Cl⁻ oxidise hota hai: chlorine nikalti hai.

Fig. 2.13Brine ka electrolysis: cathode par hydrogen, anode par chlorine, aur solution mein sodium hydroxide bachta hai.

Sodium chloride solution mein do inert electrodes wala ek cell DC supply se juda hai, left mein cathode aur right mein anode. Pehla step maujood species ke labels dikhata hai: sodium ions, chloride ions aur paani. Doosre step mein cathode highlight hai, do candidate reductions ke saath: minus 2.71 V par sodium ions cross kiye hue, aur paani se hydrogen par tick; wahan hydrogen bubbles uthte hain. Teesre step mein anode highlight hai: 1.23 V par paani se oxygen ko overpotential ki wajah se slow mark kiya gaya hai, aur 1.36 V par chloride se chlorine par tick hai; chlorine bubbles uthte hain. Chauthe step mein cathode ke paas ke solution ko sodium hydroxide label kiya gaya hai.

Concentration bhi jeetne wale ko badal sakta hai, kyunki asli competitors Nernst potentials hain, standard wale nahi. Platinum electrodes ke beech sulphuric acid ko electrolyse karo to anode product is par depend karta hai ki acid kitna strong hai:

\begin{aligned} 2 H_{2} O (l) \overset{}{\to} O_{2} (g) \\ + 4 H^{+} (a q) + 4 e^{-} \\ E_{O_{2} / H_{2} O}^{\circ} = 1.23 V \end{aligned}

(2.38)

\begin{aligned} 2 S O_{4}^{2 -} (a q) \\ \overset{}{\to} S_{2} O_{8}^{2 -} (a q) \\ + 2 e^{-} \\ E^{\circ} \approx 2.0 V \end{aligned}

(2.39)

Dilute acid oxygen deta hai. Bahut concentrated acid, jismein sulphate zyada aur free paani kam hai, iski jagah peroxodisulphate ion $S_{2} O_{8}^{2 -}$ deta hai; kabhi yahi hydrogen peroxide banane ka industrial raasta tha. Inert electrodes wale aqueous solutions ke liye kuch rules of thumb zyadatar cases cover kar lete hain:

Cathode. Kam reactive metals ke ions, jaise $A g^{+}$ aur $C u^{2 +}$ , reduce hokar metal bante hain. Bahut reactive metals ke ions ( $K^{+}$ , $N a^{+}$ , $M g^{2 +}$ , $A l^{3 +}$ ) solution mein rehte hain, aur paani reduce hokar hydrogen banta hai.
Anode. Halide ions ( $C l^{-}$ , $B r^{-}$ , $I^{-}$ ) oxidise hokar halogen bante hain, khaaskar concentrated hone par. $S O_{4}^{2 -}$ aur $N O_{3}^{-}$ jaise oxoanions nahi hote; unki jagah paani oxidise hokar oxygen banta hai.
Reactive anode. Copper, silver ya nickel anode ke saath anode ka metal khud ghul jaata hai.

Example

Copper sulphate, do tareekon se

Predict karo ki aqueous copper(II) sulphate ko (a) platinum electrodes ke beech aur (b) copper electrodes ke beech electrolyse karne par kya products bante hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveElectrolysis bench

Ek bench jismein left mein cathode aur right mein anode ke beech ek cell bana hai. Ek menu mein nau set-ups hain, jinmein molten sodium chloride, brine, platinum ya copper electrodes ke saath copper sulphate, platinum ya silver electrodes ke saath silver nitrate, dilute sulphuric acid, sodium sulphate solution aur potassium iodide solution shaamil hain. Har electrode ke liye panel competing half-reactions ko unke standard potentials ke saath list karta hai, jeetne wale ko mark karta hai aur samjhata hai kyun, jahan zaroori ho wahan overpotential bhi. Sliders current ko 0.1 se 5 amperes aur time ko 1 se 120 minutes tak set karte hain. Run button ghadi chalte hue bubbles ya badhte deposit ko animate karta hai, aur readouts charge, electrons ke moles aur 298 kelvin aur 1 bar par har product ka mass ya gas volume dete hain.

Fig. 2.14Ek electrolyte aur electrodes chuno, har electrode par muqabla padho, phir current aur time set karke Run dabao aur dekho ki har product kitna banta hai.

Practice · 3 questions

Intext questions 2.10–2.12: answers ke saath

NumericalEasyNCERT Intext Question 2.10
Ek phone charger cable 30 minute tak steady 2.0 A le jaati hai. Usse kitne electrons guzarte hain?
Solution dikhao
$Q = I t = 2.0 A \times 1800 s = 3600 C$ .
Electrons ki sankhya $= 3600 / (1.602 \times 10^{- 19}) = 2.25 \times 10^{22}$ . (Yaani, $3600 / 96 485 = 0.0373$ mol electrons.)
2.11
Quick answerEasyNCERT Intext Question 2.11
Inmein se kaunse metals industry mein electrolysis se banaye jaate hain?
sodiumaluminiumironmagnesiumlithiummercury
Saare sahi options chuno
Answer dekho
Answer
(A) sodium
(B) aluminium
(D) magnesium
(E) lithium
Sodium, aluminium, magnesium aur lithium itne reactive hain ki carbon se reduce nahi ho sakte, isliye unhe molten salts ko electrolyse karke banaya jaata hai. (Calcium aur potassium bhi.) Iron blast furnace mein carbon monoxide se banta hai, aur mercury uske ore, cinnabar ( $H g S$ ), ko roast karke.
NumericalMediumNCERT Intext Question 2.12
Chlorate ions ko chloride mein reduce kiya ja sakta hai: $C l O_{3}^{-} + 6 H^{+} + 6 e^{-}$ $\overset{}{\to} C l^{-} + 3 H_{2} O$ . 0.50 mol $C l O_{3}^{-}$ ko reduce karne ke liye kitne coulombs charge chahiye?
C
Solution dikhao
Har $C l O_{3}^{-}$ 6 electrons leta hai (chlorine +5 se −1 par jaata hai), isliye 0.50 mol ko 3.0 mol electrons chahiye, yaani 3.0 F.
$Q = 3.0 \times 96 485 = 2.89 \times 10^{5} C$ .

2.6 Batteries

Torch, TV remote, phone, car aur ghar ka inverter, sabko alag mizaaj wala ek portable galvanic cell chahiye. Do badi families hain. Primary batteries apni reaction ek baar chalati hain aur phir khatam ho jaati hain (aur unhe recycle karna chahiye, kachre mein nahi phenkna); secondary batteries ko charger se ulta chalakar dobara use kiya ja sakta hai. Family koi bhi ho, designers ki wish list same hai: per gram aur per cubic centimetre jitni ho sake utni energy, aakhir tak steady rehne wala voltage, jeb mein safety aur kam price. (Strictly, battery series mein jude cells ka set hai, lekin roz ki bolchaal mein ek cell ko bhi battery kehte hain.)

2.6.1 Primary Batteries

Wall clock ya TV remote ka 1.5 V dry cell ek primary cell hai, jise Georges Leclanché ne 1860s mein banaya tha (Fig. 2.15). Ise kaat kar kholo (yeh kaam kisi bade ka hai: andar ka paste corrosive hai) to dikhega ki bahar ka zinc can khud anode hai, isiliye purane cells zinc ke khaa jaane par leak karte hain. Beech mein ek graphite rod hai jo current collect karti hai, jiske chaaron taraf manganese dioxide aur carbon powder ka kaala mixture bhara hai, jo asli cathode reactant hai. Ammonium chloride aur zinc chloride ka geela paste electrolyte ka kaam karta hai; 'dry' ka matlab sirf itna hai ki girne ke liye koi free liquid nahi hai.

Fig. 2.15Dry (Leclanché) cell ka section. Zinc can anode hai; graphite rod manganese dioxide cathode se current collect karti hai.NCERT Fig. 2.8 · p. 54

Ek dry cell ko lambai mein kaata gaya hai. Bahar ka zinc can, jis par anode aur negative likha hai, deewarein aur base banata hai. Iske andar moist ammonium chloride aur zinc chloride paste ki ek layer hai, phir carbon powder mile manganese dioxide ka mota core, aur beech mein ek graphite rod upar brass cap tak jaati hai, jis par cathode aur positive likha hai. Ek seal upar band karti hai aur ek jacket bahar lipta hai.

Andar jo hota hai woh complicated hai, lekin half-reactions ka ek simplified pair use samjha deta hai:

anode

\begin{aligned} Z n (s) \\ \overset{}{\to} Z n^{2 +} (a q) + 2 e^{-} \end{aligned}

cathode

\begin{aligned} M n O_{2} (s) \\ + N H_{4}^{+} (a q) + e^{-} \\ \overset{}{\to} M n O (O H) (s) \\ + N H_{3} (a q) \end{aligned}

Cathode par bani ammonia gas ki tarah jama nahi hoti: anode se aaye zinc ions use complex ion $[Z n (N H_{3})_{4}]^{2 +}$ ki tarah baandh lete hain. Asal mein reduce hone wala element manganese hai, jo $M n O_{2}$ mein +4 se $M n O (O H)$ mein +3 par aata hai.

Watches aur hearing aids ko aise voltage par bahut chhota current chahiye jo gire nahi. Mercury cell, ek button cell, isi ke liye bana tha: mercury mein ghula zinc (amalgam) anode ki tarah, carbon mila mercury(II) oxide cathode ki tarah, aur dono ke beech KOH aur ZnO ka paste:

anode

\begin{aligned} Z n (H g) + 2 O H^{-} \\ \overset{}{\to} Z n O (s) \\ + H_{2} O (l) + 2 e^{-} \end{aligned}

cathode

\begin{aligned} H g O (s) \\ + H_{2} O (l) + 2 e^{-} \\ \overset{}{\to} H g (l) + 2 O H^{-} \end{aligned}

overall

\begin{aligned} Z n (H g) + H g O (s) \\ \overset{}{\to} Z n O (s) + H g (l) \end{aligned}

Overall equation dekho: usmein koi dissolved ion nahi hai. Aisa kuch bhi $Q$ mein nahi aata jiska concentration badal sake, isliye Nernst equation ke hisaab se voltage, lagbhag 1.35 V, cell ki poori life mein lagbhag constant rehta hai.

2.6.2 Secondary Batteries

Car ke bonnet ke neeche wale box mein, ya power cut mein pankhe chalane wale inverter mein, lead storage battery hoti hai, jo aaj bhi bade paimane par use hone wala sabse purana rechargeable design hai (Gaston Planté ne pehli 1859 mein banayi thi). Iske har cell mein spongy lead ki ek plate aur lead dioxide se bhari lead grid hoti hai, dono kaafi strong sulphuric acid mein, mass ke hisaab se lagbhag 35–40 %. Ise secondary (rechargeable) cell isliye kehte hain kyunki charger iski reaction ko ulta chalakar dono plates ko phir se bana sakta hai; achhi tarah rakhi battery aise saikdon cycles jhel leti hai. Discharge par:

anode

\begin{aligned} P b (s) + S O_{4}^{2 -} (a q) \\ \overset{}{\to} P b S O_{4} (s) + 2 e^{-} \end{aligned}

cathode

\begin{aligned} P b O_{2} (s) \\ + S O_{4}^{2 -} (a q) \\ + 4 H^{+} (a q) + 2 e^{-} \\ \overset{}{\to} P b S O_{4} (s) \\ + 2 H_{2} O (l) \end{aligned}

overall

\begin{aligned} P b (s) + P b O_{2} (s) \\ + 2 H_{2} S O_{4} (a q) \\ \overset{}{\to} 2 P b S O_{4} (s) \\ + 2 H_{2} O (l) \end{aligned}

Dono electrodes lead sulphate ban jaate hain, aur sulphuric acid ki jagah paani aa jaata hai, isliye battery ke run down hone par acid kamzor aur kam dense hota jaata hai. Mechanic hydrometer se acid ki density naapkar state of charge check kar sakta hai. Charging par external supply dono electrode reactions ko ulta kar deti hai: ek plate par lead sulphate wapas lead banta hai, doosri par lead dioxide, aur acid phir se ban jaata hai (Fig. 2.16). Har cell lagbhag 2 V deta hai, isliye 12 V ki car battery mein series mein chhe cells hote hain.

2/3Dono plates PbSO₄ ban jaati hain aur acid kamzor hota hai (lagbhag 1.12 g cm⁻³).

Fig. 2.16Ek lead storage cell: discharge hone par dono plates par lead sulphate jam jaata hai aur acid dilute ho jaata hai; charging dono changes ko ulta kar deti hai.NCERT Fig. 2.10 · p. 55

Ek single cell mein sulphuric acid mein ek lead plate, negative mark ki hui, aur ek lead dioxide plate, positive mark ki hui, hai, aur ek chhota hydrometer float acid ki density dikhata hai. Pehle step mein plates saaf hain aur acid density 1.28 grams per cubic centimetre padhti hai. Doosre step mein ek lamp juda hai aur jal raha hai, electrons lead plate se bahar behte hain, safed lead sulphate dono plates par jam jaata hai aur density lagbhag 1.12 tak gir jaati hai. Teesre step mein lamp ki jagah charger lagta hai aur current ko ulti taraf chalata hai; dono plates se lead sulphate gayab ho jaata hai aur density wapas 1.28 ho jaati hai.

Nickel–cadmium (Ni–Cd) cells banana lead cells se mehenga hai lekin woh zyada charge–discharge cycles jhelte hain; kabhi yahi cordless drills aur emergency lights chalate the. Inki overall discharge reaction hai:

\begin{aligned} C d (s) \\ + 2 N i (O H)_{3} (s) \\ \overset{}{\to} C d O (s) \\ + 2 N i (O H)_{2} (s) \\ + H_{2} O (l) \end{aligned}

Kai books nickel electrode ko $N i O (O H)$ likhti hain; dono tarah nickel +3 se +2 par aur cadmium 0 se +2 par jaata hai. Cadmium bhi zehreela hai, aur Ni–Cd cells ki jagah ab zyadatar nickel–metal hydride aur lithium-ion cells aa gaye hain.

2.7 Fuel Cells

Kuch shehri buses ab hydrogen par chalti hain aur peeche sirf thoda sa paani chhodti hain. Unka engine ek fuel cell (ईंधन सेल) hai: ek galvanic cell jo kabhi run down nahi hota, kyunki fuel aur oxidant lagaataar pipe se andar aate hain aur products pipe se bahar jaate hain. Coal ya gas power station se compare karo, jo lambe raaste se electricity tak pahunchta hai, jalne, steam, turbine aur generator se hokar, har stage par energy khota hai aur fuel ki energy ka sirf 35–40 % deta hai (saath mein dhuan aur carbon dioxide). Fuel cell heat engine ko skip kar deta hai. Hydrogen favourite fuel hai, lekin natural gas, methanol aur ethanol bhi is thande, electrochemical tareeke se 'jalaaye' ja sakte hain, oxygen ya hawa ko oxidant banakar.

Space mein udaya gaya classic design alkaline hydrogen–oxygen cell hai (Fig. 2.17). Iske dono electrodes porous carbon ke hain, isliye har gas ek taraf se andar sokh sakti hai jabki garam, concentrated sodium (ya potassium) hydroxide solution doosri taraf se electrode ko geela karta hai. Bare carbon par hydrogen aur oxygen bahut dheere react karte hain, isliye electrodes par catalyst chhidka jaata hai, finely divided platinum ya palladium. Half-reactions hain:

cathode

\begin{aligned} O_{2} (g) \\ + 2 H_{2} O (l) + 4 e^{-} \\ \overset{}{\to} 4 O H^{-} (a q) \end{aligned}

anode

\begin{aligned} 2 H_{2} (g) + 4 O H^{-} (a q) \\ \overset{}{\to} 4 H_{2} O (l) + 4 e^{-} \end{aligned}

overall

\begin{aligned} 2 H_{2} (g) + O_{2} (g) \\ \overset{}{\to} 2 H_{2} O (l) \end{aligned}

2/3Electrons motor se hokar behte hain; OH⁻ ions electrolyte paar karke anode tak jaate hain.

Fig. 2.17Alkaline fuel cell mein hydroxide ions electrolyte se charge le jaate hain jabki electrons external circuit mein work karte hain.NCERT Fig. 2.12 · p. 56

Cell horizontal layers ki tarah bana hai: upar hydrogen channel, ek porous carbon anode, garam sodium hydroxide electrolyte ki layer, ek porous carbon cathode aur neeche oxygen channel. Left mein ek wire anode se ek chhoti motor se hokar cathode tak jaati hai. Pehle step mein hydrogen aur oxygen apne channels mein behti hain. Doosre step mein electrons anode se motor se hokar cathode tak jaate hain, jabki hydroxide ions electrolyte mein cathode se anode ki taraf upar jaate hain. Teesre step mein anode par paani banta hai aur right mein steam ki tarah bahar jaata hai.

Har Apollo command module mein aise teen cells the, aur electricity ke alawa woh crew ko peene ka paani bhi dete the.

Fuel cell kitna efficient ho sakta hai? Sabse zyada available electrical work $- Δ_{r} G$ hai, jabki hydrogen jalane par $- Δ_{r} H$ heat ki tarah niklega. 298 K par hydrogen aur oxygen se liquid paani banne ke liye $Δ_{r} G^{\circ} / Δ_{r} H^{\circ} = 237 / 286 \approx 0.83$ , isliye ek ideal cell fuel ki 83 % energy ko electricity mein badal sakta hai. Asli fuel cells lagbhag 50–70 % tak pahunchte hain, jo phir bhi thermal plant se kahin behtar hai. Ab badi challenges hain platinum se saste catalysts, aur hydrogen ko saaf tareeke se banana aur store karna (chapter ke aakhir ka box dekho).

Quick check

3 mein se 1

Alkaline hydrogen–oxygen fuel cell mein anode par kya hota hai?

O_{2}

reduce hokar

O H^{-}

banta hai

H_{2} O

toot kar

H_{2}

aur

O_{2}

banta hai

H_{2}

oxidise hota hai,

O H^{-}

use karta hai aur paani banata haiplatinum oxidise hota hai

Answer dekho

Answer (C) $H_{2}$ oxidise hota hai, $O H^{-}$ use karta hai aur paani banata hai

Anode par hydrogen electrons khoti hai:

2 H_{2} + 4 O H^{-} \overset{}{\to} 4 H_{2} O + 4 e^{-}

. Oxygen cathode par reduce hoti hai. Platinum sirf catalyst hai.

2.8 Corrosion

Delhi ke Qutb complex ka iron pillar lagbhag 1600 saal se khule mein khada hai aur us par rust ka nishaan bhi mushkil se hai, jabki ek monsoon bahar chhoda gaya steel ka gate parat chhodne lagta hai. Aisa kyun? Zyadatar refined metals un ores se kam stable hote hain jinse woh bane, aur hawa aur nami milne par dheere-dheere wapas usi taraf jaate hain: copper ki chhaton aur bronze statues par basic copper carbonate ki hari parat jam jaati hai, silver ki paayal silver sulphide ki film se kaali pad jaati hai, aur steel par parat-daar laal-bhoora rust aa jaata hai. Yeh badlaav, corrosion (संक्षारण), pipelines, railway tracks, ships aur concrete ke andar ke steel ko kha jaata hai, aur iska kharcha har saal world GDP ka lagbhag 3 % aanka gaya hai. (Delhi ka pillar ek patli protective film ki wajah se bacha hai jo kaafi sookhe mausam mein uske unusually phosphorus-rich iron par bani.)

Rusting ko paani aur oxygen dono chahiye, aur ise samajhne ka sabse saaf tareeka hai ise metal par hi bana ek galvanic cell maanna (Fig. 2.18). Koi jagah, shayad ek scratch, ek tanaav wala mod, ya paani ki boond ke neeche oxygen ki kami wala hissa, anode ban jaati hai, jahan iron oxidise hota hai:

anode

\begin{aligned} 2 F e (s) \\ \overset{}{\to} 2 F e^{2 +} (a q) \\ + 4 e^{-} \\ E_{F e^{2 +} / F e}^{\circ} = - 0.44 V \end{aligned}

Anode par nikle electrons khud iron se hokar, jo wire ka kaam karta hai, aisi jagah tak jaate hain jahan oxygen bharpoor ho. Wahan, cathode par, oxygen hydrogen ions ke saath electrons le leti hai. Baarish ka paani kabhi bilkul pure nahi hota: ghuli carbon dioxide use halka acidic (carbonic acid) bana deti hai, aur traffic aur factories ke paas ghuli sulphur dioxide aur acid jod deti hai:

cathode

\begin{aligned} O_{2} (g) + 4 H^{+} (a q) \\ + 4 e^{-} \overset{}{\to} 2 H_{2} O (l) \\ E_{O_{2} / H_{2} O}^{\circ} = 1.23 V \end{aligned}

overall

\begin{aligned} 2 F e + O_{2} + 4 H^{+} \\ \overset{}{\to} 2 F e^{2 +} + 2 H_{2} O \\ E_{cell}^{\circ} = 1.67 V \end{aligned}

$E_{cell}^{\circ} = 1.67 V$ ke saath cell strongly spontaneous hai. Rust khud ek step baad banta hai: $F e^{2 +}$ ions anode se door chale jaate hain, ghuli oxygen se milte hain aur oxidise hokar iron(III) bante hain, jo hydrated iron(III) oxide, $F e_{2} O_{3} \cdot x H_{2} O$ , ki tarah alag hota hai aur saath mein $H^{+}$ wapas bana deta hai. Kyunki rust porous hai aur parat ki tarah utar jaata hai, woh neeche ke metal ko kabhi seal nahi karta. Namak paani ki film ko kahin behtar ionic conductor banakar sab kuch tez kar deta hai, isiliye samundar ke paas gaadiyon mein jaldi rust lagta hai.

4/4Fe²⁺ oxygen se milkar aur oxidise hota hai: rust ki ek ring banti hai.

Fig. 2.18Rusting ek electrochemical cell hai: anode par iron ghulta hai, electrons metal se hokar behte hain, aur cathode par oxygen reduce hoti hai.NCERT Fig. 2.13 · p. 57

Ek cross-section mein iron plate par paani ki boond rakhi dikhti hai. Pehle step mein hawa se oxygen zyadatar boond ke patle kinare ke paas ghulti hai. Doosre step mein boond ke centre ke neeche ek gaddha banta hai, label anode, jahan iron electrons khota hai aur iron(II) ions paani mein jaate hain; arrows dikhate hain ki electrons iron se hokar kinare ki taraf jaate hain. Teesre step mein boond ke kinare ko cathode label kiya gaya hai, jahan oxygen aur hydrogen ions electrons lekar paani banate hain. Chauthe step mein iron(II) ions paani mein oxygen se milte hain aur kinare ke paas rust, yaani hydrated iron(III) oxide, ki ek ring banti hai.

Protection ka har tareeka corrosion cell ko kahin na kahin tod deta hai: woh paani aur oxygen ko metal se door rakhta hai, ya kisi doosre metal ko anode bana deta hai.

Barrier coatings hawa aur paani ko door rakhti hain: paint, grease ya oil, plastic films aur epoxy paints jaisi chemical coatings. Yeh tabhi tak bachati hain jab tak tooti na hon.
Metal coatings. Tin plating, jo food cans ke liye use hoti hai, iron ko seal kar deti hai, lekin tin iron se kam reactive hai, isliye scratch lagte hi khula iron anode ban jaata hai aur aur bhi tezi se rust hota hai. Zinc coating (galvanising) scratch lagne par bhi bachati hai, kyunki zinc zyada reactive hai aur iron se pehle oxidise hota hai.
Sacrificial (cathodic) protection. Magnesium ya zinc jaise zyada reactive metal ka ek block iron se wire se joda jaata hai, ya ship ke hull par bolt kiya jaata hai. Woh anode ban jaata hai aur iron ki jagah khud corrode hota hai, iron cathode rehta hai, aur khatam hone par block bas badal diya jaata hai. Zameen mein dabi pipelines aur water-heater tanks isi tarah bachaye jaate hain.

Ghar par try karo

Kaunsi nail pehle rust hoti hai?

Tumhe chahiye

iron ki 5 nails
sandpaper
zinc ki ek strip ya galvanised wire, aur bare copper wire ka ek tukda
namak, nal ka paani, khaane ka tel
labels wale 5 saaf glass jars

Pehle guess karo

Tumhe kya lagta hai, kaunsi nail sabse tez rust hogi?

nail 1, nal ke paani meinnail 3, tel ke neechenail 4, zinc mein liptinail 5, copper mein lipti

Paanch iron nails ko sandpaper se saaf karo aur unpar 1 se 5 tak label lagao. Nail 4 par zinc ki ek strip (hardware shop se) ya galvanised wire ka tukda kaskar lapeto, aur nail 5 par bare copper wire ka ek tukda.
Har nail ko apne labelled jar mein rakho: nail 1 nal ke paani mein; nail 2 namak ke paani mein (ek cup mein ek chammach namak); nail 3 ubaal kar thande kiye paani mein, upar khaane ke tel ki layer daalkar; nails 4 aur 5 namak ke paani mein.
Inhe ek hafte tak chhod do, roz check karo. Note karo ki rust pehli baar kab dikhta hai aur kitna banta hai.

Table 2.8Rusting observations: day 2 par koi bhi rust aur day 7 tak kitna, note karo

Nail	Observations
1: nal ka paani
2: namak ka paani
3: ubla paani, tel
4: namak, zinc
5: namak, copper

Tumhari entries apne aap save hoti hain.

Nail 5, jo namak ke paani mein copper ke contact mein hai, sabse pehle aur sabse zyada rust hoti hai, kyunki iron iron–copper cell ka anode hai. Nail 2 nail 1 se tez rust hoti hai, kyunki namak paani ko behtar ionic conductor bana deta hai. Nail 3 mushkil se rust hoti hai: ubaalne se ghuli oxygen nikal gayi aur tel aur oxygen ko ghulne se rokta hai. Nail 4 chamakti rehti hai jabki zinc uski jagah corrode hota hai, kyunki zinc anode hai: yeh jar mein sacrificial protection hai.

Real life mein

Hydrogen economy

2023 mein India ne apna National Green Hydrogen Mission launch kiya, jiska target 2030 tak har saal lagbhag paanch million tonne 'green' hydrogen banana hai. Aaj bani zyadatar hydrogen 'grey' hai: woh natural gas se aati hai aur carbon dioxide chhodti hai. Green hydrogen iski jagah dhoop ya hawa se mili electricity se paani ko electrolyse karke banti hai, isliye use banane mein carbon dioxide nahi nikalti. Phir woh fertiliser plants, refineries aur steelworks ko di ja sakti hai, ya fuel cell mein wapas electricity banayi ja sakti hai, jahan uska akela product paani hai. Dikkat cost ki hai: green hydrogen ab bhi grey se kaafi mehengi hai, kuch had tak isliye kyunki electrolysers platinum aur iridium jaise kam milne wale catalyst metals par nirbhar hain. Electrolysers aur fuel cells dono ke liye saste, zyada chalne wale electrode materials ek bada research goal hain.

Practice · 3 questions

Intext questions 2.13–2.15: answers ke saath

Short answerMediumNCERT Intext Question 2.13
Car ka alternator uski lead storage battery ko recharge karta hai. Charging ke dauraan har plate par reaction likho, batao kya use hota hai aur kya banta hai, aur acid ka kya hota hai.
Solution dikhao
AnswerDono plates ka $P b S O_{4}$ wapas Pb aur $P b O_{2}$ ban jaata hai; paani use hota hai aur sulphuric acid phir se banta hai.
Charger ke negative terminal se judi plate par (discharge ka anode), lead sulphate reduce hota hai: $P b S O_{4} (s) + 2 e^{-}$ $\overset{}{\to} P b (s) + S O_{4}^{2 -} (a q)$ .
Doosri plate par woh oxidise hokar wapas lead dioxide banta hai: $P b S O_{4} (s) + 2 H_{2} O (l)$ $\overset{}{\to} P b O_{2} (s)$ $+ S O_{4}^{2 -} (a q)$ $+ 4 H^{+} (a q) + 2 e^{-}$ .
Overall: $2 P b S O_{4} (s) + 2 H_{2} O (l)$ $\overset{}{\to} P b (s) + P b O_{2} (s)$ $+ 2 H_{2} S O_{4} (a q)$ . Lead sulphate aur paani use hote hain; lead, lead dioxide aur sulphuric acid bante hain, isliye acid phir se strong aur zyada dense ho jaata hai. Alternator ki electrical energy chemical energy ki tarah store hoti hai.
Short answerMediumNCERT Intext Question 2.14
Hydrogen ke alawa do fuels ke naam batao jo fuel cells mein use ho sakte hain, aur acidic electrolyte mein unmein se ek ki anode half-reaction likho.
Solution dikhao
AnswerMethanol aur methane; jaise $C H_{3} O H + H_{2} O$ $\overset{}{\to} C O_{2} + 6 H^{+} + 6 e^{-}$ .
Methanol aur methane common choices hain (ethanol aur natural gas bhi). Methanol mein carbon −2 se +4 par jaata hai, isliye chhe electrons nikalte hain: $C H_{3} O H (l) + H_{2} O (l)$ $\overset{}{\to} C O_{2} (g)$ $+ 6 H^{+} (a q) + 6 e^{-}$ .
Methane mein carbon −4 se +4 par jaata hai: $C H_{4} (g) + 2 H_{2} O (l)$ $\overset{}{\to} C O_{2} (g)$ $+ 8 H^{+} (a q) + 8 e^{-}$ .
Short answerMediumNCERT Intext Question 2.15
Baarish ke paani ki ek boond steel ki railing par padi hai. Galvanic cell ki tarah samjhao ki boond ke beech ke neeche gaddha kyun banta hai jabki rust uske kinare ke paas ring ki tarah dikhta hai.
Solution dikhao
AnswerCentre anode hai (kam oxygen); kinara cathode hai (oxygen-rich); rust wahan banta hai jahan Fe²⁺ oxygen se milta hai.
Oxygen boond mein zyadatar uske patle kinare se ghulti hai, isliye kinara cathode ban jaata hai, jahan $O_{2} + 4 H^{+} + 4 e^{-}$ $\overset{}{\to} 2 H_{2} O$ .
Oxygen ki kami wale centre ke neeche iron anode hai: $F e \overset{}{\to} F e^{2 +} + 2 e^{-}$ , isliye wahan metal ghis jaata hai aur gaddha banta hai. Electrons steel se hokar kinare tak jaate hain, aur paani circuit poora karne ke liye ions le jaata hai.
$F e^{2 +}$ ions bahar ki taraf drift karte hain aur ghuli oxygen se oxidise hokar iron(III) bante hain, jo $F e_{2} O_{3} \cdot x H_{2} O$ ki tarah alag hota hai, isliye rust kinare ke paas dikhta hai, gaddhe ke upar nahi.

Recap

Summary aur key points

Electrochemical cell mein ek electrolyte mein do electrodes hote hain. Galvanic cell ek spontaneous redox reaction se electrical work karwata hai; electrolytic cell electrical energy se ek non-spontaneous reaction chalata hai (Section 2.1).
Oxidation anode par aur reduction cathode par hota hai. Galvanic cell mein anode negative hai; electrons wire se anode se cathode tak behte hain, aur salt bridge solutions ko neutral rakhta hai (Section 2.2).
Standard electrode potentials standard hydrogen electrode (0 V) ke against measure kiye gaye reduction potentials hain, aur $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ . Oxidising agent table mein apne neeche ke kisi bhi reducing agent ke saath react karta hai (Section 2.2.1).
Nernst equation, 298 K par $E_{cell} = E_{cell}^{\circ} - (0.0591 / n) \log Q$ , kisi bhi concentration par potential deti hai (Section 2.3).
Equilibrium par $E_{cell} = 0$ , isliye 298 K par $E_{cell}^{\circ} = (0.0591 / n) \log K_{c}$ (Section 2.3.1).
$Δ_{r} G = - n F E_{cell}$ aur $Δ_{r} G^{\circ} = - n F E_{cell}^{\circ} = - R T \ln K$ . $E$ intensive hai; $Δ_{r} G$ extensive hai (Section 2.3.2).
$R = ρ l / A$ , $G = 1 / R = κ A / l$ aur $Λ_{m} = κ / c$ . Metals electrons se conduct karte hain; solutions ions se, aur garam hone par behtar (Section 2.4).
Conductivity ko AC se, known cell constant $G^{*} = l / A = R κ$ wale conductivity cell mein, Wheatstone bridge ya meter se measure kiya jaata hai (Section 2.4.1).
Dilution par $κ$ girta hai lekin $Λ_{m}$ badhta hai: strong electrolytes mein dheere ( $Λ_{m} = Λ_{m}^{\circ} - A \sqrt{c}$ ), weak mein tezi se (Section 2.4.2).
Kohlrausch's law: $Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}$ . Isse weak electrolytes ki $Λ_{m}^{\circ}$ milti hai, phir $α = Λ_{m} / Λ_{m}^{\circ}$ aur $K_{a}$ (Law).
Faraday's laws: jamne wali amount charge $Q = I t$ ke proportional hai, aur ek faraday ( $96 485 C m o l^{- 1}$ ) charge $z$ wale ion ka $1 / z$ mole jamata hai (Section 2.5).
Electrolysis ke products electrolyte par, dono electrodes par compete karte paani par, electrodes par aur overpotential par depend karte hain (Section 2.5.1).
Primary cells (dry cell, mercury cell) ek baar use hote hain; secondary cells (lead storage, Ni–Cd) recharge hote hain; fuel cells tab tak chalte hain jab tak fuel milta rahe (Section 2.6, Section 2.7).
Corrosion metal surface par bana ek electrochemical cell hai; coatings, galvanising aur sacrificial anodes ise rokte hain (Section 2.8).

Flash cards · 11

Yaad rakhne wale formulas

1 / 11

Practice · 23 questions

Practice questions answers ke saath (NCERT “Exercises”)

2.1
Quick answerEasyNCERT Exercises 2.1
Aluminium, lead, nickel, silver aur zinc ko sabse strong se sabse weak displacer tak rank karo: tumhari list ka har metal apne baad wale har metal ko solution se bahar dhakel sake.
1. zinc
2. nickel
3. lead
4. silver
5. aluminium
Answer dekho
Answer
aluminium
zinc
nickel
lead
silver
$E^{\circ}$ : Al −1.66, Zn −0.76, Ni −0.25, Pb −0.13, Ag +0.80 V. Potential jitna kam, reducing agent utna strong, isliye har metal apne baad walon ko displace karta hai.
2.2
Quick answerEasyNCERT Exercises 2.2
Diya hai $E^{\circ}$ : $L i^{+} / L i$ −3.04 V, $N a^{+} / N a$ −2.71 V, $F e^{2 +} / F e$ −0.44 V, $C u^{2 +} / C u$ +0.34 V, $A g^{+} / A g$ +0.80 V. Metals ko sabse weak reducing agent se sabse strong tak order mein lagao.
1. Ag
2. Cu
3. Na
4. Li
5. Fe
Answer dekho
Answer
Ag
Cu
Fe
Na
Li
Sabse zyada $E^{\circ}$ (silver) sabse weak reducing agent hai; sabse kam (lithium) sabse strong.
2.3
Short answerMediumNCERT Exercises 2.3
Ek galvanic cell reaction $2 A l (s) + 3 N i^{2 +} (a q)$ $\overset{}{\to} 2 A l^{3 +} (a q) + 3 N i (s)$ chalata hai.
1. Cell ko standard notation mein likho.
  Solution dikhao
  Answer $A l (s) ∣ A l^{3 +} (a q)$ $∣ ∣ N i^{2 +} (a q) ∣ N i (s)$
  Aluminium oxidise hota hai, isliye woh anode hai aur left mein jaata hai: $A l (s) ∣ A l^{3 +} (a q)$ $∣ ∣ N i^{2 +} (a q) ∣ N i (s)$ .
2. Kaunsa electrode negative terminal hai?
  aluminium electrodenickel electrode
  Answer dekho
  Answer (A) aluminium electrode
  Aluminium anode hai: oxidation us par electrons chhod deta hai.
3. Wire mein current kaun le jaata hai, aur cell mein kaun?
  Solution dikhao
  AnswerWire mein electrons; solutions aur salt bridge mein ions.
  Electrons wire se Al se Ni ki taraf behte hain. Andar $A l^{3 +}$ ions left solution mein aate hain aur $N i^{2 +}$ right wale se nikalte hain; salt bridge se anions aluminium side ki taraf aur cations nickel side ki taraf jaate hain.
4. Har electrode par reaction likho aur $E_{cell}^{\circ}$ nikaalo.
  Solution dikhao
  Answer $E_{cell}^{\circ} = 1.41 V$
  Anode: $A l \overset{}{\to} A l^{3 +} + 3 e^{-}$ . Cathode: $N i^{2 +} + 2 e^{-} \overset{}{\to} N i$ . Chhe electrons balance karne ke liye 2 aur 3 se multiply karne par cell reaction milti hai.
  $E_{cell}^{\circ} = - 0.25 - (- 1.66) = 1.41 V$ .
2.4
NumericalMediumNCERT Exercises 2.4
Table 2.2 use karke har reaction ke liye 298 K par $E_{cell}^{\circ}$ , $Δ_{r} G^{\circ}$ aur $K_{c}$ nikaalo.
1. $2 A l (s) + 3 S n^{2 +} (a q)$ $\overset{}{\to} 2 A l^{3 +} (a q)$ $+ 3 S n (s)$
  Solution dikhao
  Answer1.52~V; −880~kJ~mol⁻¹; $K \approx 10^{154}$
  $E^{\circ} = - 0.14 - (- 1.66) = 1.52 V$ , aur $n = 6$ .
  $Δ_{r} G^{\circ} = - 6 \times 96 485 \times 1.52 = - 8.80 \times 10^{5} J m o l^{- 1} = - 880 k J m o l^{- 1}$ .
  $\log K_{c} = 6 \times 1.52 / 0.0591 = 154.3$ , isliye $K_{c} \approx 2 \times 10^{154}$ .
2. $C l_{2} (g) + 2 B r^{-} (a q)$ $\overset{}{\to} 2 C l^{-} (a q) + B r_{2} (l)$
  Solution dikhao
  Answer0.27~V; −52.1~kJ~mol⁻¹; $K \approx 1.4 \times 10^{9}$
  $E^{\circ} = 1.36 - 1.09 = 0.27 V$ ; $n = 2$ .
  $Δ_{r} G^{\circ} = - 2 \times 96 485 \times 0.27 = - 5.21 \times 10^{4} J m o l^{- 1} = - 52.1 k J m o l^{- 1}$ .
  $\log K_{c} = 9.14$ , isliye $K_{c} \approx 1.4 \times 10^{9}$ .
2.5
NumericalMediumNCERT Exercises 2.5
Neeche diye har cell ke liye Nernst equation set up karo aur 298 K par emf nikaalo.
1. $N i (s) ∣ N i^{2 +} (0.0020 M)$ $∣ ∣ C u^{2 +} (0.050 M)$ $∣ C u (s)$
  V
  Solution dikhao
  $E = E^{\circ} - \frac{0.0591}{2} \log \frac{[N i^{2 +}]}{[C u^{2 +}]}$ , jahan $E^{\circ} = 0.59 V$ aur $Q = 0.040$ .
  $E = 0.59 + 0.041 = 0.631 V$ .
2. $Z n (s) ∣ Z n^{2 +} (0.10 M)$ $∣ ∣ H^{+} (1 M)$ $∣ H_{2} (g, 1 b a r) ∣ P t (s)$
  V
  Solution dikhao
  $E = 0.76 - \frac{0.0591}{2} \log \frac{0.10 \times 1}{1^{2}} = 0.76 + 0.030 = 0.790 V$ .
3. $P b (s) ∣ P b^{2 +} (0.020 M)$ $∣ ∣ H^{+} (0.010 M)$ $∣ H_{2} (g, 1 b a r) ∣ P t (s)$
  V
  Solution dikhao
  $Q = [P b^{2 +}] p_{H_{2}} / [H^{+}]^{2} = 0.020 / (0.010)^{2} = 200$ .
  $E = 0.13 - 0.0296 \times \log 200 = 0.13 - 0.068 = 0.062 V$ .
4. $P t (s) ∣ I - (0.020 M)$ $∣ I_{2} (s) ∣ ∣ H^{+} (0.010 M)$ $∣ H_{2} (g, 1 b a r) ∣ P t (s)$
  V
  Solution dikhao
  Jaisa likha hai waisi cell reaction: $2 I^{-} + 2 H^{+}$ $\overset{}{\to} I_{2} + H_{2}$ , $E^{\circ} = 0.00 - 0.54 = - 0.54 V$ , $n = 2$ .
  $Q = 1 / ([I^{-}]^{2} [H^{+}]^{2})$ , aur $[I^{-}]^{2} [H^{+}]^{2} = (0.020)^{2} (0.010)^{2} = 4.0 \times 10^{- 8}$ , isliye $Q = 2.5 \times 10^{7}$ .
  $E = - 0.54 - 0.0296 \times 7.40 = - 0.759 V$ . Negative sign ka matlab hai ki cell ulta chalta hai: hydrogen iodine ko reduce karti hai.
NumericalMediumNCERT Exercises 2.6
Hearing aids ke zinc–air button cells alkaline paste mein reaction $2 Z n (s) + O_{2} (g)$ $\overset{}{\to} 2 Z n O (s)$ chalate hain. Half-cell potentials hain $Z n O (s) + H_{2} O (l) + 2 e^{-}$ $\overset{}{\to} Z n (s) + 2 O H^{-}$ , $E^{\circ} = - 1.26 V$ , aur $O_{2} (g) + 2 H_{2} O (l) + 4 e^{-}$ $\overset{}{\to} 4 O H^{-}$ , $E^{\circ} = + 0.40 V$ . Likhi gayi equation ke liye $E_{cell}^{\circ}$ aur $Δ_{r} G^{\circ}$ nikaalo.
Solution dikhao
Answer $E^{\circ} = 1.66 V$ ; $Δ_{r} G^{\circ} = - 641 k J m o l^{- 1}$
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.40 - (- 1.26) = 1.66 V$ .
Do zinc atoms mein se har ek do electrons khota hai, isliye $n = 4$ : $Δ_{r} G^{\circ} = - 4 \times 96 485 \times 1.66 = - 6.41 \times 10^{5} J m o l^{- 1} = - 641 k J m o l^{- 1}$ (ya per mole zinc $- 320 k J$ ).
Long answerMediumNCERT Exercises 2.7
Ek dost kehta hai, 'Salt solution ko dilute karne se woh ek saath kharab conductor bhi banta hai aur behtar conductor bhi.' Conductivity aur molar conductivity define karo, aur unse samjhao ki tumhare dost ka matlab kya hai.
Solution dikhao
Conductivity $κ$ solution ke ek unit cube (aamne-saamne ke faces ke beech 1 cm ya 1 m cube) ka conductance hai, S cm⁻¹ ya S m⁻¹ mein. Molar conductivity $Λ_{m} = κ / c$ us poore solution ka conductance hai jismein ek mole electrolyte ho, ek unit doori par rakhe electrodes ke beech, S cm² mol⁻¹ ya S m² mol⁻¹ mein.
Dilution par har unit cube mein kam ions hote hain, isliye $κ$ girta hai: per cube solution kharab conductor hai. Lekin ab ek mole bada volume $V$ gherta hai aur uske ions ek doosre ko kam rokte hain (weak electrolyte zyada ionise bhi hota hai), isliye $Λ_{m} = κ V$ badhta hai: per mole woh behtar conductor hai. Strong electrolytes mein yeh badhat dheemi aur $\sqrt{c}$ mein linear hai; weak electrolytes mein kam concentration par yeh steep hai.
Quick answerEasyNCERT Exercises 2.8
0.150 mol L⁻¹ potassium nitrate solution ka 298 K par $κ = 0.0176 S c m^{- 1}$ hai. Uski $Λ_{m}$ kya hai?
S cm² mol⁻¹
Answer dekho
Answer 117.3 S cm² mol⁻¹
$Λ_{m} = 1000 \times 0.0176 / 0.150 = 117 S c m^{2} m o l^{- 1}$ .
Quick answerEasyNCERT Exercises 2.9
0.0200 mol L⁻¹ KCl ( $κ = 2.77 \times 10^{- 3} S c m^{- 1}$ 298 K par) se bhare conductivity cell ka resistance 420 Ω hai. cm⁻¹ mein cell constant nikaalo.
cm⁻¹
Answer dekho
Answer 1.16 cm⁻¹
$G^{*} = 420 \times 2.77 \times 10^{- 3} = 1.16 c m^{- 1}$ ( $116 m^{- 1}$ ).
DiagramHardNCERT Exercises 2.10
Table 298 K par potassium nitrate solutions ka $κ$ deti hai. Har ek ke liye $Λ_{m}$ nikaalo, $\sqrt{c}$ ke against $Λ_{m}$ ka graph banao, aur usse $Λ_{m}^{\circ}$ ka andaaza lagao.
Table 2.9298 K par KNO₃ solutions
$c$ / (mol L⁻¹) $κ$ / (10⁻² S m⁻¹)
0.000500 0.714
0.00100 1.418
0.00500 6.924
0.0100 13.58
0.0200 26.48
Solution dikhao
Answer $Λ_{m}$ 142.8 se 132.4 S~cm²~mol⁻¹ tak girti hai; $Λ_{m}^{\circ} \approx 145 S c m^{2} m o l^{- 1}$
S cm⁻¹ mein $κ$ ke saath $Λ_{m} = 1000 κ / c$ se 142.8, 141.8, 138.5, 135.8 aur 132.4 S cm² mol⁻¹ milte hain. Concentrations ke square roots 0.0224, 0.0316, 0.0707, 0.100 aur 0.141 hain.
Points lagbhag −87 S cm² mol⁻¹ (mol L⁻¹) $^{- 1 / 2}$ slope wali ek seedhi line ke kareeb hain. Use $\sqrt{c} = 0$ tak extend karne par $Λ_{m}^{\circ} \approx 144.7$ milta hai, lagbhag 145 S cm² mol⁻¹ (ion table se $73.5 + 71.5 = 145.0$ se compare karo).
NumericalMediumNCERT Exercises 2.11
298 K par 0.0200 mol L⁻¹ formic acid ( $H C O O H$ ) ki conductivity $7.30 \times 10^{- 4} S c m^{- 1}$ hai. Table 2.7 use karke uski molar conductivity, degree of dissociation aur $K_{a}$ nikaalo.
mol L⁻¹
Solution dikhao
Answer $Λ_{m} = 36.5$ ; $α = 0.0903$ ; $K_{a} = 1.79 \times 10^{- 4}$
$Λ_{m} = 1000 \times 7.30 \times 10^{- 4} / 0.0200 = 36.5 S c m^{2} m o l^{- 1}$ .
$Λ_{m}^{\circ} = 349.6 + 54.6 = 404.2 S c m^{2} m o l^{- 1}$ , isliye $α = 36.5 / 404.2 = 0.0903$ .
$K_{a} = \frac{0.0200 \times (0.0903)^{2}}{1 - 0.0903} = 1.79 \times 10^{- 4} m o l L^{- 1}$ .
2.12
NumericalEasyNCERT Exercises 2.12
Inmein se har ek ke liye kitne coulombs charge chahiye?
1. 0.50 mol $N i^{2 +}$ ko Ni mein reduce karna
  C
  Answer dekho
  Answer 96485 C
  $0.50 \times 2 = 1.0$ F $= 96 485 C$ .
2. 2.0 mol $A u^{3 +}$ ko Au mein reduce karna
  C
  Answer dekho
  Answer 579000 C
  $2.0 \times 3 = 6.0$ F $= 5.79 \times 10^{5} C$ .
3. acid mein 1.0 mol $N O_{3}^{-}$ ko NO mein reduce karna
  C
  Answer dekho
  Answer 289000 C
  Nitrogen +5 se +2 par jaata hai: 3 F $= 2.89 \times 10^{5} C$ .
2.13
NumericalEasyNCERT Exercises 2.13
Faradays mein kitna charge pass karna padega taaki mile:
1. molten $M g C l_{2}$ se 12.0 g magnesium ( $M = 24.3 g m o l^{- 1}$ )?
  F
  Answer dekho
  Answer 0.988 F
  $12.0 / 24.3 = 0.494$ mol, × 2 = 0.988 F.
2. molten LiCl se 13.9 g lithium ( $M = 6.94 g m o l^{- 1}$ )?
  F
  Answer dekho
  Answer 2 F
  $13.9 / 6.94 = 2.00$ mol, × 1 = 2.00 F.
2.14
NumericalMediumNCERT Exercises 2.14
Inmein se har oxidation anode par kiya jaata hai. Har ek ko kitne coulombs charge chahiye?
1. 1.0 mol $M n^{2 +}$ se $M n O_{4}^{-}$ ?
  C
  Answer dekho
  Answer 482000 C
  Mn +2 se +7 par jaata hai: 5 F $= 4.82 \times 10^{5} C$ .
2. 1.0 mol $H_{2} O_{2}$ se $O_{2}$ ?
  C
  Answer dekho
  Answer 193000 C
  $H_{2} O_{2} \overset{}{\to} O_{2}$ $+ 2 H^{+} + 2 e^{-}$ : 2 F $= 1.93 \times 10^{5} C$ .
NumericalMediumNCERT Exercises 2.15
Ek steel bracket ko zinc sulphate solution mein steady 2.5 A par zinc-plate kiya jaata hai. 40 minute mein us par kitna zinc chadhta hai? ( $M = 65.4 g m o l^{- 1}$ .)
g
Solution dikhao
$Q = 2.5 \times 2400 = 6000 C$ , jo $6000 / 96 485 = 0.0622$ mol electrons hai.
$Z n^{2 +} + 2 e^{-} \overset{}{\to} Z n$ , isliye 0.0311 mol zinc, yaani $0.0311 \times 65.4 = 2.03 g$ .
2.16
NumericalHardNCERT Exercises 2.16
Ek plating shop ek nickel bath ( $N i S O_{4}$ ), ek silver bath ( $A g N O_{3}$ ) aur ek tin bath ( $S n S O_{4}$ ) ko series mein ek 2.0 A supply par chalati hai. Run tab rukta hai jab 0.880 g nickel jam chuka hota hai ( $M$ : Ni 58.7, Ag 107.9, Sn 118.7 g mol⁻¹).
1. Current kitne minute tak beha?
  min
  Solution dikhao
  $n (N i) = 0.880 / 58.7 = 0.0150$ mol, jise 0.0300 mol electrons chahiye: $Q = 0.0300 \times 96 485 = 2.89 \times 10^{3} C$ .
  $t = Q / I = 1447 s = 24.1$ minutes.
2. Kitna mass silver jama?
  g
  Solution dikhao
  Wahi 0.0300 mol electrons 0.0300 mol silver jamate hain: $0.0300 \times 107.9 = 3.24 g$ .
3. Kitna mass tin jama?
  g
  Solution dikhao
  $S n^{2 +}$ ko do electrons chahiye, isliye 0.0150 mol tin: $0.0150 \times 118.7 = 1.78 g$ .
2.17
Quick answerMediumNCERT Exercises 2.17
Standard conditions mein milane par inmein se kaunse pairs react karenge? Table 2.2 use karo.
$Z n (s)$ aur $F e^{2 +} (a q)$ $A g (s)$ aur $C u^{2 +} (a q)$ $C l_{2} (g)$ aur $F e^{2 +} (a q)$ $I_{2} (s)$ aur $B r^{-} (a q)$ acidified $M n O_{4}^{-}$ aur $C l^{-} (a q)$
Saare sahi options chuno
Answer dekho
Answer
(A) $Z n (s)$ aur $F e^{2 +} (a q)$
(C) $C l_{2} (g)$ aur $F e^{2 +} (a q)$
(E) acidified $M n O_{4}^{-}$ aur $C l^{-} (a q)$
$E_{cell}^{\circ}$ : Zn/Fe²⁺ +0.32 V (haan); Ag/Cu²⁺ −0.46 V (nahi); Cl₂/Fe²⁺ +0.59 V (haan); I₂/Br⁻ −0.55 V (nahi); MnO₄⁻/Cl⁻ +0.15 V (haan, lab mein chlorine isi tarah banti hai).
2.18
Short answerMediumNCERT Exercises 2.18
In electrolyses mein har electrode par kya banta hai?
1. nickel electrodes ke saath aqueous $N i S O_{4}$
  Solution dikhao
  AnswerCathode par Ni jamta hai; nickel anode ghulta hai.
  Cathode: $N i^{2 +} + 2 e^{-} \overset{}{\to} N i$ (−0.25 V, −0.83 V wale paani se jeetta hai). Anode: nickel paani se kahin zyada aasaani se oxidise hota hai, isliye $N i \overset{}{\to} N i^{2 +} + 2 e^{-}$ . Nickel anode se cathode tak transfer hota hai, jaise electroplating mein.
2. platinum electrodes ke saath aqueous KBr
  Solution dikhao
  AnswerCathode par $H_{2}$ ; anode par $B r_{2}$ ; KOH bachta hai.
  Cathode: $K^{+}$ (−2.93 V) paani se compete nahi kar sakta, isliye $2 H_{2} O + 2 e^{-}$ $\overset{}{\to} H_{2} + 2 O H^{-}$ . Anode: $B r^{-}$ (1.09 V) paani (1.23 V) se zyada aasaani se oxidise hota hai, aur bromine milti hai. Solution mein potassium hydroxide bachta hai.
3. platinum electrodes ke saath aqueous $N a_{2} S O_{4}$
  Solution dikhao
  AnswerCathode par $H_{2}$ ; anode par $O_{2}$ (paani toot jaata hai).
  Na $N a^{+}$ react karta hai, na $S O_{4}^{2 -}$ . Paani cathode par reduce aur anode par oxidise hota hai, aur 2 : 1 volume ratio mein hydrogen aur oxygen milti hain; salt bas paani ko conduct karne laayak banata hai.
4. graphite electrodes ke saath molten $P b B r_{2}$
  Solution dikhao
  AnswerCathode par Pb; anode par $B r_{2}$ .
  Paani nahi hai, isliye sirf ions hi reduce aur oxidise hote hain: $P b^{2 +} + 2 e^{-} \overset{}{\to} P b$ (liquid lead neeche jama hota hai) aur $2 B r^{-} \overset{}{\to} B r_{2} + 2 e^{-}$ (bhoori vapour).
Quick answerEasy
Salt bridge ka main kaam hai…
dono solutions ke beech electrons le jaanadono solutions ko electrically neutral rakhna aur circuit poora karnacathode par reduce hone wale ions denadono metals ko chhoone se rokna
Answer dekho
Answer (B) dono solutions ko electrically neutral rakhna aur circuit poora karna
Har solution mein jama hote charge ko balance karne ke liye ions bridge se bahar aate hain. Electrons kabhi solutions ya bridge se hokar nahi jaate.
Quick answerMedium
5.0 A ka current copper sulphate solution se copper jamata hai. 1.00 g jamane mein kitne minute lagte hain? ( $M = 63.5 g m o l^{- 1}$ .)
min
Answer dekho
Answer 10.1 min
$1.00 / 63.5 = 0.01575$ mol Cu ko 0.0315 mol e⁻ = 3039 C chahiye; $t = 3039 / 5.0 = 608 s = 10.1$ min.
E3
Quick answerEasy
Har device ko uske kaam se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
dry cell : primary cell, lagbhag 1.5 V
lead storage battery : rechargeable, per cell lagbhag 2 V
fuel cell : jab tak reactants daale jaate hain tab tak chalta hai
conductivity cell : solution ka resistance measure karta hai
Sirf conductivity cell energy ka source nahi hai: yeh AC ke saath use hone wala measuring vessel hai.
Quick answerEasy
Ek faraday woh charge hai jo ek mole le jaate hain.
Answer dekho
Answer electrons
$F = N_{A} e = 96 485 C m o l^{- 1}$ .
Quick answerMedium
Electrolytic cell mein anode supply ke positive terminal se juda hota hai, aur wahan oxidation hota hai.
SahiGalat
Answer dekho
Answer Sahi
Supply anode se electrons kheenchti hai, isliye wahan species oxidise hoti hain. Oxidation hamesha anode par hota hai; galvanic cell se sirf terminal ka sign badalta hai.

$c$ / (mol L⁻¹)	$κ$ / (10⁻² S m⁻¹)
0.000500	0.714
0.00100	1.418
0.00500	6.924
0.0100	13.58
0.0200	26.48

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Intext Question 2.1	Intext questions 2.1–2.3 · Q2.1
NCERT Intext Question 2.2	Intext questions 2.1–2.3 · Q2.2
NCERT Intext Question 2.3	Intext questions 2.1–2.3 · Q2.3
NCERT Intext Question 2.4	Intext questions 2.4–2.6 · Q2.4
NCERT Intext Question 2.5	Intext questions 2.4–2.6 · Q2.5
NCERT Intext Question 2.6	Intext questions 2.4–2.6 · Q2.6
NCERT Intext Question 2.7	Intext questions 2.7–2.9 · Q2.7
NCERT Intext Question 2.8	Intext questions 2.7–2.9 · Q2.8
NCERT Intext Question 2.9	Intext questions 2.7–2.9 · Q2.9
NCERT Intext Question 2.10	Intext questions 2.10–2.12 · Q2.10
NCERT Intext Question 2.11	Intext questions 2.10–2.12 · Q2.11
NCERT Intext Question 2.12	Intext questions 2.10–2.12 · Q2.12
NCERT Intext Question 2.13	Intext questions 2.13–2.15 · Q2.13
NCERT Intext Question 2.14	Intext questions 2.13–2.15 · Q2.14
NCERT Intext Question 2.15	Intext questions 2.13–2.15 · Q2.15
NCERT Exercises 2.1	Exercises · Q2.1
NCERT Exercises 2.2	Exercises · Q2.2
NCERT Exercises 2.3	Exercises · Q2.3
NCERT Exercises 2.4	Exercises · Q2.4
NCERT Exercises 2.5	Exercises · Q2.5
NCERT Exercises 2.6	Exercises · Q2.6
NCERT Exercises 2.7	Exercises · Q2.7
NCERT Exercises 2.8	Exercises · Q2.8
NCERT Exercises 2.9	Exercises · Q2.9
NCERT Exercises 2.10	Exercises · Q2.10
NCERT Exercises 2.11	Exercises · Q2.11
NCERT Exercises 2.12	Exercises · Q2.12
NCERT Exercises 2.13	Exercises · Q2.13
NCERT Exercises 2.14	Exercises · Q2.14
NCERT Exercises 2.15	Exercises · Q2.15
NCERT Exercises 2.16	Exercises · Q2.16
NCERT Exercises 2.17	Exercises · Q2.17
NCERT Exercises 2.18	Exercises · Q2.18

Quick check: MCQs

2.1 Electrochemical Cells3 questions
2.2 Galvanic Cells3 questions
2.2.1 Electrode Potential ka Measurement3 questions
2.3 Nernst Equation3 questions
2.3.2 Electrochemical Cell aur Reaction ki Gibbs Energy3 questions
2.4 Electrolytic Solutions ka Conductance3 questions
2.4.1 Ionic Solutions ki Conductivity ka Measurement3 questions
2.4.2 Concentration ke saath Conductivity aur Molar Conductivity ka Variation4 questions
2.5 Electrolytic Cells aur Electrolysis3 questions
2.5.1 Electrolysis ke Products3 questions
2.6.2 Secondary Batteries4 questions
2.7 Fuel Cells3 questions
2.8 Corrosion3 questions

Glossary · 15

Key terms (English + Hindi)

cell constantसेल स्थिरांकcell sthirank: Conductivity cell ka ratio $l / A$ , jo standard KCl solution ke saath $G^{*} = R κ$ se nikaala jaata hai.Chapter mein pehli baar yahan aaya →
conductivityचालकताchalakta: Resistivity ka reciprocal, $κ = 1 / ρ$ : material ke ek unit cube ka conductance, S m⁻¹ ya S cm⁻¹ mein.Chapter mein pehli baar yahan aaya →
corrosionसंक्षारणsankshaaran: Aas-paas ke maahaul se metal surface ka dheema electrochemical oxidation, jaise iron par rust lagna.Chapter mein pehli baar yahan aaya →

electrode potentialइलेक्ट्रोड विभवelectrode vibhav: Electrode aur uske aas-paas ke electrolyte ke beech ka potential difference.Chapter mein pehli baar yahan aaya →
electrolytic cellवैद्युत अपघटनी सेलvaidyut apghatani cell: Aisa cell jismein ek external electrical supply non-spontaneous reaction chalati hai.Chapter mein pehli baar yahan aaya →
emf of a cellविद्युत वाहक बलvidyut vahak bal: Jab koi current na liya jaye, tab cell ke electrodes ke beech ka potential difference.Chapter mein pehli baar yahan aaya →

Faraday constantफैराडे स्थिरांकFaraday sthirank: Ek mole electrons ka charge, $F = 96 485$ C mol⁻¹.Chapter mein pehli baar yahan aaya →
fuel cellईंधन सेलeendhan cell: Aisa galvanic cell jiska fuel aur oxidant lagaataar diya jaata hai, jaise hydrogen–oxygen cell.Chapter mein pehli baar yahan aaya →

galvanic cellगैल्वेनी सेलgalvani cell: Aisa cell jismein ek spontaneous redox reaction electrons ko external circuit mein ghumati hai, jisse chemical energy electrical work ban jaati hai.Chapter mein pehli baar yahan aaya →

limiting molar conductivityसीमांत मोलर चालकताseemant molar chalakta: Concentration zero ke kareeb aane par molar conductivity, $Λ_{m}^{\circ}$ .Chapter mein pehli baar yahan aaya →

molar conductivityमोलर चालकताmolar chalakta: Conductivity ko molar concentration se divide karke, $Λ_{m} = κ / c$ , S m² mol⁻¹ ya S cm² mol⁻¹ mein.Chapter mein pehli baar yahan aaya →

overpotentialअधिविभवadhivibhav: Electrode potentials se mili value ke upar ka extra voltage, jo kisi sust electrode reaction ko useful rate par chalane ke liye chahiye.Chapter mein pehli baar yahan aaya →

salt bridgeलवण सेतुlavan setu: Jelly mein jame inert electrolyte ki tube jo do half-cells ko jodti hai aur dono solutions ko electrically neutral rakhti hai.Chapter mein pehli baar yahan aaya →
standard electrode potentialमानक इलेक्ट्रोड विभवmanak electrode vibhav: Half-cell ka reduction potential jab saari species apne standard states mein hon, SHE ke against measure kiya gaya; symbol $E^{\circ}$ .Chapter mein pehli baar yahan aaya →
standard hydrogen electrodeमानक हाइड्रोजन इलेक्ट्रोडmanak hydrogen electrode: 1 bar hydrogen ke saath 1 M $H^{+}$ mein platinised platinum; iska potential 0 V define kiya gaya hai.Chapter mein pehli baar yahan aaya →

2.1 Electrochemical Cells

2.2 Galvanic Cells

2.2.1 Electrode Potential ka Measurement

2.3 Nernst Equation

2.3.1 Nernst Equation se Equilibrium Constant

2.3.2 Electrochemical Cell aur Reaction ki Gibbs Energy

2.4 Electrolytic Solutions ka Conductance

2.4.1 Ionic Solutions ki Conductivity ka Measurement

2.4.2 Concentration ke saath Conductivity aur Molar Conductivity ka Variation

2.5 Electrolytic Cells aur Electrolysis

2.5.1 Electrolysis ke Products

2.6 Batteries

2.6.1 Primary Batteries

2.6.2 Secondary Batteries

2.7 Fuel Cells

2.8 Corrosion

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Quick check: MCQs

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