Redox Reactions: Class 11 Chemistry Chapter 7 Hinglish mein

Chemistry Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

85 min
1 simulation
37 practice questions
27 MCQs

Is chapter mein aap kya seekhenge

Oxygen–hydrogen, electron-transfer aur oxidation-number, teeno views se redox reactions pehchano, aur samjhao ki oxidation aur reduction hamesha saath-saath kyun hote hain
Kisi reaction mein oxidising aur reducing agents ke naam batao, aur displacement reactions aur standard electrode potentials se unki strength compare karo
Rules ke ek consistent set se oxidation numbers assign karo, fractional (average) values ka matlab samjho aur Stock notation use karo
Redox reactions ko combination, decomposition, displacement ya disproportionation mein classify karo
Acidic ya basic solution mein redox equations ko oxidation-number method aur half-reaction method se balance karo
Samjhao ki redox titrations apne end points kaise dikhati hain, aur calculations mein unki stoichiometry use karo
Redox couple, Daniell cell ka kaam karna aur standard electrode potential ka matlab describe karo

NCERT Chemistry Class 11, Chapter 7, pp. 235–255 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Maachis ki teeli jalao, monsoon bhar iron ke gate par rust lagte dekho, kata hua apple plate par chhod do ya phone ki battery khatam hone do. Har case mein electrons ek substance se doosre substance mein ja rahe hain. Aise changes jinmein ek substance oxidise hota hai aur doosra reduce, unhe redox reactions (अपचयोपचय अभिक्रियाएँ) kehte hain. Yeh word reduction aur oxidation ko jodkar bana hai.

Redox chemistry chupchaap rozmarra ki zindagi ka bada hissa chalati hai. LPG, coal aur petrol jalane par oxidation se energy nikalti hai. Aluminium aur sodium apne compounds se electrolysis se nikale jaate hain, jo electricity se chalne wala redox change hai, aur caustic soda aur chlorine bhi isi tarah brine se bante hain. Tumhare cells zinda rehne ke liye glucose ko oxidise karte hain, hari pattiyan carbon dioxide ko reduce karti hain, aur bleaching, corrosion aur har battery ki jad mein redox hi hai. Hydrogen ko clean fuel banane ki behas ho ya ozone layer ki, dono redox reactions par hi tiki hain.

Chemists aaj ke matlab tak ek hi kadam mein nahi pahunche. Pehle oxidation ka matlab tha oxygen lena, phir electrons khona, aur aakhir mein oxidation number ka badhna. Hum isi raaste par chalenge, phir is idea se redox reactions ko families mein baantenge, unki equations balance karenge, titrations karenge aur ek simple cell banayenge.

7.1 Redox reactions ka classical idea: oxidation aur reduction

Oxidation ka sabse purana matlab sabse seedha hai: koi substance oxygen se judta hai. Copper wire ko flame mein garam karo to woh kaala pad jaata hai; lamp mein spirit jalao to woh carbon dioxide aur paani ban jaata hai:

\begin{aligned} 2 C u + O_{2} \overset{}{\to} 2 C u O \\ C_{2} H_{5} O H + 3 O_{2} \\ \overset{}{\to} 2 C O_{2} + 3 H_{2} O \end{aligned}

Metals ke liye oxygen ki bhookh un rocks ko bhi samjhati hai jinhe hum khodkar nikalte hain. Geological time mein surface ke paas ke metals hawa ki oxygen (volume se lagbhag 21 %) se jud gaye, isliye haematite ( $F e_{2} O_{3}$ ) aur bauxite (zyadatar hydrated $A l_{2} O_{3}$ ) jaise ores oxides hain.

Doosra clue un reactions se mila jo hydrogen hata dete hain. Ethanol vapour ko garam copper ke upar se guzaarne par woh hydrogen kho deta hai aur ethanal ban jaata hai, $C_{2} H_{5} O H \overset{}{\to} C H_{3} C H O + H_{2} .$ Chemists ne hydrogen hataane ko bhi oxidation ginna shuru kiya, aur jalte fuels bhi is pattern mein fit hote hain, kyunki unka hydrogen aakhir mein oxygen se jud jaata hai.

Phir chemists ne dekha ki doosre electronegative elements bhi oxygen jaisa behave karte hain. Chlorine mein jalta aluminium, $2 A l + 3 C l_{2} \overset{}{\to} 2 A l C l_{3},$ ya sulphur ke saath garam kiya zinc, $Z n + S \overset{}{\to} Z n S,$ metal ko utni hi pakki tarah oxidise karta hai jitna oxygen karti. Kisi electropositive partner ko hataana bhi oxidation ginta hai: jab chlorine water potassium iodide ko iodine mein badalta hai, $2 K I + C l_{2} \overset{}{\to} 2 K C l + I_{2},$ to iodide apna potassium kho deta hai aur oxidise hota hai.

Reduction iska ulta roop bankar aaya. Shuru mein iska matlab tha oxygen hataana, furnace wala purana matlab jismein ore ko 'reduce' karke metal banaya jaata tha, jaise roasted zinc ore ko coke ke saath garam karna: $Z n O + C \overset{}{\to} Z n + C O$ (Fig. 7.1). Aaj ismein electronegative element hataana, hydrogen jodna, jaise $C H_{3} C H O + H_{2} \overset{}{\to} C_{2} H_{5} O H,$ ya electropositive element jodna bhi aata hai.

Compare

Oxidation aur reduction, classical view

Oxidation

oxygen ya koi doosra electronegative element jodta hai
hydrogen ya koi electropositive element (metal) hataata hai

Reduction

oxygen ya koi doosra electronegative element hataata hai
hydrogen ya koi electropositive element jodta hai

3/3Ek hi event mein zinc oxide reduce hua aur carbon oxidise.

Fig. 7.1Ek transfer, do naam: oxide reduce hota hai aur carbon oxidise hota hai.

Teen circles zinc, oxygen aur carbon dikhate hain. Step 1: oxygen zinc se zinc oxide ke roop mein bonded hai, aur right side par ek alag carbon atom hai. Step 2: ek dashed arrow oxygen atom ko carbon ki taraf jaate dikhata hai. Step 3: zinc akela reh jaata hai, uske tag par likha hai lost oxygen yaani reduced, aur carbon monoxide ke tag par likha hai gained oxygen yaani oxidised.

Inmein se koi bhi reaction dobara dekho, wahi pattern dikhega: ek substance jo chhodta hai, doosra use le leta hai. Is reaction ko dekho: $2 F e C l_{3} + S n C l_{2} \overset{}{\to} 2 F e C l_{2} + S n C l_{4},$ jismein iron(III) chloride chlorine khota hai (reduced), theek usi samay tin(II) chloride use leta hai (oxidised). Oxidation kabhi reduction ke bina nahi hota, isliye is jode ko ek hi naam diya gaya hai: redox.

Example 7.1NCERT Problem 7.1 · p. 236

Classical tareeke se oxidation aur reduction pehchanna

Har reaction mein batao kaun sa substance oxidise hota hai aur kaun sa reduce, sirf oxygen, hydrogen aur electronegativity wale ideas use karke.

(i) $P b O + C O \overset{}{\to} P b + C O_{2}$ (ii) $2 H_{2} S + S O_{2} \overset{}{\to} 3 S + 2 H_{2} O$ (iii) $6 L i + N_{2} \overset{}{\to} 2 L i_{3} N$

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

Part (iii) mein electronegativity ke baare mein faisla karna pada, aur kai reactions mein yeh faisla mushkil hota hai. Ek zyada saaf test yeh poochne se milta hai ki electrons kahan jaate hain.

7.2 Electron transfer reactions ke roop mein redox reactions

Classical rules kaam karte hain, lekin unke liye yaad rakhna padta hai ki kaun sa partner 'oxygen jaisa' hai. Products ek gehri kahani batate hain. Calcium chloride, magnesium oxide aur lithium nitride ionic solids hain, jinhe behtar tareeke se aise likhte hain: $C a^{2 +} (C l -)_{2},$ $M g^{2 +} O^{2 -}$ aur $(L i^{+})_{3} N^{3 -} .$ Inhe banane ke liye metal atoms ko non-metal atoms ko electrons dene padte hain.

Hum har change ko do half reactions (अर्ध अभिक्रियाएँ) ke roop mein likh sakte hain, ek jo electrons chhodta hai aur ek jo unhe leta hai (Fig. 7.2). Chlorine mein jalte calcium ke liye:

Do half reactions

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

Dono lines ko jodne se electrons cancel ho jaate hain aur bachta hai $C a + C l_{2} \overset{}{\to} C a C l_{2} .$ Jo half electrons khota hai woh oxidation hai; jo unhe leta hai woh reduction hai. Calcium electrons deta hai, isliye woh chlorine ko reduce karta hai, matlab calcium reducing agent hai. Chlorine electrons leta hai, calcium ko oxidise karta hai aur oxidising agent hai. Electron wali picture har classical case samjha deti hai: oxygen, chlorine aur sulphur electrons ke bhookhe hain, aur metals aasani se electrons de dete hain.

3/3Calcium ban jaata hai $C a^{2 +},$ yaani oxidised; chlorines ab hain $C l^{-},$ yaani reduced.

Fig. 7.2Calcium do electrons khota hai (oxidation); har chlorine atom ek electron leta hai (reduction).

Ek calcium atom do chlorine atoms ke beech hai, har chlorine ka outer shell dashed line se bana hai. Step 1: calcium mein do outer electrons dikhte hain aur har chlorine mein ek khaali jagah. Step 2: curved arrows calcium se har chlorine tak ek electron le jaate hain. Step 3: calcium sikudkar calcium ion ban gaya hai jis par oxidised, minus two electrons likha hai, aur har chlorine badhkar chloride ion ban gaya hai jis par reduced, plus one electron likha hai.

Definition

Oxidation

Kisi atom, ion ya molecule ka electrons khona. Jo species electrons khoti hai woh reducing agent hai.

Example $Z n \overset{}{\to} Z n^{2 +} + 2 e^{-}$

Definition

Reduction

Kisi atom, ion ya molecule ka electrons lena. Jo species electrons leti hai woh oxidising agent hai.

Example $C u^{2 +} + 2 e^{-} \overset{}{\to} C u$

Compare

Do tarah ke reagents

Oxidising agent (oxidant)

electrons leta hai
khud reduce hota hai
jaise: $O_{2},$ $C l_{2},$ $M n O_{4}^{-}$

Reducing agent (reductant)

electrons deta hai
khud oxidise hota hai
jaise: Mg, Zn, $H_{2},$ C

Bahut se students yaad rakhne ke liye OIL RIG use karte hain: Oxidation Is Loss, Reduction Is Gain (electrons ka). Agents ke naam zyada log galat karte hain: reducing agent woh hai jo khud oxidise hota hai.

Example 7.2NCERT Problem 7.2 · p. 237

Half reactions se redox change saabit karna

Lithium room temperature par bhi nitrogen se react karta hai: $6 L i + N_{2} \overset{}{\to} 2 L i_{3} N .$ Dikhao ki yeh redox reaction hai aur reducing agent ka naam batao.

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

7.2.1 Competitive electron transfer reactions

Agar do substances electrons ke liye compete karein, to kaun jeetega? Ek simple bench test iska jawaab deta hai. Aage padhne se pehle ise karke dekho.

Try karke dekho

Kaun sa metal apne electrons deta hai?

Tumhe chahiye

zinc granules
nangi copper wire, sandpaper se saaf ki hui
1 M copper sulphate
1 M zinc sulphate
0.1 M silver nitrate
stand mein teen test tubes

Pehle guess karo

Tumhe kin tubes mein metal par nayi coating aane ki ummeed hai?

Sirf AA aur CB aur Cteeno

Har solution ka 5 mL teen label kiye test tubes mein daalo: copper sulphate (A), zinc sulphate (B) aur dilute silver nitrate (C).
A mein ek saaf zinc granule daalo. Saaf ki hui copper wire ki ek coil B mein aur doosri C mein daalo.
Paanch minute tak dekho, phir ek ghante baad dobara dekho. Metal par koi coating aur solution ke colour mein koi badlav note karo.

Tumne kya dekha?

Tube	Prediction	Kya hua
A: $C u S O_{4}$ mein Zn
B: $Z n S O_{4}$ mein Cu
C: $A g N O_{3}$ mein Cu

Tumhari entries apne aap save hoti hain.

A mein zinc par spongy reddish-brown copper ki parat chadh jaati hai aur blue colour dheere-dheere halka padta hai. C mein copper par chamakte grey-white silver crystals ugne lagte hain aur solution halka blue ho jaata hai. Tube B mein koi badlav nahi hota. Matlab zinc copper se zyada aasani se electrons chhodta hai, aur copper silver se zyada aasani se. Yeh teeno tests, aur kai aur, tum Fig. 7.4 wali bench par dobara kar sakte ho.

Tube A mein zinc atoms metal ki surface ko chhoote copper ions ko seedhe electrons de dete hain (Fig. 7.3). Zinc atoms zinc ions bankar solution mein chale jaate hain, jabki copper ions copper atoms bankar strip par hi reh jaate hain:

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

(7.15)

3/4Zinc atom nikal jaata hai, ab woh hai $Z n^{2 +};$ surface par uski jagah ek copper atom aa jaata hai.

Fig. 7.3Metal ki surface par zinc atoms copper ions ko electrons dete hain.NCERT Fig. 7.1 · p. 238

Zinc surface ka close-up, jismein copper ions ke blue solution ke neeche grey zinc atoms ki do rows hain; right side par ek legend zinc atoms, copper ions, copper atoms aur zinc ions ke naam batata hai. Step 1: copper ions solution mein aazaadi se ghoomte hain. Step 2: ek copper ion surface ko chhoota hai aur ek arrow zinc atom se us tak do electrons jaate dikhata hai. Step 3: woh zinc atom dashed zinc ion bankar solution mein upar chala jaata hai, aur uski jagah ek brown copper atom baith jaata hai. Step 4: ab surface ki teen jagahon par copper hai, do aur zinc ions solution mein hain aur blue colour halka pad gaya hai.

Kya ulta ho sakta hai? Tube B mein kuch bhi dikhta nahi, aur sabse sensitive chemical tests bhi solution mein lagbhag koi copper ions nahi dhoondh paate. Reaction (7.15) ka equilibrium bahut zyada right side ki taraf hai: 298 K par iska equilibrium constant hota hai lagbhag $10^{37},$ ek aisa number jise calculate karna tum Class 12 mein seekhoge.

Copper aage silver ions ko electrons deta hai, jaisa tube C ne dikhaya: $C u + 2 A g^{+} \overset{}{\to} C u^{2 +} + 2 A g .$ Yahan bhi products bahut zyada favoured hain. Lekin har pair itna ek-tarfa nahi hota. Lead(II) nitrate solution mein tin metal sirf thoda react karta hai, $S n + P b^{2 +} \overset{}{⇌} S n^{2 +} + P b,$ aur equilibrium par dono tarah ke ions milti-julti maatra mein bache rehte hain ( $K \approx 2$ ), kyunki tin aur lead apne electrons lagbhag barabar mazbooti se pakadte hain.

Metals ko is hisaab se line mein lagane par ki woh kitni aasani se electrons chhodte hain, ek activity series ya electrochemical series milti hai; hamari teen tubes pehle hi yeh dikha chuki hain: $Z n > C u > A g .$ Agar muqabla aisa set karein ki electrons surface ke aar-paar seedhe jaane ke bajaye ek wire se hokar jaayein, to galvanic cell banta hai, jo chemical energy ko electrical energy mein badalta hai (Section 7.4, aur detail mein Class 12 mein).

InteractiveDisplacement bench

Buttons se chhe metal strips (magnesium, zinc, iron, nickel, copper, silver) mein se ek aur saat solutions (magnesium, zinc, iron, nickel aur copper ke sulphates, silver nitrate, dilute hydrochloric acid) mein se ek chuna jaata hai. Pehle tum predict karte ho ki kuch hoga ya nahi, phir strip dip karte ho: jab reaction hota hai to displaced metal ki coating ya hydrogen ke bubbles dikhte hain aur solution ka colour badalta hai. Ek panel balanced ionic equation, dono standard electrode potentials, cell voltage aur equilibrium constant ka andaaza deta hai, tumhari predictions ka score rakhta hai aur test kiye metals ko reducing power ke order mein list karta hai.

Fig. 7.4Ek metal aur ek solution chuno, prediction karo, phir Dip the strip dabao aur dekho ki metal par coating aati hai ya nahi, equation kya hai aur E° values ise kaise decide karti hain.

7.3 Oxidation number

Jab product ionic ho to electron transfer aasani se dikhta hai. Ab jalte hydrogen ke baare mein socho, $2 H_{2} + O_{2} \overset{}{\to} 2 H_{2} O .$ Paani ek covalent molecule hai, isliye koi bhi electron poori tarah hydrogen ko chhodkar nahi gaya. Phir bhi O–H bonds ke shared pairs oxygen ke zyada paas rehte hain, jo zyada electronegative atom hai: hydrogen ne kuch electron density chhodi hai aur oxygen ne kuch li hai. Electron density khiski hai, lekin koi electron poori tarah haath nahi badla. Yahi baat $H_{2} + B r_{2} \overset{}{\to} 2 H B r$ par bhi laagu hoti hai, aur is reaction ke nitrogen par bhi: $N_{2} + 3 H_{2} \overset{}{\to} 2 N H_{3} .$

Aise aadhe-adhoore shifts ko whole numbers se track karne ke liye chemists oxidation numbers use karte hain. Inke peeche ka rule jaan-boojhkar kiya gaya ek exaggeration hai: har bond mein maan lo ki shared electrons poori tarah zyada electronegative atom ke hain. Tab har atom par jo charge hoga, wahi uska oxidation number hai. Paani mein H = +1 aur O = −2; hydrogen bromide mein H = +1 aur Br = −1; ammonia mein N = −3 aur H = +1. Yeh maan lena nuksaan nahi karta, aur redox changes ko bahut saaf tareeke se describe karta hai.

Definition

Oxidation number

Kisi substance mein ek atom par woh charge jo tab hota jab har shared electron pair poori tarah uske bond ke zyada electronegative atom ko de diya jaaye (ek jaise atoms ke beech ke pairs barabar baante jaate hain).

Example $H_{2} O$ mein oxygen −2 hai aur har hydrogen +1.

4/4H ne 1 electron khoya (+1), O ne 2 liye (−2) aur Cl ne 1 khoya (+1); sum zero hai.

Fig. 7.5Har shared pair zyada electronegative atom ko do, phir gino.

H–O–Cl molecule, jismein bonding pairs atoms ke beech black dots aur lone pairs grey dots hain; neeche electronegativities 2.2, 3.4 aur 3.2 likhi hain. Step 1: dono bonding pairs beech mein hain. Step 2: H–O pair khisak kar oxygen par chala jaata hai. Step 3: O–Cl pair bhi oxygen par khisak jaata hai, kyunki oxygen chlorine se zyada electronegative hai. Step 4: badges hydrogen ke liye +1, oxygen ke liye −2 aur chlorine ke liye +1 dikhate hain, jinka total zero hai.

Har baar structure banana dheema hoga, isliye hum rules ki ek chhoti list use karte hain jo wahi answers deti hai (Fig. 7.5 dikhata hai ki chlorine positive kyun ho sakta hai).

Free elements zero hote hain. Fe, $N_{2},$ $B r_{2},$ $O_{3},$ $P_{4}$ ya $S_{8}$ ke har atom ka oxidation number 0 hai.
Sab milkar overall charge ke barabar hota hai: neutral compound ke liye zero, ion ke liye ion ka charge. Unknown value nikalne ke liye yahi rule use karte hain.
Kuch values kabhi nahi badalti. Fluorine hamesha −1 hai; apne sabhi compounds mein group 1 metals +1, group 2 metals +2 aur aluminium +3 hote hain; $S^{2 -}$ jaisa monatomic ion apna hi charge rakhta hai.
Hydrogen +1 hota hai jab woh non-metals se juda ho, lekin binary metal hydrides mein −1, jaise KH aur $M g H_{2} .$
Oxygen −2 hota hai, siwaay wahan jahan woh khud se ya fluorine se juda ho: peroxides ( $H_{2} O_{2},$ $B a O_{2}$ ) mein −1, superoxides mein $- \frac{1}{2}$ hota hai, jaise $C s O_{2},$ $O F_{2}$ mein +2 aur dioxygen difluoride mein +1, yaani $O_{2} F_{2} .$
Chlorine, bromine aur iodine halides mein −1 hote hain, lekin oxygen ya kisi halke halogen se jude hon to positive: $H C l O_{4}$ mein Cl +7 hai aur chlorine trifluoride mein +3, yaani $C l F_{3} .$

Jaise, $S O_{4}^{2 -}$ mein rule 2 se $x + 4 (- 2) = - 2,$ isliye sulphur +6 hai.

Jab kisi formula mein ek element ke kai atoms hon, to rules unki average value dete hain. Aksar yahi har atom ki value bhi hoti hai, lekin hamesha nahi, jaisa is section ke aakhir wala box dikhata hai.

Period 3 ke elements ko line mein lagao to ek pattern dikhta hai (Fig. 7.6). Sabse zyada oxidation number har group mein ek step badhta hai, sodium ke +1 se chlorine ke +7 tak, jabki non-metals negative values tak bhi jaate hain, silicon ke −4 se chlorine ke −1 tak. Aam taur par ek main-group element groups 1 aur 2 mein apne group number tak, aur groups 13 se 17 mein (group number − 10) tak jaata hai. Metals apne compounds mein kabhi negative nahi hote, non-metals dono taraf ja sakte hain, aur transition metals aksar kai positive states dikhate hain: iron +2 aur +3, manganese +2 se +7 tak kuch bhi.

2/3Non-metals negative values bhi dikhate hain, jaise $M g_{2} S i$ (−4), $N a_{3} P$ (−3), $N a_{2} S$ (−2) aur NaCl (−1) mein.

Fig. 7.6Period 3 mein sabse zyada aur sabse kam oxidation numbers.NCERT p. 241

Na, Mg, Al, Si, P, S aur Cl ke bars, jinke neeche unke group numbers hain. Step 1: zero ke upar solid bars sabse zyada oxidation numbers dikhate hain, sodium ke +1 se ek-ek step badhkar chlorine ke +7 tak. Step 2: zero ke neeche dashed bars non-metals ki sabse kam values dikhate hain: silicon ke liye −4, phosphorus ke liye −3, sulphur ke liye −2 aur chlorine ke liye −1. Step 3: silicon, phosphorus, sulphur aur chlorine ke paas brackets dikhate hain ki sabse zyada aur sabse kam values mein hamesha 8 ka fark hota hai.

Oxidation state ko iska synonym maana jaata hai: $S O_{3}$ mein sulphur +6 oxidation state mein hai. Ek se zyada state dikhane wale metals ke liye German chemist Alfred Stock ne 1919 mein suggest kiya ki oxidation number ko naam mein hi daal diya jaaye; kuch saal baad apnaye gaye form mein yeh metal ke baad brackets mein ek Roman numeral hota hai. Yahi Stock notation hai. 'Ferrous chloride' iron(II) chloride, $F e C l_{2},$ ban jaata hai aur 'ferric chloride' iron(III) chloride, $F e C l_{3};$ cuprous aur cupric oxides copper(I) oxide, $C u_{2} O,$ aur copper(II) oxide, CuO, ban jaate hain. Numeral turant bata deta hai ki kaun sa form zyada oxidised hai.

Example 7.3NCERT Problem 7.3 · p. 241

Stock notation mein compounds ke naam

Har compound ka Stock name batao: $C u_{2} S,$ $P b O_{2},$ $V_{2} O_{5},$ $M n_{2} O_{7},$ $C o C l_{2},$ $K_{2} P t C l_{6} .$

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

Oxidation numbers har redox word ko aisi cheez bana dete hain jise arithmetic se check kiya ja sake. Table dono views ko saath-saath rakhti hai.

Table 7.1Electron view aur oxidation-number view mein redox terms

	Electron view	Oxidation-number view
Oxidation	electrons khoye	number badhta hai
Reduction	electrons liye	number ghatta hai
Oxidising agent	electrons leta hai; khud reduce hota hai	iska apna number ghatta hai
Reducing agent	electrons deta hai; khud oxidise hota hai	iska apna number badhta hai
Redox reaction	electrons transfer hote hain	kuch numbers badalte hain

Example 7.4NCERT Problem 7.4 · p. 241

Oxidant aur reductant dhoondhna

Kaala copper(II) oxide ammonia gas mein garam karne par chamakdaar copper ban jaata hai: $3 C u O + 2 N H_{3} \overset{}{\to} 3 C u + N_{2} + 3 H_{2} O .$ Dikhao ki yeh redox hai aur oxidant aur reductant ke naam batao.

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

Quick check

4 mein se 1

K_{2} C r O_{4}

mein chromium ka oxidation number kya hai?

Answer dekho

Answer 6

2 (+ 1) + x + 4 (- 2) = 0

se milta hai

x = + 6.

Kya aap jaante hain?NCERT Box: fractional oxidation number · p. 245

Jab oxidation number fraction ho

Rules kabhi-kabhi fraction dete hain. Magnetite, $F e_{3} O_{4},$ mein iron aata hai $+ \frac{8}{3};$ azide ion, $N_{3}^{-},$ mein nitrogen ka average hai $- \frac{1}{3};$ propane, $C_{3} H_{8},$ mein carbon ka average hai $- \frac{8}{3} .$ Koi bhi atom electron ka ek-tihaai hissa nahi kho sakta, to phir chal kya raha hai?

Fraction is baat ka ishaara hai ki ek hi element ke atoms alag-alag environments mein hain. Magnetite asal mein $F e O \cdot F e_{2} O_{3}$ hai: har formula mein ek iron(II) aur do iron(III) (Fig. 7.7). Azide ion teen nitrogens ki ek line hai; rules ke hisaab se har bond ke electrons baantne par end wale atoms −1 aur beech wala +1 aata hai, jinka average hai $- \frac{1}{3} .$ Propane mein dono end carbons −3 hain aur beech wala −2. Red lead, $P b_{3} O_{4},$ bhi isi tarah do lead(II) aur ek lead(IV) hai. Asli values sirf structure batata hai; fractional oxidation number ek average hai.

2/2Ek iron(II) aur do iron(III): +2, +3 aur +3 ka average hai $+ \frac{8}{3} .$

Fig. 7.7Fraction whole-number states ka average hai.

Fe label wale teen circles ek Fe3O4 unit ke iron atoms dikhate hain. Step 1: teeno ek jaise dikhte hain aur har ek par plus eight thirds likha hai, jo value rules dete hain. Step 2: pehla circle dashed ho jaata hai aur FeO part wale bracket ke neeche uspar +2 likha hai, jabki baaki do par Fe2O3 part wale bracket ke neeche +3 likha hai; ek line dikhati hai ki two plus three plus three over three equals eight thirds.

7.3.1 Redox reactions ke types

Oxidation numbers se hum redox reactions ko chaar families mein baant sakte hain. Har case mein 'redox' ka test bas yeh hai ki koi oxidation number badalta hai ya nahi.

Combination reactions

Do substances ka judkar ek banna, $A + B \to C,$ tab redox hota hai jab unmein se kam se kam ek free element ho, kyunki kisi doosri cheez se judne par element ka zero oxidation number badalna hi padega. Oxygen mein har tarah ka jalna isi family mein aata hai. Aluminium foil ki protective parat $4 A l + 3 O_{2} \overset{}{\to} 2 A l_{2} O_{3}$ se banti hai (Al 0 → +3, O 0 → −2). LPG ka propane $C_{3} H_{8} + 5 O_{2}$ $\overset{}{\to} 3 C O_{2} + 4 H_{2} O$ ke roop mein jalta hai: carbon $- \frac{8}{3}$ ke average se +4 tak badhta hai aur oxygen −2 par girti hai. Do compounds ka judna zaroori nahi ki redox ho: $C a O + C O_{2} \overset{}{\to} C a C O_{3}$ mein koi oxidation number nahi badalta.

Decomposition reactions

Jab ek compound toot kar simple substances banata hai, to products dekho: agar unmein koi free element hai, to koi na koi oxidation number zaroor badla hai, aur decomposition redox hai. Saltpetre ko garam karne par, $2 K N O_{3} \overset{}{\to} 2 K N O_{2} + O_{2},$ nitrogen +5 se +3 par aata hai aur kuch oxygen −2 se 0 tak jaati hai. Purani car airbags is reaction se phoolti thi: $2 N a N_{3} \overset{}{\to} 2 N a + 3 N_{2},$ jismein sodium +1 se 0 par girta hai aur nitrogen $- \frac{1}{3}$ se 0 tak badhta hai. Lekin bahut se decompositions redox nahi hote: $M g C O_{3} \overset{}{\to} M g O + C O_{2}$ mein koi oxidation number nahi badalta.

Displacement reactions

Displacement mein ek free element kisi compound ke element se jagah badal leta hai: $X + Y Z \to X Z + Y .$ Aane wala element metal bhi ho sakta hai aur non-metal bhi.

Metal displacement. Zyada reactive metal kam reactive metal ko uske compound se bahar dhakel deta hai, isliye yeh reactions metals nikalne mein kaam aate hain. Thermite reaction, $F e_{2} O_{3} + 2 A l \overset{}{\to} A l_{2} O_{3} + 2 F e,$ safed-garam pighla hua iron deta hai aur railway rails ko weld karne mein use hota hai. Titanium ko uske chloride se sodium ke zariye nikala ja sakta hai, $T i C l_{4} + 4 N a \overset{}{\to} T i + 4 N a C l,$ aur copper sulphate solution mein rakhi iron ki keel par copper ki reddish-brown parat chadh jaati hai. Jo metal aakhir mein ions banta hai, woh hamesha pair ka zyada strong reducing agent hota hai.

Non-metal displacement: hydrogen. Hydrogen chhodne ke liye kisi metal ko kitna manana padta hai, yeh seedha uski reducing power ka naap hai (Table 7.2). Dilute acids ke saath bubbles magnesium mein sabse tez bante hain, zinc mein dheere aur iron mein aur bhi dheere, theek unki activity ke order mein. Dilute sulphuric acid ke saath zinc, $Z n + H_{2} S O_{4}$ $\overset{}{\to} Z n S O_{4} + H_{2},$ laboratory mein hydrogen ka aam source hai.

Table 7.2Metals hydrogen kahan se lete hain (sabse reactive pehle)

Hydrogen source	Metals	Example
thanda paani	group 1 metals; Ca, Sr, Ba	$2 K + 2 H_{2} O$ $\overset{}{\to} 2 K O H + H_{2}$
steam	Mg, Zn, Fe	$3 F e + 4 H_{2} O (g)$ $\overset{}{\to} F e_{3} O_{4} + 4 H_{2}$
sirf dilute acid	Ni, Sn (dheere)	$S n + 2 H C l$ $\overset{}{\to} S n C l_{2} + H_{2}$
inmein se koi nahi	Cu, Ag, Au	hydrogen nahi, dilute HCl ke saath bhi nahi

Non-metal displacement: halogens. Laboratory ka ek pasandeeda test is family ko kaam karte dikhata hai. Halide solution mein chlorine water daalo, thode se non-polar solvent ke saath hilao, aur solvent layer dekho: orange-brown matlab bromine free hua, violet matlab iodine. Chlorine ne halide ions se electrons le liye hain, jaise $C l_{2} + 2 B r^{-} \overset{}{\to} 2 C l^{-} + B r_{2} .$ Yeh layer test isliye kaam karta hai kyunki group 17 mein neeche jaate hue oxidising power ghatti hai: chlorine bromine aur iodine ko free karta hai, bromine sirf iodine ko, aur iodine kisi ko nahi. Sabse strong fluorine to khud paani par hi hamla karke oxygen chhod dega, isliye use is tarah kabhi use nahi kiya jaata. Industry sea water aur brines se bromine isi chlorine reaction se nikalti hai.

Disproportionation reactions

Disproportionation mein ek element, jo ek hi oxidation state se shuru hota hai, usi reaction mein oxidise bhi hota hai aur reduce bhi. Yeh sirf aise element ke liye possible hai jiske teen ya zyada accessible oxidation states hon, aur tabhi jab woh beech wali state se shuru ho. Zakhm saaf karne ke liye rakhi hydrogen peroxide ki bottle isi tarah dheere-dheere kamzor padti hai: $2 H_{2} O_{2} \overset{}{\to} 2 H_{2} O + O_{2} .$ Peroxide ki oxygen −1 se shuru hoti hai; kuch $O_{2}$ mein 0 tak upar jaati hai aur kuch paani mein −2 tak neeche.

Chlorine yeh idea oxidation-state ladder par saaf dikhata hai (Fig. 7.8). Thande dilute alkali mein, $C l_{2} + 2 O H^{-}$ $\overset{}{\to} C l^{-} + C l O^{-} + H_{2} O$ : aadhe chlorine atoms −1 par girte hain aur aadhe +1 tak badhte hain. Hypochlorite ion household bleach ka active hissa hai, jo coloured daagon ko oxidise karke colourless products bana deta hai. Garam alkali mein hypochlorite khud aage chloride aur chlorate ( $C l O_{3}^{-},$ +5) mein disproportionate hota hai. Bromine aur iodine bhi aisa hi karte hain, garam alkali mein phosphorus aur sulphur bhi, aur paani mein copper(I) bhi: $2 C u^{+} \overset{}{\to} C u + C u^{2 +} .$

Halogens mein fluorine exception hai. Koi bhi cheez usse electrons nahi kheench sakti, isliye uski koi positive oxidation state hai hi nahi aur upar chadhne ke liye koi oonchi seedhi nahi. Alkali ke saath woh bas fluoride mein reduce hota hai, aur uski jagah hydroxide ion ki oxygen oxidise hoti hai, aur banta hai oxygen difluoride, $O F_{2} .$

2/4 $C l_{2}$ (0) toot kar $C l O^{-}$ (+1) aur $C l^{-}$ (−1) banata hai: bleach wala reaction.

Fig. 7.8Disproportionation ke liye beech ki seedhi chahiye.

Chlorine ki oxidation states ki ek khadi ladder, neeche −1 se upar +7 tak, jiski seedhiyon ke paas chloride, chlorine, hypochlorite, chlorite, chlorate aur perchlorate likhe hain. Step 1: sirf ladder. Step 2: 0 par chlorine highlight hai, ek solid arrow upar +1 wale hypochlorite tak jaata hai jis par oxidised likha hai aur ek dashed arrow neeche −1 wale chloride tak jis par reduced likha hai, label cold alkali. Step 3: +1 par hypochlorite highlight hai, arrows upar +5 wale chlorate tak aur neeche chloride tak, label hot alkali. Step 4: upar perchlorate aur neeche chloride box mein hain, notes ke saath ki woh sirf neeche ja sakte hain ya sirf upar.

Example 7.5NCERT Problem 7.5 · p. 244

Kaun si species disproportionate ho sakti hain?

$N H_{3},$ $H N O_{2},$ $N O_{3}^{-}$ aur $H_{2} O_{2}$ mein se kaun disproportionate nahi ho sakta? Jo ho sakte hain, unki equation likho.

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

Example 7.6NCERT Problem 7.6 · p. 244

Redox reactions ko classify karna

Har redox reaction ki family ka naam batao: (a) $2 M g + C O_{2} \overset{}{\to} 2 M g O + C$ (b) $2 A g_{2} O \overset{}{\to} 4 A g + O_{2}$ (c) $2 F e + 3 C l_{2} \overset{}{\to} 2 F e C l_{3}$ (d) $4 K C l O_{3} \overset{}{\to} 3 K C l O_{4} + K C l$ (halka garam karne par)

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

Example 7.7NCERT Problem 7.7 · p. 246

Ek hi acid, alag behaviour

Hausmannite, $M n_{3} O_{4},$ aur magnetite, $F e_{3} O_{4},$ dono ek metal ki +2 aur +3 states ke mixed oxides hain. Concentrated hydrochloric acid ke saath garam karne par ek hari-si chlorine gas chhodta hai aur doosra sirf chlorides ka solution deta hai. Kaun sa kaun hai, aur kyun?

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

7.3.2 Redox reactions ko balance karna

Redox equations aksar balance karne mein mushkil lagti hain kyunki ek saath kai species badalti hain. Do methods inhe kaabu mein laate hain. Oxidation number method oxidation numbers ke badhne aur ghatne ko track karta hai; half-reaction method (jise ion–electron method bhi kehte hain) oxidation aur reduction parts ko alag-alag balance karta hai. Dono ek hi principle par tike hain: reducing agent jitne electrons khota hai, oxidising agent ko utne hi electrons milne chahiye (Fig. 7.9). Jo method tumhe zyada saaf lage, woh use karo.

2/3Paanch oxalates aur do permanganates: dono taraf 10 electrons.

Fig. 7.9Halves jodne se pehle khoye gaye electrons liye gaye electrons ke barabar hone chahiye.

Ek beam balance, jiske left mein oxidation half (oxalate se carbon dioxide) aur right mein reduction half (permanganate se manganese(II)) hai. Step 1: left pan mein 2 electron counters hain aur right pan mein 5, aur beam right ki taraf jhuka hai. Step 2: halves ko 5 aur 2 se multiply karne ke baad dono pans mein 10 counters hain aur beam seedha hai. Step 3: pans khaali hain kyunki electrons cancel ho gaye, aur neeche balanced overall equation dikhti hai.

(a) Oxidation number method

Step by step

Oxidation number method, ek-ek step karke

Step 1 / 5

Skeleton likho

React karne wali aur banne wali har cheez ke sahi formulas use karo. Example, acid mein: $M n O_{4}^{-} + F e^{2 +}$ $\overset{}{\to} M n^{2 +} + F e^{3 +} .$

1 / 5

Example 7.8NCERT Problem 7.8 · p. 246

Acid mein oxidation number method: breathalyser

Purane roadside breath testers acidified orange dichromate use karte the, jo ethanol ko acetic acid mein oxidise karte hue green ho jaata hai. Acidic solution mein $C r_{2} O_{7}^{2 -} + C_{2} H_{5} O H$ $\overset{}{\to} C r^{3 +} + C H_{3} C O O H$ ko balance karo.

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

Example 7.9NCERT Problem 7.9 · p. 247

Base mein oxidation number method

Qualitative analysis mein hara chromium(III) hydroxide alkali mein hydrogen peroxide se peele chromate mein oxidise hota hai. Basic solution mein $C r (O H)_{3} + H_{2} O_{2}$ $\overset{}{\to} C r O_{4}^{2 -} + H_{2} O$ ko balance karo.

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

(b) Half-reaction method

Yeh method oxidation aur reduction ko do alag equations maanta hai, har ek ko poori tarah balance karta hai, aur uske baad hi dono ko jodta hai. Acidic solution mein:

Skeleton ionic equation ko ek oxidation half aur ek reduction half mein baanto.
Har half mein pehle 'key' atoms theek karo: O aur H ke alawa sab.
Oxygen ko water molecules se theek karo, phir hydrogen ko $H^{+}$ ions se.
Charge ko electrons se theek karo, jo hamesha zyada positive side par jaate hain.
Halves ko same electron count tak scale karo, unhe jodo, dono sides par aane wali cheezein cancel karo, aur atoms aur charge check karo.

Basic solution mein ek extra step chahiye. Hydroxide ions ke beech free hydrogen ions, yaani $H^{+},$ tik nahi sakte, isliye acid ki tarah balance karne ke baad unhe neutralise karo: jitne $H^{+}$ hain utne $O H^{-}$ dono sides par rakho, har $H^{+}$ / $O H^{-}$ pair ko water molecule bana do, aur dono sides par aane wale paani ko kaat do. Yeh raha permanganate se oxalate ki titration wale reaction par yeh method; ise ek-ek line karke kholo:

Permanganate oxalate ko oxidise karta hai

\begin{aligned} 2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} \\ + 16 H^{+} \overset{}{\to} 2 M n^{2 +} \\ + 10 C O_{2} + 8 H_{2} O \end{aligned}

InteractiveHalf-reaction balancer

Ek menu acidic ya basic solution mein chhe redox reactions deta hai, jaise iron(II) ke saath permanganate, iodide ke saath dichromate, nitrate ke saath copper aur sulphite ke saath permanganate. Next aur Back buttons half-reaction method ko ek-ek step karke chalate hain: skeleton, oxidation numbers ke saath dono halves, baaki atoms balance karna, oxygen ko water se, hydrogen ko hydrogen ions se, base mein hydroxide ions par switch, electrons, halves ko barabar karke jodna, aur aakhir mein ek table jo har element aur charge check karti hai.

Fig. 7.10Ek reaction chuno aur Next step dabao, half-reaction method se use balance karne ke liye, basic solution mein switch karne ke saath.

Example 7.10NCERT Problem 7.10 · p. 248

Basic solution mein half reactions

Garam sodium hydroxide mein zinc dust nitrate ions ko ammonia mein reduce karta hai, yeh nitrate ka ek laboratory test hai. Half-reaction method se basic solution mein $Z n + N O_{3}^{-}$ $\overset{}{\to} Z n O_{2}^{2 -} + N H_{3}$ ko balance karo.

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

Quick check

4 mein se 1

Half-reaction ke steps (acidic solution) ko order mein lagao.

O balance karne ke liye H₂O jodo
O aur H ke alawa atoms balance karo
Charge balance karne ke liye electrons jodo
Electrons barabar karo aur jodo
H balance karne ke liye H⁺ jodo
Do halves mein baanto

Answer dekho

Answer

Do halves mein baanto
O aur H ke alawa atoms balance karo
O balance karne ke liye H₂O jodo
H balance karne ke liye H⁺ jodo
Charge balance karne ke liye electrons jodo
Electrons barabar karo aur jodo

Pehle O aur H ke alawa atoms, phir O water se, H hydrogen ions se, aur uske baad hi charge electrons se.

7.3.3 Titrations ke aadhaar ke roop mein redox reactions

Redox titration kisi oxidant ya reductant ki maatra naapti hai, theek waise jaise acid–base titration acid ya base ko naapti hai. Fark yeh hai ki end point kaise dikhta hai: pH indicator ki jagah tumhe aisi cheez chahiye jo khud redox change par react kare. Teen tareeke aam hain.

Apne aap coloured reagent. Potassium permanganate gehra purple hota hai, lekin acid mein jo manganese(II) ion banta hai woh lagbhag colourless hai. Jab tak iron(II) ya oxalate jaisa koi reductant bacha hai, har boond turant colourless ho jaati hai. Jab reductant khatam ho jaata hai, to agli boond halka pink chhod deti hai jo tikta hai (Fig. 7.11). Solution ko rangne ke liye itna kam permanganate chahiye, lagbhag $10^{- 6} m o l L^{- 1},$ ki dikhne wala end point lagbhag theek equivalence point par hota hai, yaani woh pal jab daala gaya oxidant reductant se unke reacting mole ratio mein match karta hai. Permanganate apna indicator khud hai.

3/3Ek boond extra halka pink chhod deti hai jo tikta hai.

Fig. 7.11Permanganate apna end point khud batata hai.

Purple potassium permanganate solution se bhari ek burette, lagbhag colourless iron(II) solution wale conical flask ke upar khadi hai. Step 1: ek boond girti hai aur chhota purple daag banata hai jo turant gayab ho jaata hai. Step 2: end point ke paas purple daag bada hai aur fade hone se pehle kuch der tikta hai. Step 3: flask ka poora solution halka pink reh jaata hai, jo end point dikhata hai.

Alag se daala redox indicator. Potassium dichromate orange se chromium(III) ke green mein badalta to hai, lekin itna dheere ki end point judge karna mushkil hai. Isliye diphenylamine jaisa indicator daala jaata hai: dichromate ki pehli thodi si excess use gehre violet mein oxidise kar deti hai.

Iodine aur starch (iodometry). Kai oxidants ko indirectly naapa jaata hai. Excess potassium iodide daala jaata hai, aur oxidant utni hi maatra ke iodide ko iodine mein badal deta hai, jo triiodide ion ke roop mein ghula rehta hai, yaani $I_{3}^{-} .$ Phir iodine ko sodium thiosulphate se titrate kiya jaata hai, $I_{2} + 2 S_{2} O_{3}^{2 -}$ $\overset{}{\to} 2 I^{-} + S_{4} O_{6}^{2 -} .$ Aakhir ke paas thoda starch daalne se gehra blue starch–iodine colour aata hai, jo aakhri iodine khatam hote hi ekdum gayab ho jaata hai. Copper(II) isi tarah estimate kiya jaata hai, is reaction se: $2 C u^{2 +} + 4 I^{-}$ $\overset{}{\to} 2 C u I + I_{2},$ aur bleaching powder ya liquid bleach ka 'available chlorine' bhi, is reaction se: $C l O^{-} + 2 I^{-} + 2 H^{+}$ $\overset{}{\to} C l^{-} + I_{2} + H_{2} O .$

Try karke dekho

Vitamin C banaam iodine

Tumhe chahiye

povidone-iodine antiseptic (first-aid kit wala brown liquid)
ek vitamin C tablet, pisa hua
woh paani jismein chawal dhoye ya ubaale gaye hon (ismein starch hota hai)
do saaf glass aur ek chammach

Pehle guess karo

Vitamin C milane par gehre blue colour ka kya hoga?

gayab ho jaata haiaur gehra ho jaata haipink ho jaata haikuch nahi badalta

Do saaf glass aadhe paani se bharo aur har ek mein antiseptic ki 5 boondein milao: halka yellow-brown colour aayega.
Har glass mein ek chammach rice water daalo aur milao. Colour note karo.
Pisi hui vitamin C tablet sirf doosre glass mein milao, aur ek minute tak dekho.

Tumne jo colours dekhe

Glass	Rice water ke saath	Vitamin C ke baad
1		(kuch nahi daala)
2

Tumhari entries apne aap save hoti hain.

ExampleNCERT p. 249

Permanganate titration se tablet mein iron

Ek iron–folic acid tablet ko dilute sulphuric acid mein ghola jaata hai taaki uska saara iron ho jaaye $F e^{2 +} .$ Pakka halka pink aane tak solution mein 0.0200 M $K M n O_{4}$ ka 10.8 mL lagta hai. Tablet mein kitna iron tha? (Fe = 55.85 g mol⁻¹)

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

7.3.4 Oxidation number ke concept ki limitations

Oxidation numbers ek tool hain, electrons asal mein kya karte hain uski picture nahi. Yeh ek convenient fiction par tike hain, bonding electrons ka poora transfer, isliye yeh kuch nahi batate ki electrons sach mein kitna khiskte hain. Jab do bonded atoms ki electronegativity lagbhag barabar ho to yeh uncertain ho jaate hain, aur jaisa fractional values wale box ne dikhaya, yeh alag states wale atoms ko chhupa sakte hain. Aajkal chemists aksar oxidation ko zyada general tareeke se kisi atom ke aas-paas electron density mein kami aur reduction ko badhotri ke roop mein describe karte hain, jo un changes ko bhi cover karta hai jinmein electrons sirf thode khiskte hain.

7.4 Redox reactions aur electrode processes

Zinc rod ko copper sulphate solution mein daalo to Section 7.2.1 wala reaction metal ki surface par chalta hai: electrons seedhe zinc atoms se copper ions tak jaate hain, aur nikli energy solution ko bas thoda garam karti hai. Is energy ko electricity ke roop mein ikattha karne ke liye zinc ko copper ions se milna hi nahi chahiye. Electrons ko lamba raasta, ek wire se hokar, bhejna padega.

Isliye hum dono partners ko alag rakhte hain: ek beaker mein zinc sulphate solution mein zinc rod, doosre mein copper sulphate solution mein copper rod. Ab har beaker mein ek pair hai, ek metal jo apne hi ions ko chhoo raha hai. Aisa pair redox couple kehlata hai. Ise pehle oxidised form aur phir boundary dikhane wale slash ke baad reduced form likhkar dikhate hain: $Z n^{2 +}$ /Zn aur $C u^{2 +}$ /Cu.

Definition

Redox couple

Ek species ke oxidised form aur usi species ke reduced form ka pair, jo ek half reaction se jude hon; oxidised/reduced likha jaata hai.

Example $C u^{2 +}$ /Cu; ya jab dono forms ions hon to $F e^{3 +}$ / $F e^{2 +}$ .

Ab rods ko voltmeter se hokar ek wire se jodo, aur solutions ko ek salt bridge (लवण सेतु) se jodo: ulti U-tube jismein agar mein jamaayi potassium nitrate ya potassium chloride ki jelly bhari ho, jo ions ko guzarne deti hai lekin solutions ko alag rakhti hai. Yeh arrangement Daniell cell hai (Fig. 7.12). Bridge ke bina kuch nahi hota; bridge ke saath voltmeter lagbhag 1.1 V padhta hai aur teen cheezein saath-saath hoti hain:

Zinc rod par zinc atoms oxidise hote hain, $Z n \overset{}{\to} Z n^{2 +} + 2 e^{-} .$ Yeh rod negative electrode, yaani anode, hai.
Electrons external wire se copper rod tak jaate hain, jahan copper ions reduce hote hain, $C u^{2 +} + 2 e^{-} \overset{}{\to} C u .$ Yeh rod positive electrode, yaani cathode, hai. Conventional current ulti taraf, copper se zinc ki taraf, behta hai.
Cell ke andar ions charge le jaate hain. Negative ions bridge se hokar zinc side ki taraf jaate hain, jahan $Z n^{2 +}$ badh raha hai, aur positive ions copper side ki taraf, jahan $C u^{2 +}$ khatam ho raha hai. Dono solutions neutral rehte hain, isliye reaction chalta reh sakta hai.

3/4Electrons wire se zinc se copper ki taraf behte hain; ions bridge se hokar chalte hain.

Fig. 7.12Electrons ko wire se bhejne par zinc–copper reaction electricity ka source ban jaata hai.NCERT Fig. 7.3 · p. 250

Do beakers: left mein colourless zinc sulphate solution mein zinc rod, right mein blue copper sulphate solution mein copper rod, jo voltmeter se hokar ek wire se jude hain. Step 1: salt bridge nahi hai aur meter 0.00 volts padhta hai. Step 2: ek U-shaped salt bridge solutions ko jodta hai aur meter 1.10 volts padhta hai. Step 3: arrows dikhate hain ki electrons wire se zinc se copper ki taraf jaate hain, current ulti direction mein, anions bridge se zinc side ki taraf aur cations copper side ki taraf; zinc par anode, negative, oxidation likha hai aur copper par cathode, positive, reduction. Step 4: chalne ke baad zinc rod patli, copper rod moti aur blue solution halka ho gaya hai.

SimulationCell ke bahar: wire mein electrons

Ek simple circuit jismein Daniell cell ki jagah 1.10 volt ka cell, ek switch, ek ammeter aur bulb label wala 11 ohm ka resistor hai. Play dabane par electrons negative terminal chhodte, bulb se guzarte aur positive terminal par lautte dikhte hain, aur current ammeter par dikhta hai. Sliders cell voltage aur resistance badalte hain, aur switch circuit ko kholta ya band karta hai.

Fig. 7.13Ek 1.10 V Daniell cell 11 Ω ka bulb chalata hai. Electrons ko negative terminal chhodte dekhne ke liye Play dabao.

Try karo

Play dabao aur electrons ko follow karo: woh negative terminal (zinc anode) chhodte hain, bulb se guzarte hain aur positive terminal (copper cathode) mein ghuste hain.
Ammeter padho: 0.10 A. Yaani har second 0.10 C, isliye anode par har second lagbhag 3 × 10¹⁷ zinc atoms ghulte hain.
Bulb ka resistance 22 Ω tak badhao: current aadha ho jaata hai, isliye zinc aadhi speed se ghulta hai. Switch kholo to reaction ruk jaata hai, theek waise jaise salt bridge nikaal lene par.

Rods ke beech voltage ka matlab hai ki har electrode ka apna electrode potential hai, yaani metal aur uske solution ke beech ka potential difference. Kyunki yeh concentration aur temperature par depend karta hai, values ko standard conditions mein compare kiya jaata hai: har dissolved species unit concentration (1 mol L⁻¹) par, har gas 1 bar par, aur 298 K. Tab potential kehlata hai standard electrode potential, $E^{\circ} .$ Akele ek electrode ka potential naapa nahi ja sakta, isliye sabko ek hi reference, standard hydrogen electrode, ke against naapa jaata hai, jiska $E^{\circ}$ convention se 0.00 V fix hai.

Tables har couple ko reduction ke roop mein list karti hain, oxidised form + electrons → reduced form (Table 7.3). Bada positive $E^{\circ}$ matlab oxidised form electrons ko tezi se pakadta hai, isliye woh strong oxidising agent hai, isse bhi behtar: $H^{+} .$ Bada negative $E^{\circ}$ matlab reduced form aasani se electrons chhodta hai, isliye woh hydrogen gas se behtar reducing agent hai. List mein sabse upar fluorine hai aur sabse neeche lithium. Cell ke liye standard conditions mein voltage dono potentials ka difference hai; Daniell cell ke liye $0.34 - (- 0.76) = 1.10 V,$ yeh result tum Class 12 mein baar-baar use karoge.

Standard cell voltage

\begin{aligned} E_{cell}^{\circ} & = E_{cathode}^{\circ} \\ - E_{anode}^{\circ} \end{aligned}

jahan

$E_{cathode}^{\circ}$: us couple ka standard potential jo reduce hota hai
$E_{anode}^{\circ}$: us couple ka standard potential jo oxidise hota hai

Table 7.3298 K par standard electrode potentials (sabse strong oxidising agent upar, sabse strong reducing agent neeche; paani mein ions)

Reduction half reaction	$E^{\circ}$ / V
$F_{2} + 2 e^{-} \overset{}{\to} 2 F^{-}$	+2.87
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \overset{}{\to} M n^{2 +} + 4 H_{2} O$	+1.51
$C l_{2} + 2 e^{-} \overset{}{\to} 2 C l^{-}$	+1.36
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \overset{}{\to} 2 C r^{3 +} + 7 H_{2} O$	+1.33
$O_{2} + 4 H^{+} + 4 e^{-} \overset{}{\to} 2 H_{2} O$	+1.23
$B r_{2} + 2 e^{-} \overset{}{\to} 2 B r^{-}$	+1.09
$A g^{+} + e^{-} \overset{}{\to} A g$	+0.80
$F e^{3 +} + e^{-} \overset{}{\to} F e^{2 +}$	+0.77
$I_{2} + 2 e^{-} \overset{}{\to} 2 I^{-}$	+0.54
$C u^{2 +} + 2 e^{-} \overset{}{\to} C u$	+0.34
$2 H^{+} + 2 e^{-} \overset{}{\to} H_{2}$	0.00
$P b^{2 +} + 2 e^{-} \overset{}{\to} P b$	−0.13
$S n^{2 +} + 2 e^{-} \overset{}{\to} S n$	−0.14
$N i^{2 +} + 2 e^{-} \overset{}{\to} N i$	−0.25
$F e^{2 +} + 2 e^{-} \overset{}{\to} F e$	−0.44
$Z n^{2 +} + 2 e^{-} \overset{}{\to} Z n$	−0.76
$2 H_{2} O + 2 e^{-} \overset{}{\to} H_{2} + 2 O H^{-}$	−0.83
$A l^{3 +} + 3 e^{-} \overset{}{\to} A l$	−1.66
$M g^{2 +} + 2 e^{-} \overset{}{\to} M g$	−2.37
$N a^{+} + e^{-} \overset{}{\to} N a$	−2.71
$K^{+} + e^{-} \overset{}{\to} K$	−2.93
$L i^{+} + e^{-} \overset{}{\to} L i$	−3.04

2/3 $C u^{2 +}$ (upar) Zn (neeche) ko oxidise karta hai: $Z n + C u^{2 +} \overset{}{\to} Z n^{2 +} + C u$ hota hai.

Fig. 7.14Oxidised form ladder par apne neeche wale kisi bhi reduced form se react karta hai.

Ek khadi ladder, scale ke hisaab se nahi, paanch couples ke saath: fluorine/fluoride +2.87 volts par, copper(II)/copper +0.34 par, hydrogen ion/hydrogen 0.00 par ek box mein, zinc(II)/zinc −0.76 par aur lithium ion/lithium −3.04 par; oxidised forms left column mein aur reduced forms right mein. Right side ke arrows dikhate hain ki oxidising power upar ki taraf aur reducing power neeche ki taraf badhti hai. Step 2: copper(II) ion aur zinc metal highlight hain aur ek solid arrow unhe jodta hai: woh react karte hain. Step 3: zinc ion aur copper metal highlight hain aur cross wala ek dashed arrow unhe jodta hai: koi reaction nahi.

ExampleNCERT Table 7.1 · p. 251

Standard potentials se reaction predict karna

Standard conditions mein, (a) kya acidified permanganate chloride ions ko oxidise kar sakta hai? (b) kya iron(III) ions bromide ions ko oxidise kar sakte hain? Table use karo.

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

Recap

Summary aur key points

Redox reactions mein ek oxidation aur ek reduction ka jodaa hota hai. Classical roop se oxidation matlab oxygen lena ya hydrogen khona, aur reduction iska ulta (Section 7.1).
Electrons ki bhaasha mein oxidation matlab electrons khona aur reduction matlab lena; reducing agent oxidise hota hai aur oxidising agent reduce (Section 7.2).
Metals electrons ke liye compete karte hain: zinc $C u^{2 +}$ ko reduce karta hai aur copper reduce karta hai $A g^{+},$ ulta kabhi nahi, aur isse activity series milti hai (Section 7.2.1).
Oxidation numbers har bond ko ionic maanne se aate hain; simple rules inhe dete hain, aur inka sum overall charge ke barabar hota hai (Section 7.3).
Fractional oxidation number alag-alag asli states wale atoms ka average hai, jaise $F e_{3} O_{4}$ mein Fe(II) aur Fe(III) (the box on fractions).
Stock notation metal ka oxidation number Roman numeral se dikhata hai, jaise iron(III) chloride mein (Example 7.3).
Redox reactions chaar families mein aate hain: combination, decomposition, displacement aur disproportionation (Section 7.3.1).
Disproportionation ke liye intermediate state wala element chahiye, isliye fluorine ise kabhi nahi dikhata (Fig. 7.8).
Equations ko khoye aur liye electrons barabar karke balance karte hain, chahe oxidation numbers se ya half reactions se, acid mein $H^{+}$ / $H_{2} O$ aur base mein $O H^{-}$ / $H_{2} O$ ke saath (Section 7.3.2).
Redox titrations ka end point permanganate ke apne colour se, diphenylamine jaise redox indicator se, ya starch–iodine colour se dikhta hai (Section 7.3.3).
Oxidation numbers hisaab-kitaab ka ek tareeka hain; oxidation zyada general roop se electron density mein kami hai (Section 7.3.4).
Daniell cell mein couples $Z n^{2 +}$ /Zn aur $C u^{2 +}$ /Cu ek wire aur ek salt bridge se jude hote hain; hydrogen electrode ke against naape gaye standard electrode potentials oxidising aur reducing agents ki ranking karte hain (Section 7.4).

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 Exercises 7.1	Practice Q1
NCERT Exercises 7.2	Practice Q2
NCERT Exercises 7.3	Practice Q3
NCERT Exercises 7.4	Practice Q4
NCERT Exercises 7.5	Practice Q5
NCERT Exercises 7.6	Practice Q6
NCERT Exercises 7.7	Practice Q7
NCERT Exercises 7.8	Practice Q8
NCERT Exercises 7.9	Practice Q9
NCERT Exercises 7.10	Practice Q10
NCERT Exercises 7.11	Practice Q11
NCERT Exercises 7.12	Practice Q12
NCERT Exercises 7.13	Practice Q13
NCERT Exercises 7.14	Practice Q14
NCERT Exercises 7.15	Practice Q15
NCERT Exercises 7.16	Practice Q16
NCERT Exercises 7.17	Practice Q17
NCERT Exercises 7.18	Practice Q18
NCERT Exercises 7.19	Practice Q19
NCERT Exercises 7.20	Practice Q20
NCERT Exercises 7.21	Practice Q21
NCERT Exercises 7.22	Practice Q22
NCERT Exercises 7.23	Practice Q23
NCERT Exercises 7.24	Practice Q24
NCERT Exercises 7.25	Practice Q25
NCERT Exercises 7.26	Practice Q26
NCERT Exercises 7.27	Practice Q27
NCERT Exercises 7.28	Practice Q28
NCERT Exercises 7.29	Practice Q29
NCERT Exercises 7.30	Practice Q30

Quick check: MCQs

7.1 Redox reactions ka classical idea: oxidation aur reduction3 questions
7.2.1 Competitive electron transfer reactions4 questions
7.3 Oxidation number4 questions
Disproportionation reactions4 questions
(b) Half-reaction method4 questions
7.3.4 Oxidation number ke concept ki limitations4 questions
7.4 Redox reactions aur electrode processes4 questions

Glossary · 15

Key terms (English + Hindi)

displacement reactionविस्थापन अभिक्रियाvisthapan abhikriya: Aisa reaction jismein ek element doosre element ko uske compound se bahar kar deta hai: X + YZ → XZ + Y.Chapter mein pehli baar yahan aaya →
disproportionationअसमानुपातनasamanupatan: Aisa redox reaction jismein ek element, jo ek hi oxidation state se shuru hota hai, oxidise bhi hota hai aur reduce bhi.Chapter mein pehli baar yahan aaya →

electrochemical series: Standard electrode potential ke order mein likhe couples, jo oxidising aur reducing agents ki ranking karte hain.Chapter mein pehli baar yahan aaya →

half reactionअर्ध अभिक्रियाardh abhikriya: Redox reaction ka oxidation part ya reduction part, jo akela likha jaaye aur jismein electrons dikhaye gaye hon.Chapter mein pehli baar yahan aaya →

oxidationऑक्सीकरणoxikaran: Kisi species ka electrons khona, jo oxidation number ke badhne ke roop mein dikhta hai (classical roop se, oxygen lena ya hydrogen khona).
oxidation numberऑक्सीकरण संख्याoxikaran sankhya: Woh charge jo kisi atom par hota agar har shared electron pair poori tarah zyada electronegative atom ka hota.
oxidising agentऑक्सीकारकoxikarak: Aisi species jo electrons leti hai aur isliye doosri species ko oxidise karti hai; khud reduce hoti hai.Chapter mein pehli baar yahan aaya →

redox couple: Half reaction mein ek species ke oxidised aur reduced forms, oxidised/reduced likhe jaate hain, jaise $Z n^{2 +}$ / $Z n .$ Chapter mein pehli baar yahan aaya →
redox reactionअपचयोपचय अभिक्रियाapachayopachay abhikriya: Aisa reaction jismein ek species oxidise hoti hai aur doosri reduce, isliye oxidation numbers badalte hain.Chapter mein pehli baar yahan aaya →
redox titration: Oxidant aur reductant ke beech titration, jiska end point colour change se dikhta hai.Chapter mein pehli baar yahan aaya →
reducing agentअपचायकapachayak: Aisi species jo electrons deti hai aur isliye doosri species ko reduce karti hai; khud oxidise hoti hai.Chapter mein pehli baar yahan aaya →
reductionअपचयनapachayan: Kisi species ka electrons lena, jo oxidation number ke ghatne ke roop mein dikhta hai (classical roop se, oxygen khona ya hydrogen lena).

salt bridgeलवण सेतुlavan setu: Electrolyte gel ki tube jo do half-cells ko jodti hai; ions isse hokar chalte hain, aur dono solutions bina mile neutral rehte hain.Chapter mein pehli baar yahan aaya →
standard electrode potentialमानक इलेक्ट्रोड विभवmaanak electrode vibhav: Unit concentrations, 1 bar aur 298 K par kisi half-cell ka reduction potential, hydrogen electrode (0.00 V) ke against naapa gaya.Chapter mein pehli baar yahan aaya →
Stock notation: Metal ka oxidation number brackets mein Roman numeral se likhna, jaise iron(III) chloride mein.Chapter mein pehli baar yahan aaya →

Redox Reactions: Class 11 Chemistry Chapter 7 Hinglish mein

Is chapter mein aap kya seekhenge

7.1 Redox reactions ka classical idea: oxidation aur reduction

7.2 Electron transfer reactions ke roop mein redox reactions

7.2.1 Competitive electron transfer reactions

7.3 Oxidation number

7.3.1 Redox reactions ke types

Combination reactions

Decomposition reactions

Displacement reactions

Disproportionation reactions

7.3.2 Redox reactions ko balance karna

(a) Oxidation number method

(b) Half-reaction method

7.3.3 Titrations ke aadhaar ke roop mein redox reactions

7.3.4 Oxidation number ke concept ki limitations

7.4 Redox reactions aur electrode processes

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye