Electrochemistry: Class 12 Chemistry · Chapter 2

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

125 min
1 simulation
38 practice questions
41 MCQs

What you will learn in Electrochemistry

Describe galvanic and electrolytic cells, write cell notation and identify the anode, cathode and direction of electron flow
Use standard electrode potentials to find cell emf and to predict whether a redox reaction is spontaneous
Apply the Nernst equation to find cell and electrode potentials at any concentration
Link the standard potential of a cell to the standard Gibbs energy change and the equilibrium constant of its reaction
Define resistivity, conductivity, molar conductivity and cell constant, and explain how conductivity is measured and how it varies with concentration
State Kohlrausch's law and use it to find limiting molar conductivities and dissociation constants of weak electrolytes
Apply Faraday's laws and predict the products of electrolysis
Explain how batteries and fuel cells work and describe corrosion as an electrochemical process and how to prevent it

Original companion text aligned to NCERT Chemistry Class 12, Chapter 2, pp. 31–60. Not affiliated with or endorsed by NCERT.

Your phone battery runs a chemical reaction forwards to light the screen, and the charger drives the same reaction backwards overnight. That two-way traffic between chemical energy and electrical energy is electrochemistry (विद्युत रसायन).

You will meet it all through this chapter: in the lithium-ion cell in your pocket, in the brine cells that make chlorine for water treatment, in the rust on a railing, and even in your own nerves, where ions crossing cell membranes carry signals. We build the ideas in NCERT's order: cells and electrode potentials, the Nernst equation, how solutions conduct, electrolysis, batteries, fuel cells and corrosion.

2.1 Electrochemical Cells

In Class 11 you met the Daniell cell: a zinc rod in zinc sulphate solution and a copper rod in copper sulphate solution, joined by a wire on the outside and a salt bridge on the inside. Electrons given up by zinc travel through the wire to copper ions, so the 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)

does electrical work instead of just warming the beaker. With both ions at 1 mol L⁻¹, the cell gives 1.10 V.Concentration stands in here for each ion's activity, a kind of 'effective concentration' that matches the real one closely only in dilute solutions. A device like this, in which a reaction that runs by itself pushes electrons round a circuit, is a galvanic (voltaic) cell (गैल्वेनी सेल).

Now connect an adjustable power supply so that it pushes against the cell, and raise its voltage $E_{ext}$ slowly from zero (Fig. 2.1). Below 1.10 V the cell still wins: electrons leave the zinc, zinc dissolves and copper is plated out, only more slowly. At exactly 1.10 V the two pushes cancel, the current stops and so does the reaction. Above 1.10 V the supply wins. Current flows the other way, zinc ions are turned back into zinc metal and the copper dissolves. The same hardware is now an electrolytic cell: electrical energy is forcing a non-spontaneous reaction to run.

3/3Supply above 1.10 V: current reverses and the cell becomes electrolytic.

Fig. 2.1Raising an opposing voltage first slows the Daniell cell, then stops it, then drives it backwards.NCERT Fig. 2.2 · p. 32

A loop circuit joins a Daniell cell, marked 1.10 V, on the left to an adjustable supply on the right; a dial shows the supply voltage. In the first step the dial reads 0.80 V, an arrow shows electrons leaving the zinc side, and notes say zinc dissolves and copper deposits. In the second step the dial reads 1.10 V, the arrow is gone and a note says no current and no reaction. In the third step the dial reads 1.40 V, the electron arrow points the other way and notes say zinc deposits and copper dissolves, so the device is now an electrolytic cell.

Definition

Galvanic (voltaic) cell

A cell in which a spontaneous redox reaction produces electrical energy.

ExampleA torch cell lighting a bulb.

Definition

Electrolytic cell

A cell in which an external source of electrical energy drives a non-spontaneous reaction.

ExampleA battery being recharged, or a spoon being silver-plated.

Compare

Two kinds of electrochemical cell

Galvanic cell

chemical energy → electrical energy
the reaction is spontaneous ( $Δ_{r} G < 0$ )
anode is the negative terminal
dry cell, fuel cell, a battery in use

Electrolytic cell

electrical energy → chemical energy
an external supply drives a non-spontaneous reaction
anode is joined to the positive terminal
electroplating, refining copper, a battery on charge

2.2 Galvanic Cells

A galvanic cell makes a redox reaction pay out its Gibbs energy as electrical work instead of heat. Drop zinc dust into copper sulphate solution and the same reaction runs, but the beaker only gets warm, because electrons jump straight from zinc atoms to the copper ions touching them. Keep the two partners apart and force the electrons through an external wire, and that wire can light an LED or spin a small motor.

Every cell reaction can be split into two half-reactions, one at each electrode. For the Daniell cell they are

reduction, at the copper electrode

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

(2.2)

oxidation, at the zinc electrode

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

(2.3)

Add them and the electrons cancel, giving back reaction ((2.1)). Each half-reaction runs in its own compartment, a half-cell, also called a redox couple and written oxidised form/reduced form: $C u^{2 +} / C u$ , $Z n^{2 +} / Z n$ . Pair any two half-cells and you have a new cell, so the possible combinations run into thousands. The wiring is always alike (Fig. 2.2): a wire with a meter (and a switch) joins the two electrodes, and a salt bridge (लवण सेतु) joins the two solutions. A bridge is unnecessary only when both electrodes can share one electrolyte, as in the lead storage battery of Section 2.6.2.

3/3 $Z n^{2 +}$ enters the left solution and $C u^{2 +}$ leaves the right one; the bridge sends anions left and cations right.

Fig. 2.2A Daniell cell. Electrons travel through the wire; ions carry the charge through the solutions and the salt bridge.NCERT Fig. 2.1 · p. 32

Two beakers stand side by side, zinc in zinc sulphate on the left and copper in copper sulphate on the right, with an upside-down U-shaped salt bridge dipping into both and a voltmeter on the wire joining the rods. The first step labels the parts. The second step marks the zinc as the anode where oxidation happens, negative, and the copper as the cathode where reduction happens, positive, with electrons moving through the wire from zinc to copper. The third step shows ions: zinc ions leaving the zinc rod, copper ions settling on the copper rod, and the salt bridge sending nitrate ions towards the zinc side and potassium ions towards the copper side.

The salt bridge is a tube of an unreactive electrolyte, such as $K N O_{3}$ or $K C l$ , set in agar jelly. As zinc ions pour into the left beaker and copper ions leave the right one, the left solution would soon carry a positive excess and the right a negative one, and the flow would stop. Ions drifting out of the bridge prevent that. Anions move towards the zinc side and cations towards the copper side, keeping both solutions neutral and completing the circuit.

Why should a voltage appear at all? Dip a zinc rod into zinc sulphate solution. A few zinc atoms slip into the water as $Z n^{2 +}$ , each leaving two electrons behind in the metal, while a few $Z n^{2 +}$ ions do the reverse. For zinc the first process has the edge, so the rod ends up with a slight excess of electrons and a sheath of positive ions gathers in the liquid next to it: a charged double layer, like a microscopic capacitor. For copper in copper sulphate the balance tips the other way, and the rod becomes slightly positive. The potential difference across this double layer is the electrode potential of the half-cell.

Electrode potentials are always tabulated for the half-reaction written as a reduction (the IUPAC convention), which is why older books call them 'reduction potentials'. Measured with every species in its standard state, that is, dissolved species at 1 mol L⁻¹, gases at 1 bar and solids and liquids pure, the value is the standard electrode potential, $E^{\circ}$ .

Definition

Standard electrode potential

The potential of a half-cell, written as a reduction, when all its species are in their standard states, measured against the standard hydrogen electrode.

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

Which electrode is which? Follow the electrons. Oxidation releases them, so the electrode where it happens, the anode, collects a surplus and is the negative terminal of a galvanic cell. Reduction uses them up at the cathode, which is therefore positive. Once the circuit is closed, electrons leave the anode through the wire and arrive at the cathode; the conventional current, by its old sign convention, is drawn the opposite way.

Writing out 'a zinc rod in zinc sulphate solution, joined by a salt bridge to a copper rod in copper sulphate solution' every time soon gets tedious, so chemists put a cell on one line:

write the anode half-cell first, at the left, and the cathode half-cell last, at the right;
a single vertical line, |, marks a boundary between two phases, such as metal and solution;
a double vertical line, ||, marks the salt bridge;
concentrations or pressures go in brackets after each species.

Read this way, the potential of a cell is the reduction potential of the right-hand electrode minus that of the left-hand one: cathode minus anode. For a cell written the way it actually runs, the answer comes out positive:

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

(2.7)

For example, take a cell in which lead reduces silver ions, $P b (s) + 2 A g^{+} (a q)$ $\overset{}{\to} P b^{2 +} (a q) + 2 A g (s)$ . Lead is oxidised, so it is the anode and goes on the left: $P b (s) ∣ P b^{2 +} (a q) ∣ ∣ A g^{+} (a q) ∣ A g (s)$ . With every ion at 1 mol L⁻¹, $E_{cell}^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{P b^{2 +} / P b}^{\circ} = 0.80 V - (- 0.13 V) = 0.93 V$ . The 2 in front of $A g^{+}$ does not double its electrode potential. A potential is energy per unit charge, so it does not depend on how many electrons the equation is written for.

What does a voltmeter on the terminals actually show? The potential difference between them is the cell potential, but a meter reads the full $E_{cathode} - E_{anode}$ only while the cell supplies almost no current. Once a current flows, part of the voltage is used up driving charge through the cell's own internal resistance, and the reading drops (Fig. 2.3). The zero-current value is the emf of the cell (विद्युत वाहक बल), and it is what tables and calculations mean by $E_{cell}$ .

SimulationEmf versus terminal voltage

A simple circuit with a 1.10 volt cell that has 4 ohms of internal resistance, a switch, an ammeter and a 10 ohm resistor labelled lamp. A voltmeter can be placed across the cell or across the lamp, and readouts show the current and the terminal voltage. Pressing Play shows electrons moving from the negative terminal round the circuit. Sliders change the cell voltage and the resistance.

Fig. 2.3A 1.10 V cell with 4 Ω of internal resistance drives a 10 Ω lamp. Set the voltmeter across the cell, then open and close the switch.

Try this

Put the voltmeter across the cell and open the switch. It reads 1.10 V: with no current, you are measuring the emf.
Close the switch. The reading drops to about 0.79 V, because 0.079 A flowing through the 4 Ω inside the cell uses up 0.31 V.
Raise the lamp's resistance to 100 Ω. The current shrinks and the reading climbs back towards 1.10 V. Why must an emf be measured at (almost) zero current?

Try this at home

A lemon cell

You’ll need

2 lemons
2 galvanised nails and 1 plain iron nail
2 strips of copper (or thick copper wire)
a digital multimeter
connecting wires with crocodile clips

Predict first

Which metal will be the negative terminal of a copper–zinc lemon cell?

copperzincneither: the lemon juice is the terminal

Push a galvanised (zinc-coated) nail and a strip of clean copper into a lemon, about 2 cm apart, without letting them touch.
Set a digital multimeter to its 2 V DC range. Touch the red lead to the copper and the black lead to the nail, and note the reading.
Join two lemon cells in series (copper of one to the nail of the other) and measure across the free ends.
Replace the galvanised nail in one lemon with a plain iron nail and measure again.

Table 2.1Lemon cell readings (write your prediction next to each before you measure)

Set-up	Reading (V)
Cu + Zn, one lemon
two lemons in series
Cu + Fe, one lemon

Your entries are saved automatically.

Copper and zinc give about 0.9 V: a little under the 1.10 V you might expect, because lemon juice is not a 1 mol L⁻¹ solution of either ion and the cell has a high internal resistance. Two lemons in series give about twice as much. Copper with plain iron gives less, roughly 0.4–0.6 V, because iron and copper are closer together in electrode potential than zinc and copper. The meter shows a positive reading only when its red lead is on the more positive metal, which is copper in every case.

Quick check

1 of 3

In the cell

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

, electrons in the external wire flow…

from Ag to Nithrough the salt bridge from Ni to Agfrom Ni to Agfirst one way, then the other

Show answer

Answer (C) from Ni to Ag

Nickel, on the left, is the anode: it is oxidised and releases electrons, which travel through the wire to the silver cathode. The salt bridge carries ions, never electrons.

2.2.1 Measurement of Electrode Potential

A voltmeter has two terminals, so it can never measure the potential of one half-cell alone; only the difference between two is measurable. The fix is the one we use for heights. Mountains are measured above sea level, an agreed zero. Electrode potentials are measured against the standard hydrogen electrode (SHE), which is assigned a potential of exactly 0 V at every temperature:

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

The SHE needs a surface where hydrogen gas and hydrogen ions can trade electrons quickly. That surface is a platinum foil roughened with a coat of finely divided platinum ('platinum black'), which gives a huge area and acts as a catalyst. The foil sits in acid with $[H^{+}] = 1 m o l L^{- 1}$ while pure hydrogen at 1 bar streams over it. In shorthand: $P t (s) ∣ H_{2} (g, 1 b a r) ∣ H^{+} (a q, 1 M)$ (Fig. 2.4).

2/3Copper is the cathode: the meter reads $+ 0.34$ V, so $E_{C u^{2 +} / C u}^{\circ} = + 0.34$ V.

Fig. 2.4Pairing a half-cell with the SHE gives its standard electrode potential directly, sign included.NCERT Fig. 2.3 · p. 34

On the left, a beaker of 1 M acid holds a glass hood around a platinised platinum foil, with hydrogen at 1 bar bubbling in. A wire runs through a voltmeter to a second half-cell on the right. The first step shows the hydrogen electrode alone with its labels. In the second step the right half-cell is copper in 1 M copper ions; the meter reads plus 0.34 V and electrons flow from the hydrogen electrode to the copper. In the third step the right half-cell is zinc in 1 M zinc ions; the meter reads minus 0.76 V because electrons now flow from the zinc to the hydrogen electrode.

To find the standard potential of any other couple, build a cell with the SHE on the left and the unknown half-cell, at standard conditions, on the right. Then $E_{cell}^{\circ} = E_{R}^{\circ} - E_{L}^{\circ} = E_{R}^{\circ} - 0 = E_{R}^{\circ}$ : the reading is the electrode potential. With copper, the meter shows $+ 0.34$ V and copper is the cathode, so $E_{C u^{2 +} / C u}^{\circ} = + 0.34 V$ . With zinc, the cell actually runs the other way: zinc is oxidised and hydrogen gas is released at the platinum. Written with the SHE on the left, the emf is therefore negative, $- 0.76$ V, and $E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V$ .

Try it at the bench: drop zinc granules and copper turnings into two tubes of dilute hydrochloric acid. The zinc fizzes; the copper sits unchanged. The signs predicted this. Zinc's negative $E^{\circ}$ says it parts with electrons more readily than hydrogen does, so $H^{+}$ can take them and hydrogen bubbles off. Copper's positive $E^{\circ}$ says $C u^{2 +}$ wants electrons more than $H^{+}$ does, so $H^{+}$ cannot oxidise copper. (Copper does dissolve in nitric acid, but there the nitrate ion is the oxidising agent.)

Once both values are known, the Daniell cell needs no new measurement: $E_{cell}^{\circ} = 0.34 V - (- 0.76 V) = 1.10 V$ .

Some couples have no metal of their own, because both partners are ions, gases or dissolved molecules. They borrow an inert electrode, usually platinum (sometimes gold or graphite), purely as a meeting place: the metal itself is not changed, it simply passes electrons between the dissolved partners and the wire. Besides the SHE, examples are a chlorine electrode, $P t (s) ∣ C l_{2} (g) ∣ C l^{-} (a q)$ , and an iron(III)/iron(II) electrode, $P t (s) ∣ F e^{3 +} (a q), F e^{2 +} (a q)$ , with both ions in one solution.

Measured this way, standard potentials have been tabulated for hundreds of couples. Table 2.2 lists a selection, from the highest $E^{\circ}$ to the lowest.

Table 2.2Selected standard electrode potentials (298 K). Dissolved ions are aqueous; (s) marks a solid and (g) a gas.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

Think of each row as a tug-of-war for electrons, scored by $E^{\circ}$ . The higher the row, the harder its left-hand species pulls, so the best oxidising agents sit at the top left ( $F_{2}$ , acidified $M n O_{4}^{-}$ , $C l_{2}$ ). The lower the row, the more easily its right-hand species lets electrons go, so the best reducing agents sit at the bottom right (Li, K, Na). Each strong player has a feeble partner: fluoride ion hardly gives up electrons, and $L i^{+}$ hardly takes them. Couples with negative $E^{\circ}$ are better reducing agents than $H_{2}$ ; positive ones are poorer.

This gives a quick test for any reaction between an oxidising agent on the left and a reducing agent on the right. Under standard conditions the reaction is spontaneous when the oxidising agent sits above the reducing agent in the table, because then $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ is positive (Fig. 2.5).

2/4 $C u^{2 +}$ sits above Zn, so copper ions oxidise zinc: $E^{\circ} = + 1.10$ V.

Fig. 2.5An oxidising agent reacts with any reducing agent that lies below it on the ladder.

Ten rungs run from fluorine at +2.87 V at the top to lithium at −3.04 V at the bottom, with oxidised forms on the left and reduced forms on the right. Arrows at the sides say that oxidising strength grows upwards on the left and reducing strength grows downwards on the right. In the second step the copper and zinc rungs are highlighted and a curved arrow joins zinc, low on the right, to copper ions, higher on the left, with a tick and E° = +1.10 V. In the third step copper metal and hydrogen ions are highlighted; copper ions sit above hydrogen ions, so the arrow would run uphill and a cross marks no reaction. In the fourth step chlorine and iodide are joined by a ticked arrow with E° = +0.82 V.

Measured potentials are also a laboratory tool. A pH meter is an electrochemical cell; a titration can be followed by watching the potential jump at the end point; and cell potentials give equilibrium constants and solubility products that are hard to measure any other way (Section 2.3.1).

Example

Which metals release hydrogen from acid?

Using Table 2.2, decide which of nickel, silver and aluminium should dissolve in 1 mol L⁻¹ hydrochloric acid with release of hydrogen, and find $E_{cell}^{\circ}$ for each one that does.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Intext questions 2.1–2.3: questions with answers

Short answerEasyNCERT Intext Question 2.1
You have a nickel rod, 1 M nickel sulphate solution and the usual laboratory kit. Explain how to find $E^{\circ}$ for $N i^{2 +} / N i$ , which electrode will turn out to be the anode, and what sign $E^{\circ}$ will have.
Show solution
AnswerPair it with the SHE; nickel turns out to be the anode, so $E^{\circ}$ is negative ( $- 0.25$ ~V).
Set up $P t (s) ∣ H_{2} (g, 1 b a r)$ $∣ H^{+} (a q, 1 M)$ $∣ ∣ N i^{2 +} (a q, 1 M) ∣ N i (s)$ at 298 K and measure the emf with a high-resistance voltmeter, so that almost no current flows.
Because $E_{L}^{\circ} = 0$ for the SHE, the emf of the cell as written equals $E_{N i^{2 +} / N i}^{\circ}$ . When connected, nickel dissolves and hydrogen is evolved, so nickel is really the anode, the cell as written has a negative emf, and $E_{N i^{2 +} / N i}^{\circ} = - 0.25 V$ .
Quick answerEasyNCERT Intext Question 2.2
A student leaves silver nitrate solution standing in a copper bowl for a week. What will she find?
Nothing: copper is less reactive than silverThe bowl is pitted and coated with silver, and the solution turns blueNothing: neither metal reacts with nitrate solutionsBrown gas, because silver ions oxidise the nitrate
Show answer
Answer (B) The bowl is pitted and coated with silver, and the solution turns blue
For $C u + 2 A g^{+} \overset{}{\to} C u^{2 +} + 2 A g$ , $E_{cell}^{\circ} = 0.80 - 0.34 = + 0.46 V$ , so copper reduces the silver ions: silver crystals coat the bowl, and $C u^{2 +}$ entering the solution turns it blue. Copper is the more reactive of the two, so the solution cannot be stored in copper.
2.3
Quick answerMediumNCERT Intext Question 2.3
Using Table 2.2, pick every species that can oxidise iodide ions to iodine under standard conditions. ( $E_{I_{2} / I^{-}}^{\circ} = + 0.54 V$ .)
$F e^{3 +}$ $N i^{2 +}$ $C l_{2}$ $Z n^{2 +}$ acidified $M n O_{4}^{-}$
Select all that apply
Show answer
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) and acidified $M n O_{4}^{-}$ (+1.51 V) all lie above +0.54 V. $N i^{2 +}$ (−0.25 V) and $Z n^{2 +}$ (−0.76 V) lie below, so they cannot.

2.3 Nernst Equation

A torch cell that has been switched on for an hour gives a lower voltage than a fresh one, though its chemicals are the same. What has changed is concentrations: standard potentials describe solutions at exactly 1 mol L⁻¹, and a working cell soon drifts away from that. Walther Nernst worked out the correction in 1889. For a metal M in contact with its ions, $M^{n +} (a q) + n e^{-} ⟶ M (s)$ , the electrode potential (against the SHE) at temperature $T$ is

Nernst equation for an electrode

\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)

where

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

The full form has $\ln ([reduced form] / [oxidised form])$ , but a pure solid metal counts as 1, so only $[M^{n +}]$ is left. Lowering the ion concentration makes the electrode potential more negative: with fewer ions around to be deposited, the metal's tendency to dissolve gains the upper hand.

DerivationNernst equation for the Daniell cell

Apply (2.8) to each electrode, with $n = 2$ for both:

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)

The cell potential is cathode minus anode. The two logarithms merge into one, because $\ln (1 / a) - \ln (1 / b) = \ln (b / a)$ , and the standard potentials combine into $E_{cell}^{\circ}$ :

\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)

At 298 K, changing to base-10 logarithms and putting in $R$ and $F$ gives $2.303 R T / F = 0.0591 V$ , so

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

(2.12)

Read the result physically. More copper ions (the reactant) raise the cell potential; more zinc ions (the product) lower it. It is Le Chatelier's principle measured in volts.

The same reasoning works for any cell reaction $a A + b B - > c C + d D$ that transfers $n$ electrons. It gives the general Nernst equation, where $Q$ is the reaction quotient: pure solids and liquids are left out of it, and gases enter as partial pressures in bar.

\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)

Use a single $n$ for the whole cell: the number of electrons transferred in the balanced overall equation. For $F e (s) + 2 F e^{3 +} (a q) \overset{}{\to} 3 F e^{2 +} (a q)$ , for example, $n = 2$ and $Q = [F e^{2 +}]^{3} / [F e^{3 +}]^{2}$ . Each concentration is raised to its coefficient, and iron metal does not appear at all. At 298 K the factor $2.303 R T / n F$ is $0.0591 / n$ volts, so for a two-electron reaction the cell potential changes by about 30 mV for every tenfold change in $Q$ (Fig. 2.6).

3/4The line meets zero where $Q = K$ : the cell is at equilibrium and can do no work.

Fig. 2.6The Nernst equation is a straight line of

E_{cell}

against

\log Q

, with slope

- 0.0591 / n

volts at 298 K.

Cell potential in volts is plotted upwards against log Q along the bottom, for a cell with standard potential 0.12 V. In the first step a straight line for n equal to 2 falls with slope minus 0.0296 V per unit of log Q, and a dot at log Q equal to 0 marks E equal to E°. In the second step the dot moves left to log Q equal to minus 3, where E is higher, 0.21 V. In the third step the dot sits where the line crosses zero, at log Q about 4.1, labelled Q equals K: the cell is at equilibrium. In the fourth step a second, twice as steep line for n equal to 1 is added, crossing zero at about log Q equal to 2.0.

Example 2.1NCERT Example 2.1 · p. 39

Nernst equation for an aluminium–copper cell

A cell runs the reaction $2 A l (s) + 3 C u^{2 +} (0.010 M)$ $\overset{}{\to} 2 A l^{3 +} (0.0050 M) + 3 C u (s)$ . Write the cell in standard notation and find its potential at 298 K. Take $E_{A l^{3 +} / A l}^{\circ} = - 1.66 V$ and $E_{C u^{2 +} / C u}^{\circ} = + 0.34 V$ .

Try it yourself first. Then reveal the solution one step at a time.

The hydrogen electrode makes a neat pH meter. For $H^{+} (a q) + e^{-} \overset{}{\to} \frac{1}{2} H_{2} (g)$ with hydrogen at 1 bar, $E = 0 - 0.0591 \log (1 / [H^{+}])$ , that is, $E = - 0.0591 pH$ volts at 298 K. Each unit rise in pH makes the electrode 59 mV more negative. Laboratory pH meters use a glass electrode instead, which responds to $H^{+}$ in the same way but needs no hydrogen gas.

InteractiveCell builder

A panel with two half-cells drawn as beakers joined by a salt bridge and a voltmeter. You choose the metal couple for each side from a list of ten, including the hydrogen electrode, and set each ion concentration on a logarithmic slider from 0.0001 to 1 mole per litre. The panel decides which electrode is the anode, writes the cell in standard notation with the balanced reaction, and shows the standard cell potential, the reaction quotient, the cell potential from the Nernst equation, the Gibbs energy change and the equilibrium constant. Arrows show the electron direction in the wire. The Discharge button lets the reaction run: the ion concentrations change step by step, the voltmeter falls and a small graph traces the cell potential, for equal volumes of the two solutions, until it reaches zero at equilibrium; a message then gives how much of the cathode ion is left.

Fig. 2.7Choose two half-cells and their ion concentrations to see the cell notation,

E_{cell}^{\circ}

, the Nernst-corrected

E_{cell}

Δ_{r} G

and

K

. Press Discharge to watch the cell run down to equilibrium.

Quick check

1 of 3

Which change raises the potential of a Daniell cell?

diluting the copper sulphate solutionadding more zinc sulphateusing a thicker zinc roddiluting the zinc sulphate solution

Show answer

Answer (D) diluting the zinc sulphate solution

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

. Lowering

[Z n^{2 +}]

(a product) makes

Q

smaller and

E_{cell}

larger. A thicker rod changes nothing, because a solid does not appear in

Q

2.3.1 Equilibrium Constant from Nernst Equation

Why does a cell eventually go flat? Every pair of electrons that passes through the lamp turns one more $C u^{2 +}$ ion into copper and one more zinc atom into $Z n^{2 +}$ . So $Q = [Z n^{2 +}] / [C u^{2 +}]$ keeps growing, and the Nernst equation says $E_{cell}$ keeps shrinking. The cell is spent when $E_{cell} = 0$ , and that is exactly the condition for chemical equilibrium: $Q$ has become the equilibrium constant $K_{c}$ . Putting $E_{cell} = 0$ and $Q = K_{c}$ into the Nernst equation gives

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

(2.14)

For the Daniell cell, $\log K_{c} = 2 \times 1.10 / 0.0591 = 37.2$ , so $K_{c} \approx 1.7 \times 10^{37}$ . At equilibrium, $[C u^{2 +}]$ is about $10^{- 37}$ times $[Z n^{2 +}]$ : for every practical purpose the reaction goes to completion. No chemical analysis could ever measure such a tiny concentration, yet one voltmeter reading gives the equilibrium constant.

Example 2.2NCERT Example 2.2 · p. 40

Equilibrium constants from cell potentials

Find $K_{c}$ at 298 K for (a) $N i (s) + C u^{2 +} (a q)$ $\overset{}{⇌} N i^{2 +} (a q) + C u (s)$ and (b) $S n (s) + P b^{2 +} (a q)$ $\overset{}{⇌} S n^{2 +} (a q) + P b (s)$ , using values from Table 2.2.

Try it yourself first. Then reveal the solution one step at a time.

2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction

How much electrical work can one mole of reaction deliver? Suppose the balanced equation moves $n$ moles of electrons. They carry a charge $n F$ , where $F$ is the Faraday constant (the charge of one mole of electrons), and they fall through a potential difference $E_{cell}$ , so the work is $n F E_{cell}$ . That is the largest possible amount: it is obtained only if the cell is drained infinitely slowly, so that no voltage is lost inside it (a reversible discharge). In Class 11 you saw that the maximum non-expansion work a reaction can do is $- Δ_{r} G$ . Equating the two,

Δ_{r} G = - n F E_{cell}

(2.15)

A spontaneous reaction has $Δ_{r} G < 0$ , so its cell potential is positive. The two sides of the equation are different kinds of quantity. $E_{cell}$ is intensive: a Daniell cell the size of a button and one the size of a bucket both give 1.10 V. $Δ_{r} G$ is extensive: it belongs to the equation as you write it, and scaling the equation scales $n$ and $Δ_{r} G$ together while $E_{cell}$ is untouched (Example 2.3 shows this with numbers).

With every species in its standard state, the same relation gives the standard Gibbs energy of the reaction:

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

(2.16)

Since $Δ_{r} G^{\circ} = - R T \ln K$ (Class 11 thermodynamics), a single measured standard cell potential yields $Δ_{r} G^{\circ}$ and $K$ together. The three quantities always agree in sign (Table 2.3).

Table 2.3How the signs of

E_{cell}^{\circ}

Δ_{r} G^{\circ}

and

K

go together

$E_{cell}^{\circ}$	$Δ_{r} G^{\circ}$	$K$	Meaning
positive	negative	greater than 1	forward reaction favoured
zero	zero	equal to 1	neither direction favoured
negative	positive	less than 1	reverse reaction favoured

Example 2.3NCERT Example 2.3 · p. 40

Gibbs energy of the hydrogen–oxygen reaction

The hydrogen–oxygen fuel cell (Section 2.7) runs the reaction $2 H_{2} (g) + O_{2} (g) \overset{}{\to} 2 H_{2} O (l)$ with $E_{cell}^{\circ} = 1.23 V$ . Find $Δ_{r} G^{\circ}$ for this equation, and then per mole of water formed.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A cell reaction has

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

. Which is true under standard conditions?

Δ_{r} G^{\circ} < 0

and

K > 1

Δ_{r} G^{\circ} > 0

and

K < 1

Δ_{r} G^{\circ} > 0

and

K > 1

Δ_{r} G^{\circ} = 0

and

K = 1

Show answer

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

Δ_{r} G^{\circ} = - n F E^{\circ}

is positive when

E^{\circ}

is negative, and then

\ln K = - Δ_{r} G^{\circ} / R T

is negative, so

K < 1

. The signs of

Δ_{r} G^{\circ}

and

K - 1

always disagree.

Practice · 3 questions

Intext questions 2.4–2.6: questions with answers

Quick answerEasyNCERT Intext Question 2.4
A hydrogen electrode (hydrogen at 1 bar) is dipped in a soft drink of pH 3.0. What is its potential at 298 K?
V
Show solution
For $H^{+} + e^{-} \overset{}{\to} \frac{1}{2} H_{2}$ at 1 bar, $E = 0 - 0.0591 \log (1 / [H^{+}]) = - 0.0591 pH$ . With pH 3.0, $E = - 0.177 V$ .
NumericalMediumNCERT Intext Question 2.5
What voltage does a cell running $F e (s) + 2 A g^{+} (0.0050 M)$ $\overset{}{\to} F e^{2 +} (0.10 M) + 2 A g (s)$ give at 298 K? ( $E_{cell}^{\circ} = 1.24 V$ .)
V
Show solution
$Q = [F e^{2 +}] / [A g^{+}]^{2} = 0.10 / (0.0050)^{2} = 4.0 \times 10^{3}$ .
With $n = 2$ : $E_{cell} = 1.24 - 0.0296 \log (4.0 \times 10^{3}) = 1.24 - 0.106 = 1.13 V$ .
2.6
NumericalMediumNCERT Intext Question 2.6
Copper is etched off circuit boards with iron(III) chloride solution: $C u (s) + 2 F e^{3 +} (a q)$ $\overset{}{\to} C u^{2 +} (a q) + 2 F e^{2 +} (a q)$ , for which $E_{cell}^{\circ} = 0.43 V$ at 298 K.
1. Find $Δ_{r} G^{\circ}$ in kJ mol⁻¹.
  kJ mol⁻¹
  Show solution
  $Δ_{r} G^{\circ} = - n F E^{\circ} = - 2 \times 96 485 \times 0.43 = - 8.30 \times 10^{4} J m o l^{- 1} = - 83.0 k J m o l^{- 1}$ .
2. Find the equilibrium constant $K_{c}$ .
  Show solution
  $\log K_{c} = 2 \times 0.43 / 0.0591 = 14.55$ , so $K_{c} \approx 3.6 \times 10^{14}$ . (From $\ln K = - Δ_{r} G^{\circ} / R T$ you get the same answer.)

2.4 Conductance of Electrolytic Solutions

A cell works only if charge can travel all the way round its circuit: as electrons in the wires and as ions in the solutions. How well do solutions conduct, and why? We start with quantities you know from physics.

Start with a wire. Double its length and its resistance $R$ doubles, because charge has twice as far to push; double its cross-sectional area and $R$ halves, because there are twice as many side-by-side paths. (You measured resistances with a Wheatstone bridge in physics; the ohm, Ω, is $k g m^{2} s^{- 3} A^{- 2}$ in SI base units.) Both effects fit in one line:

R = ρ \frac{l}{A}

(2.17)

where

$ρ$: resistivity, in ohm metres (Ω m)

Everything about the shape of the sample now sits in $l / A$ . What remains, $ρ$ (rho), belongs to the material: it is the resistivity (older books say specific resistance), in ohm metres, Ω m. Put $l = 1$ m and $A = 1$ m² to see what it means: $ρ$ is the resistance between opposite faces of a one-metre cube. Laboratory values are often quoted for a centimetre cube instead, and $1 Ω m = 100 Ω c m$ .

For solutions it is more natural to ask how easily charge flows, so both quantities are turned upside down. The reciprocal of $R$ is the conductance $G$ , in siemens ( $1 S = 1 Ω^{- 1}$ , a unit once called the mho), and the reciprocal of $ρ$ is the conductivity $κ$ (kappa; older name, specific conductance):

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

(2.18)

Just as $ρ$ is the resistance of a unit cube, $κ$ is the conductance of a unit cube between opposite faces: S m⁻¹ for a metre cube or S cm⁻¹ for a centimetre cube, with $1 S c m^{- 1} = 100 S m^{- 1}$ . Table 2.4 shows how hugely $κ$ varies from one material to another, across more than twenty powers of ten; for any one material it also shifts with temperature and, slightly, with pressure.

Table 2.4Approximate conductivities near 298 KNCERT 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 (along the layers)	about $2 \times 10^{5}$	non-metal conductor
seawater	about 5	ionic solution
germanium (pure)	about 2	semiconductor
0.100 M KCl	1.29	ionic solution
tap water	$10^{- 3}$ to $10^{- 1}$	ionic solution
silicon (pure)	about $4 \times 10^{- 4}$	semiconductor
ultra-pure water	$5.5 \times 10^{- 6}$	ionic (self-ionisation)
glass	$10^{- 15}$ to $10^{- 11}$	insulator
PTFE (Teflon)	below $10^{- 16}$	insulator

Read the table from the bottom. Insulators such as PTFE and glass barely conduct at all. Semiconductors such as silicon and germanium sit in the middle, and traces of another element raise their conductivity enormously; that tunability is why they run every phone and computer. Conductors fill the top: metals and alloys, joined by a few non-metals such as graphite. Beyond the top of the scale are superconductors, with no resistance at all. Mercury below 4.2 K was the first, found in 1911, and some copper-oxide ceramics now superconduct above 77 K, the boiling point of liquid nitrogen.

In a metal the current is carried by electrons alone: electronic (metallic) conductance (Fig. 2.8). The valence electrons belong to no particular atom, and under a voltage they drift through the lattice of positive ions. Each electron that enters one end of a wire pushes another out of the far end, so the metal itself never changes. How well it conducts depends on how many valence electrons each atom contributes and on how the atoms are arranged, which is why silver beats iron. Warming a metal makes it conduct less, because the more strongly vibrating ions scatter the drifting electrons more often.

Dissolve a pinch of salt in very pure water and its conductivity jumps many thousandfold. The water alone conducts only feebly, through its $10^{- 7} m o l L^{- 1}$ each of $H^{+}$ and $O H^{-}$ ; the salt adds far more carriers. In a solution, then, the current is carried by ions, cations and anions drifting in opposite directions: electrolytic (ionic) conductance. What makes one solution a better conductor than another?

How many ions: more electrolyte per litre means more carriers.
What kind of ions: a doubly charged ion carries twice the charge of a singly charged one.
How mobile they are: small bare ions are not always fast. $L i^{+}$ drags a large shell of water molecules with it and moves more slowly than the bigger $K^{+}$ .
The solvent: a more viscous solvent slows every ion down.
Temperature: warming lowers the viscosity, so, unlike metals, solutions conduct better when hot.

2/3In a solution, cations and anions drift in opposite directions.

Fig. 2.8Electrons carry the current in a metal; ions moving both ways carry it in a solution, where reactions happen at the electrodes.

The top half shows a copper wire with electrons drifting from left to right while the copper ions of the lattice stay in place. The bottom half shows a salt solution between a negative electrode on the left and a positive electrode on the right. In the first step only the wire is active and a note says the wire is unchanged. In the second step positive ions in the solution drift left towards the negative electrode while negative ions drift right towards the positive electrode. In the third step products appear at the electrodes, a deposit on the left and gas bubbles on the right, with a note that the solution changes.

Compare

Two ways to conduct

Electronic (metals, graphite)

carriers: electrons
no chemical change in the conductor
conductance falls as temperature rises

Ionic (solutions, molten salts)

carriers: cations and anions, moving opposite ways
reactions at the electrodes change the composition
conductance rises as temperature rises

Quick check

1 of 3

Warming a sodium chloride solution from 25 °C to 50 °C makes its conductivity…

fall, as in a metal wirestay the same, since the number of ions is fixedrise, because the ions move more freelyfall to zero

Show answer

Answer (C) rise, because the ions move more freely

The solvent becomes less viscous, so the same ions move faster and carry more current. In a metal the opposite happens, because lattice vibrations scatter electrons more.

2.4.1 Measurement of the Conductivity of Ionic Solutions

Measuring a solution's resistance is trickier than measuring a wire's. Try it with a battery and you electrolyse the solution (Section 2.5): gas or metal collects on the electrodes and the solution changes while you watch. So the measurement uses alternating current of about a kilohertz; each half-cycle reverses whatever the previous one started, and nothing accumulates.

A liquid also has no shape of its own, so it is held in a conductivity cell: two parallel platinum plates, coated with platinum black for a large surface, fixed a set distance apart in a glass holder. The liquid between the plates is a column of length $l$ and cross-section $A$ , so

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

All the geometry is packed into one number, the cell constant $G^{*} = l / A$ (सेल स्थिरांक), in m⁻¹ or cm⁻¹. Rather than trusting a ruler on spongy platinised plates, chemists calibrate: fill the cell with a standard KCl solution, whose $κ$ has been measured with great care, read the resistance and multiply:

G^{*} = \frac{l}{A} = R κ

In the classic set-up (Fig. 2.9) the cell is one of the four arms of a Wheatstone bridge, with fixed resistances $P$ and $Q$ and a variable resistance $S$ making up the rest. Audio-frequency AC from an oscillator drives the bridge, so a headphone placed across the middle hums while the bridge is out of balance (modern versions use an electronic detector). Turn $S$ until the hum vanishes; the bridge is then balanced and

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

(2.19)

3/3Adjust $S$ until the detector is silent: $R_{x} = S P / Q$ .

Fig. 2.9Measuring a solution's resistance: a conductivity cell in an AC Wheatstone bridge.NCERT Fig. 2.5 · p. 44

On the left, a conductivity cell has two platinised platinum plates of area A a distance l apart, with the column of solution between them shaded. On the right, a diamond-shaped bridge has arms P and Q on top and a variable resistance S and the cell R x below, with an AC oscillator across one diagonal and a detector across the other. In the first step only the cell is highlighted, with l and A marked. In the second step the bridge is unbalanced and the detector shows a buzzing signal. In the third step S has been adjusted, the detector is silent, and the balance condition R x equals S times P over Q is shown.

A handheld conductivity meter of the kind used to check drinking water does all of this inside one box and shows $κ$ directly; the principle is the same.

Once $G^{*}$ is known, the same cell gives the conductivity of any solution from its resistance:

κ = \frac{G^{*}}{R}

(2.20)

Conductivity alone is a poor way to compare electrolytes, because a more concentrated solution conducts better simply by having more ions. To compare like with like, divide by the concentration. The molar conductivity $Λ_{m}$ (capital lambda) is

Λ_{m} = \frac{κ}{c}

(2.21)

With $κ$ in S m⁻¹ and $c$ in mol m⁻³, $Λ_{m}$ comes out in S m² mol⁻¹. Concentrations are usually given in mol L⁻¹, and $1 m o l L^{- 1} = 1000 m o l m^{- 3}$ , so convert first. Chemists also use S cm² mol⁻¹, with $κ$ in S cm⁻¹; since 1 L is 1000 cm³, the working formula and the link between the two units are

Λ_{m} = \frac{1000 κ}{c}

where

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

\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

Calibrating a cell, then measuring a salt solution

A conductivity cell filled with 0.0100 mol L⁻¹ KCl, whose conductivity is $0.1413 S m^{- 1}$ , has a resistance of 312 Ω. Filled with 0.0500 mol L⁻¹ NaCl solution, the same cell reads 79.4 Ω. Find the cell constant and the conductivity and molar conductivity of the NaCl solution.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.5NCERT Example 2.5 · p. 46

Resistivity of an acid column

A glass tube of internal diameter 0.80 cm and length 20.0 cm is filled with 0.100 mol L⁻¹ hydrochloric acid. The resistance of the acid column is $1.02 \times 10^{3} Ω$ . Find the resistivity, conductivity and molar conductivity of the acid.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Why is AC rather than DC used to measure the resistance of a solution?

AC carries more charge per secondDC would electrolyse the solution and change its compositionPlatinum electrodes cannot carry DCAC makes the ions move faster

Show answer

Answer (B) DC would electrolyse the solution and change its composition

With DC, products build up at the electrodes and the solution changes during the measurement. With AC, each half-cycle undoes the last. AC does not make ions faster; they simply oscillate.

2.4.2 Variation of Conductivity and Molar Conductivity with Concentration

Add water to any electrolyte solution, strong or weak, and $κ$ drops. Recall what $κ$ measures: the conductance of a single unit cube of solution. Pour in water and the same ions are shared among more cubes, so each cube has fewer carriers.

Molar conductivity does the opposite: it rises on dilution. Imagine collecting all the solution that holds exactly one mole of electrolyte and spreading it between two long electrodes kept one unit apart, like the tray in Fig. 2.10. The plates must then have an area equal to that volume $V$ , so $G = κ A / l$ becomes

Λ_{m} = κ V, V = \frac{1}{c}

(2.22)

Why does the product grow? Diluting tenfold multiplies $V$ by exactly ten but divides $κ$ by less than ten, because the ions left in each cube are further apart, drag on one another less and so move faster (in a weak electrolyte, extra molecules ionise too). As $c$ approaches zero, $Λ_{m}$ approaches a limiting value, the limiting molar conductivity $Λ_{m}^{\circ}$ , also called the molar conductivity at infinite dilution. Strong and weak electrolytes approach it in strikingly different ways (Fig. 2.11).

2/3Twice the volume: half the ions per cube, but a slightly larger total conductance.

Fig. 2.10Dilution leaves fewer ions in each unit cube (κ falls) but spreads the same mole over a bigger area (Λm rises).

Two long electrodes one unit apart hold a tray of solution that contains exactly one mole of electrolyte, seen from above, with ions drawn as plus and minus dots and one unit cube outlined. Bar gauges on the right compare kappa, the conductance of the unit cube, with lambda m, the conductance of the whole tray. In the first step the tray is short and the cube holds eight ions. In the second step the solution is diluted twice: the tray is twice as long, the cube holds four ions, the kappa bar is about half as long and the lambda bar is a little longer. In the third step the tray is four times as long, the cube holds two ions, kappa is about a quarter and lambda is longer still.

Strong electrolytes. A strong electrolyte such as KCl is fully dissociated at every concentration, so dilution adds no new ions. Its molar conductivity still rises a little, because each ion is surrounded by a loose cloud of oppositely charged ions that drags on it, and the drag weakens as the ions spread out. At low concentrations, experiment shows that

Λ_{m} = Λ_{m}^{\circ} - A \sqrt{c}

(2.23)

This is a straight line in disguise: plot $Λ_{m}$ against $\sqrt{c}$ and the intercept is $Λ_{m}^{\circ}$ while the slope is $- A$ (Example 2.6 does exactly this). $A$ measures how strongly the ions hold each other back, so it grows with their charges: doubly charged ions gather denser clouds around them. Chemists therefore label electrolytes by charge type, cation first. $K N O_{3}$ is a 1-1 electrolyte, $B a C l_{2}$ is 2-1 and $C u S O_{4}$ is 2-2, and in a given solvent at a given temperature, electrolytes of the same type have nearly the same $A$ .

4/4The steep curve cannot be extended; $Λ_{m}^{\circ}$ (404) comes from Kohlrausch's law.

Fig. 2.11Strong and weak electrolytes behave very differently as the concentration approaches zero.NCERT Fig. 2.6 · p. 47

Molar conductivity in siemens square centimetres per mole is plotted upwards against the square root of concentration. In the first step measured points for sodium chloride lie almost on a gentle straight line just below 126. In the second step the line is extended back to the vertical axis, meeting it at 126.4, labelled the limiting molar conductivity. In the third step a curve for formic acid, a weak acid, is added: it is low at higher concentrations and rises steeply as the concentration approaches zero. In the fourth step a dashed continuation and a question mark show that the curve cannot be extended reliably, and the value 404 from Kohlrausch's law is marked on the axis instead.

Example 2.6NCERT Example 2.6 · p. 48

Finding the limiting molar conductivity of NaCl

The molar conductivity of dilute sodium chloride solutions at 298 K is given below. Show that $Λ_{m}$ varies linearly with $\sqrt{c}$ , and find $Λ_{m}^{\circ}$ and $A$ .

Table 2.5Molar conductivity of dilute NaCl

$c$ / (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

Try it yourself first. Then reveal the solution one step at a time.

Kohlrausch's law. Friedrich Kohlrausch, measuring limiting molar conductivities in the 1870s, noticed that they behave like sums. Look along any row of Table 2.6: replacing chloride by nitrate always lowers $Λ_{m}^{\circ}$ by 4.8, whatever the cation. Look down a column: changing lithium for sodium always adds 11.4, whatever the anion.

Table 2.6

Λ_{m}^{\circ}

of some salts at 298 K, in S cm² mol⁻¹

	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	–

A difference that stays the same whatever the partner means that each ion contributes its own fixed share, untouched by the other ion. Kohlrausch summed this up as a law:

LawKohlrausch's law of independent migration of ions

At infinite dilution, the molar conductivity of an electrolyte is the sum of separate contributions from its cations and its anions, each ion contributing a fixed amount that does not depend on its partner.

\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)

where

$ν_{+}, ν_{-}$: numbers of cations and anions in one formula unit
$λ_{+}^{\circ}, λ_{-}^{\circ}$: each ion's own limiting molar conductivity

Table 2.7Limiting molar conductivities

λ^{\circ}

of ions in water at 298 K, in S cm² mol⁻¹ (for doubly charged ions, per mole of 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. The formic acid curve in Fig. 2.11 tells a different story. Here dilution does more than loosen the ion clouds: it creates ions. At 0.02 mol L⁻¹ only about one molecule in eleven is ionised; a hundredfold dilution ionises more than half of them. So $Λ_{m}$ climbs ever more steeply as $c$ falls, because the degree of dissociation $α$ keeps growing, and a straight-line extrapolation to $c = 0$ would be guesswork. Nor can you simply measure your way to the limit: long before every molecule has ionised, the solution conducts barely better than the water it is made from. Instead, $Λ_{m}^{\circ}$ of a weak electrolyte is assembled from ion values with Kohlrausch's law (Example 2.8).

Once $Λ_{m}^{\circ}$ is known, conductivity gives the degree of dissociation. At infinite dilution every molecule is ionised; at concentration $c$ only the fraction $α$ is, and to a good approximation

α = \frac{Λ_{m}}{Λ_{m}^{\circ}}

(2.26)

For a weak acid HA the dissociation constant (Class 11) then follows from conductivity alone:

\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)

Using Kohlrausch's law. The next three examples put the law to work: adding up ion values (Example 2.7), assembling a weak electrolyte out of three strong ones (Example 2.8), and turning a single conductivity reading into a dissociation constant (Example 2.9).

Example 2.7NCERT Example 2.7 · p. 50

Limiting molar conductivities from ion values

Use Table 2.7 to find $Λ_{m}^{\circ}$ for barium chloride and for sodium sulphate.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.8NCERT Example 2.8 · p. 50

Limiting molar conductivity of a weak base

At 298 K, $Λ_{m}^{\circ}$ is 149.8 for $N H_{4} C l$ , 249.2 for $N a O H$ and 126.4 for $N a C l$ , all in S cm² mol⁻¹. Find $Λ_{m}^{\circ}$ for aqueous ammonia, treated as $N H_{4} O H$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 2.9NCERT Example 2.9 · p. 50

Dissociation constant of ammonia from conductivity

A 0.0100 mol L⁻¹ solution of ammonia has a conductivity of $1.13 \times 10^{- 4} S c m^{- 1}$ at 298 K. Using $Λ_{m}^{\circ} = 272.6 S c m^{2} m o l^{- 1}$ from Example 2.8, find the degree of dissociation and the base dissociation constant $K_{b}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

A KCl solution is diluted from 0.10 to 0.010 mol L⁻¹. What happens to

κ

and

Λ_{m}

both fall

κ

rises,

Λ_{m}

fallsboth rise

κ

falls,

Λ_{m}

rises

Show answer

Answer (D) $κ$ falls, $Λ_{m}$ rises

Each unit cube now holds fewer ions, so

κ

falls. But one mole is spread through ten times the volume and the ions hinder each other less, so

κ V = Λ_{m}

rises.

Practice · 3 questions

Intext questions 2.7–2.9: questions with answers

Short answerEasyNCERT Intext Question 2.7
A conductivity meter dipped in a salt solution reads lower and lower as you keep adding water. Why?
Show solution
AnswerEach unit volume holds fewer ions to carry charge.
Conductivity is the conductance of a unit cube of solution between opposite faces. Diluting spreads the ions through more solvent, so each unit cube contains fewer ions, and fewer carriers mean a smaller current for the same voltage. This outweighs the small gain in each ion's speed.
Short answerMediumNCERT Intext Question 2.8
Explain how $Λ_{m}^{\circ}$ of water could be obtained from the limiting molar conductivities of three strong electrolytes, and evaluate it using Table 2.7.
S cm² mol⁻¹
Show solution
Answer $Λ_{m}^{\circ} (H_{2} O) = 548.7 S c m^{2} m o l^{- 1}$
Water itself conducts far too little for direct measurement, so use Kohlrausch's law: $Λ_{m}^{\circ} (H_{2} O) = Λ_{m}^{\circ} (H C l) + Λ_{m}^{\circ} (N a O H) - Λ_{m}^{\circ} (N a C l)$ , in which the sodium and chloride contributions cancel.
From the ion values, this is $λ_{H^{+}}^{\circ} + λ_{O H^{-}}^{\circ} = 349.6 + 199.1 = 548.7 S c m^{2} m o l^{- 1}$ .
2.9
NumericalMediumNCERT Intext Question 2.9
A 0.0050 mol L⁻¹ acetic acid solution has $Λ_{m} = 22.9 S c m^{2} m o l^{- 1}$ . Using $λ_{H^{+}}^{\circ} = 349.6$ and $λ_{C H_{3} C O O^{-}}^{\circ} = 40.9 S c m^{2} m o l^{- 1}$ , find:
1. the degree of dissociation;
  Show solution
  $Λ_{m}^{\circ} = 349.6 + 40.9 = 390.5 S c m^{2} m o l^{- 1}$ , so $α = 22.9 / 390.5 = 0.0586$ .
2. the dissociation constant $K_{a}$ .
  mol L⁻¹
  Show solution
  $K_{a} = c α^{2} / (1 - α) = 0.0050 \times {0.0586}^{2} / 0.9414 = 1.83 \times 10^{- 5} m o l L^{- 1}$ .

2.5 Electrolytic Cells and Electrolysis

Copper for electrical wiring must be better than 99.9 % pure, and the way to get it is electrolysis. Hang a thick slab of crude copper and a thin sheet of pure copper in acidified copper sulphate solution, and connect a DC supply: slab to the positive terminal, sheet to the negative. The supply pushes electrons into the sheet, making it the cathode, where copper ions are reduced and plate out:

cathode, reduction

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

(2.28)

It pulls electrons out of the slab, making it the anode, where copper atoms are oxidised and dissolve:

anode, oxidation

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

(2.29)

Copper simply migrates from slab to sheet. The impurities part company on the way: more reactive metals such as iron and zinc dissolve but stay in solution, because their ions are harder to reduce than $C u^{2 +}$ , while silver and gold do not dissolve at all and fall to the bottom as anode mud, worth recovering for themselves. Nothing here would happen without the supply: this is an electrolytic cell doing work that no spontaneous reaction would.

For the most reactive metals electrolysis is the only practical route, because their ions sit so low in Table 2.2 that no cheap reducing agent such as carbon can reduce them. Aluminium, for instance, is made by electrolysing its oxide dissolved in molten cryolite ( $N a_{3} A l F_{6}$ ), which lets the cell run near 950 °C instead of the 2000 °C or more needed to melt the oxide alone. Sodium and magnesium come from their molten chlorides.

Quantitative aspects of electrolysis. How much copper does an hour of current deposit? Michael Faraday answered such questions by experiment in 1833–34, decades before the electron was discovered. Today his two laws read as simple bookkeeping of electrons.

LawFaraday's first law of electrolysis

The amount of substance that reacts at an electrode is directly proportional to the quantity of electricity (the charge) passed through the electrolyte.

LawFaraday's second law of electrolysis

Pass the same charge through different electrolytes, and the masses of the substances set free are in the ratio of their equivalent masses, where equivalent mass = molar mass ÷ number of electrons needed per ion.

With a modern constant-current supply you set a current $I$ for a time $t$ , and the charge passed is simply

Q = I t

where

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

(The first law was once used backwards, to measure charge. Before constant-current supplies existed, a small electrolytic cell called a coulometer was placed in the circuit, and the silver or copper it deposited was weighed.) To turn coulombs into moles we need the charge of a mole of electrons, the Faraday constant (फैराडे स्थिरांक): $F = N_{A} e = (6.022 14 \times 10^{23} m o l^{- 1}) (1.602 177 \times 10^{- 19} C) = 96 485 C m o l^{- 1}$ . A charge of this size is called one faraday (1 F). For rough work, 96 500 C mol⁻¹ will do; older books print 96 487 C mol⁻¹, which makes no difference at three significant figures.

The electrode equation fixes how many faradays each mole of product needs. One mole of $N a^{+}$ needs one mole of electrons (1 F), one mole of $C u^{2 +}$ needs 2 F and one mole of $A l^{3 +}$ needs 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}

Putting it together, a current $I$ flowing for a time $t$ deposits a mass $m$ of an element of molar mass $M$ whose ion needs $z$ electrons:

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

Industrial cells run on huge currents. A modern aluminium smelting cell carries about 300 000 A, which is about 3.1 mol of electrons every second: enough for about 28 g of aluminium per second, or roughly 2.4 tonnes a day from a single cell (Fig. 2.12 shows the second law on a bench scale).

3/3One faraday gives 1 mol Ag, ½ mol Cu and ⅓ mol Au: 107.9 g, 31.8 g and 65.7 g.

Fig. 2.12The same charge deposits amounts of metal in the ratio of their equivalent masses (Faraday's second law).

A DC supply and an ammeter drive current through three cells joined in series: silver nitrate, copper sulphate and a gold(III) salt. In the first step the circuit is shown with empty cathodes. In the second step a label reads 1 F = 96 485 coulombs and metal has built up on each cathode: most on the silver cathode, half as many moles on the copper and a third as many on the gold. In the third step bars compare the amounts: 1 mol of silver weighing 107.9 g, half a mole of copper weighing 31.8 g, and a third of a mole of gold weighing 65.7 g.

Example 2.10NCERT Example 2.10 · p. 52

Silver-plating spoons

A set of spoons is silver-plated in silver nitrate solution with a steady current of 0.80 A for 45 minutes. What mass of silver is deposited? (Molar mass of Ag = 107.9 g mol⁻¹.)

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

In the electrolytic refining of copper, the impure copper is made…

the cathode, so that impurities are reducedthe electrolytethe salt bridgethe anode, so that it dissolves

Show answer

Answer (D) the anode, so that it dissolves

The impure slab must dissolve, which is oxidation, so it is the anode (joined to the positive terminal). Pure copper deposits on the cathode, and gold and silver collect as anode mud.

2.5.1 Products of Electrolysis

Electrolyse copper sulphate and you get copper; electrolyse sodium chloride solution and you get no sodium at all. Three things decide the products: what ions are present, whether water is present to compete with them, and whether the electrodes can react. An inert electrode (platinum, gold, graphite) merely hands electrons over; a reactive electrode (copper, silver, nickel) may itself be oxidised, as the copper slab was in refining.

Think of each electrode as a contest. At the cathode, the species easiest to reduce wins: the couple with the highest reduction potential. At the anode, the species easiest to oxidise wins: the couple with the lowest reduction potential. Potentials only say what can happen, though. Some electrode reactions are so sluggish that they need an extra push, an overpotential (अधिविभव), before they go at a useful rate, and making oxygen from water is notoriously sluggish.

In molten sodium chloride there is no contest: $N a^{+}$ and $C l^{-}$ are the only candidates, so the melt yields sodium and chlorine:

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

Dissolve the salt in water and water joins both contests (Fig. 2.13). Cathode contest:

\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}

Reduction of $H^{+}$ wins by 2.71 V. The few $H^{+}$ ions are replaced from water as fast as they are used ( $H_{2} O \overset{}{⇌} H^{+} + O H^{-}$ ), so the cathode reaction is best written with water itself, leaving hydroxide behind:

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

(2.35)

Anode contest:

\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)

Here the potentials point the wrong way: water, with the lower $E^{\circ}$ , ought to win. It loses because of the oxygen overpotential, so concentrated brine gives chlorine. Adding the two electrode reactions, and doubling to clear the halves, gives the overall change:

\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}

One process, three saleable products: hydrogen, chlorine and sodium hydroxide. This chlor-alkali process is among the largest electrochemical industries in India and worldwide.

3/4O₂ evolution is slow (overpotential), so Cl⁻ is oxidised: chlorine is released.

Fig. 2.13Electrolysis of brine: hydrogen at the cathode, chlorine at the anode, sodium hydroxide left in solution.

A cell with two inert electrodes in sodium chloride solution is connected to a DC supply, cathode on the left and anode on the right. The first step labels the species present: sodium ions, chloride ions and water. In the second step the cathode is highlighted with two candidate reductions: sodium ions at minus 2.71 V crossed out, and water to hydrogen ticked; hydrogen bubbles rise there. In the third step the anode is highlighted: water to oxygen at 1.23 V is marked slow because of overpotential, and chloride to chlorine at 1.36 V is ticked; chlorine bubbles rise. In the fourth step the solution near the cathode is labelled sodium hydroxide.

Concentration can change the winner too, because the real competitors are Nernst potentials, not standard ones. Electrolyse sulphuric acid between platinum electrodes and the anode product depends on how strong the acid is:

\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 gives oxygen. Very concentrated acid, rich in sulphate and short of free water, gives the peroxodisulphate ion $S_{2} O_{8}^{2 -}$ instead; this was once the industrial route to hydrogen peroxide. For aqueous solutions with inert electrodes, a few rules of thumb cover most cases:

Cathode. Ions of less reactive metals, such as $A g^{+}$ and $C u^{2 +}$ , are reduced to the metal. Ions of very reactive metals ( $K^{+}$ , $N a^{+}$ , $M g^{2 +}$ , $A l^{3 +}$ ) stay in solution, and water is reduced to hydrogen.
Anode. Halide ions ( $C l^{-}$ , $B r^{-}$ , $I^{-}$ ) are oxidised to the halogen, especially when concentrated. Oxoanions such as $S O_{4}^{2 -}$ and $N O_{3}^{-}$ are not; water is oxidised to oxygen instead.
Reactive anode. With a copper, silver or nickel anode, the anode metal itself dissolves.

Example

Copper sulphate, two ways

Predict the products when aqueous copper(II) sulphate is electrolysed (a) between platinum electrodes and (b) between copper electrodes.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveElectrolysis bench

A bench with a cell drawn between a cathode on the left and an anode on the right. A menu offers nine set-ups, among them molten sodium chloride, brine, copper sulphate with platinum or copper electrodes, silver nitrate with platinum or silver electrodes, dilute sulphuric acid, sodium sulphate solution and potassium iodide solution. For each electrode the panel lists the competing half-reactions with their standard potentials, marks the winner and explains why, including overpotential where it matters. Sliders set the current from 0.1 to 5 amperes and the time from 1 to 120 minutes. The Run button animates bubbles or a growing deposit while a clock runs, and the readouts give the charge, the moles of electrons and the mass or gas volume of each product at 298 kelvin and 1 bar.

Fig. 2.14Choose an electrolyte and electrodes, read the contest at each electrode, then set a current and time and press Run to see how much of each product forms.

Practice · 3 questions

Intext questions 2.10–2.12: questions with answers

NumericalEasyNCERT Intext Question 2.10
A phone charger cable carries a steady 2.0 A for 30 minutes. How many electrons pass through it?
Show solution
$Q = I t = 2.0 A \times 1800 s = 3600 C$ .
Number of electrons $= 3600 / (1.602 \times 10^{- 19}) = 2.25 \times 10^{22}$ . (Equivalently, $3600 / 96 485 = 0.0373$ mol of electrons.)
2.11
Quick answerEasyNCERT Intext Question 2.11
Which of these metals are produced industrially by electrolysis?
sodiumaluminiumironmagnesiumlithiummercury
Select all that apply
Show answer
Answer
(A) sodium
(B) aluminium
(D) magnesium
(E) lithium
Sodium, aluminium, magnesium and lithium are too reactive to reduce with carbon, so they are made by electrolysing molten salts. (Calcium and potassium are too.) Iron is made in a blast furnace with carbon monoxide, and mercury by roasting its ore, cinnabar ( $H g S$ ).
NumericalMediumNCERT Intext Question 2.12
Chlorate ions can be reduced to chloride: $C l O_{3}^{-} + 6 H^{+} + 6 e^{-}$ $\overset{}{\to} C l^{-} + 3 H_{2} O$ . How much charge, in coulombs, is needed to reduce 0.50 mol of $C l O_{3}^{-}$ ?
C
Show solution
Each $C l O_{3}^{-}$ takes 6 electrons (chlorine goes from +5 to −1), so 0.50 mol needs 3.0 mol of electrons, that is, 3.0 F.
$Q = 3.0 \times 96 485 = 2.89 \times 10^{5} C$ .

2.6 Batteries

A torch, a TV remote, a phone, a car and a home inverter each need a portable galvanic cell with a different personality. Two broad families exist. Primary batteries run their reaction once and are then spent (and should be recycled, not binned); secondary batteries can be driven backwards by a charger and used again. Whatever the family, designers chase the same wish list: as much energy as possible per gram and per cubic centimetre, a voltage that holds steady until the end, safety in a pocket and a low price. (Strictly, a battery is a set of cells joined in series, but everyday speech uses the word for a single cell too.)

2.6.1 Primary Batteries

The 1.5 V dry cell in a wall clock or TV remote is a primary cell, devised by Georges Leclanché in the 1860s (Fig. 2.15). Cut one open (an adult's job: the paste inside is corrosive) and you find that the outer zinc can is itself the anode, which is why old cells leak once their zinc is eaten through. Down the centre runs a graphite rod that collects current, packed round with a black mixture of manganese dioxide and carbon powder, the real cathode reactant. A damp paste of ammonium chloride and zinc chloride serves as electrolyte; 'dry' only means there is no free liquid to spill.

Fig. 2.15A dry (Leclanché) cell in section. The zinc can is the anode; the graphite rod collects current from the manganese dioxide cathode.NCERT Fig. 2.8 · p. 54

A dry cell is cut lengthwise. The outer zinc can, labelled anode and negative, forms the walls and base. Inside it is a layer of moist ammonium chloride and zinc chloride paste, then a thick core of manganese dioxide mixed with carbon powder, with a graphite rod running down the centre to a brass cap on top, labelled cathode and positive. A seal closes the top and a jacket wraps the outside.

What happens inside is messy, but a simplified pair of half-reactions captures it:

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}

The ammonia formed at the cathode does not build up as gas: zinc ions from the anode bind it as the complex ion $[Z n (N H_{3})_{4}]^{2 +}$ . Manganese is the element actually reduced, dropping from +4 in $M n O_{2}$ to +3 in $M n O (O H)$ .

Watches and hearing aids need a tiny current at a voltage that does not sag. The mercury cell, a button cell, was built for that: zinc dissolved in mercury (an amalgam) as the anode, mercury(II) oxide mixed with carbon as the cathode, and a paste of KOH and ZnO between them:

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}

Look at the overall equation: no dissolved ion appears in it. Nothing whose concentration could change enters $Q$ , so by the Nernst equation the voltage, about 1.35 V, stays nearly constant for the whole life of the cell.

2.6.2 Secondary Batteries

Inside the box under a car bonnet, or the inverter that keeps fans running through a power cut, is the lead storage battery, the oldest rechargeable design still in mass use (Gaston Planté built the first in 1859). Each of its cells has a plate of spongy lead and a lead grid packed with lead dioxide, both standing in fairly strong sulphuric acid, about 35–40 % by mass. What makes it a secondary (rechargeable) cell is that a charger can push its reaction backwards and rebuild both plates; a well-kept battery survives hundreds of such cycles. On discharge:

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}

Both electrodes turn into lead sulphate, and sulphuric acid is replaced by water, so the acid becomes weaker and less dense as the battery runs down. A mechanic can check the state of charge by measuring the acid's density with a hydrometer. On charging, the external supply reverses both electrode reactions: lead sulphate on one plate goes back to lead, on the other to lead dioxide, and the acid is regenerated (Fig. 2.16). Each cell gives about 2 V, so a 12 V car battery has six cells in series.

2/3Both plates turn to PbSO₄ and the acid weakens (about 1.12 g cm⁻³).

Fig. 2.16A lead storage cell: discharging coats both plates with lead sulphate and dilutes the acid; charging reverses both changes.NCERT Fig. 2.10 · p. 55

A single cell holds a lead plate, marked negative, and a lead dioxide plate, marked positive, in sulphuric acid, with a small hydrometer float showing the acid density. In the first step the plates are clean and the acid density reads 1.28 grams per cubic centimetre. In the second step a lamp is connected and lit, electrons flow out of the lead plate, white lead sulphate coats both plates and the density falls to about 1.12. In the third step a charger replaces the lamp and drives current the other way; the lead sulphate disappears from both plates and the density returns to 1.28.

Nickel–cadmium (Ni–Cd) cells cost more to make than lead cells but survive more charge–discharge cycles; they once powered cordless drills and emergency lights. Their overall discharge reaction is:

\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}

Many books write the nickel electrode as $N i O (O H)$ instead; either way nickel goes from +3 to +2 and cadmium from 0 to +2. Cadmium is toxic as well, and Ni–Cd cells have largely given way to nickel–metal hydride and lithium-ion cells.

2.7 Fuel Cells

Some city buses now run on hydrogen and leave nothing behind but a trickle of water. Their engine is a fuel cell (ईंधन सेल): a galvanic cell that never runs down, because fuel and oxidant are piped in continuously and the products are piped away. Compare a coal or gas power station, which reaches electricity the long way round, through burning, steam, turbine and generator, losing energy at each stage and delivering only 35–40 % of the fuel's energy (plus smoke and carbon dioxide). A fuel cell skips the heat engine. Hydrogen is the favourite fuel, but natural gas, methanol and even ethanol can be 'burned' in this cold, electrochemical way, with oxygen or air as the oxidant.

The classic design, flown in space, is the alkaline hydrogen–oxygen cell (Fig. 2.17). Its two electrodes are porous carbon, so each gas can soak in from one side while hot, concentrated sodium (or potassium) hydroxide solution wets the electrode from the other. Hydrogen and oxygen react only sluggishly at bare carbon, so the electrodes are dusted with a catalyst, finely divided platinum or palladium. The half-reactions are:

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 run through the motor; OH⁻ ions cross the electrolyte to the anode.

Fig. 2.17In the alkaline fuel cell, hydroxide ions carry charge through the electrolyte while electrons do work in the external circuit.NCERT Fig. 2.12 · p. 56

The cell is drawn as horizontal layers: a hydrogen channel at the top, a porous carbon anode, a layer of hot sodium hydroxide electrolyte, a porous carbon cathode and an oxygen channel at the bottom. A wire on the left runs from the anode through a small motor to the cathode. In the first step hydrogen and oxygen flow into their channels. In the second step electrons travel from the anode through the motor to the cathode, while hydroxide ions move upwards through the electrolyte from the cathode to the anode. In the third step water forms at the anode and is carried out as steam on the right.

Each Apollo command module carried three of these cells, and besides electricity they supplied the crew's drinking water.

How efficient can a fuel cell be? The most electrical work available is $- Δ_{r} G$ , while burning the hydrogen would release $- Δ_{r} H$ as heat. For hydrogen and oxygen forming liquid water at 298 K, $Δ_{r} G^{\circ} / Δ_{r} H^{\circ} = 237 / 286 \approx 0.83$ , so an ideal cell could turn 83 % of the fuel's energy into electricity. Real fuel cells reach about 50–70 %, still far better than a thermal plant. The main challenges now are cheaper catalysts than platinum, and making and storing the hydrogen cleanly (see the box at the end of the chapter).

Quick check

1 of 3

In the alkaline hydrogen–oxygen fuel cell, what happens at the anode?

O_{2}

is reduced to

O H^{-}

H_{2} O

is split into

H_{2}

and

O_{2}

H_{2}

is oxidised, using up

O H^{-}

and forming waterplatinum is oxidised

Show answer

Answer (C) $H_{2}$ is oxidised, using up $O H^{-}$ and forming water

At the anode hydrogen loses electrons:

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

. Oxygen is reduced at the cathode. The platinum is only a catalyst.

2.8 Corrosion

The iron pillar at the Qutb complex in Delhi has stood in the open for about 1600 years with hardly a trace of rust, yet a steel gate left out through one monsoon starts to flake. What is going on? Most refined metals are less stable than the ores they came from, and given air and moisture they slowly head back: copper roofs and bronze statues gain a green crust of basic copper carbonate, silver anklets darken with a film of silver sulphide, and steel grows flaky red-brown rust. This drift, corrosion (संक्षारण), eats into pipelines, railway tracks, ships and the steel inside concrete, and its cost has been estimated at about 3 % of world GDP each year. (The Delhi pillar survives thanks to a thin protective film that formed on its unusually phosphorus-rich iron in a fairly dry climate.)

Rusting needs both water and oxygen, and the clearest way to understand it is as a galvanic cell built on the metal itself (Fig. 2.18). Some spot, perhaps a scratch, a strained bend, or a patch starved of oxygen under a water drop, becomes the anode, where iron is oxidised:

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}

Electrons released at the anode travel through the iron itself, which acts as the wire, to a spot where oxygen is plentiful. There, at the cathode, oxygen takes up the electrons together with hydrogen ions. Rainwater is never quite pure: dissolved carbon dioxide makes it faintly acidic (carbonic acid), and near traffic and factories dissolved sulphur dioxide adds more acid:

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}

With $E_{cell}^{\circ} = 1.67 V$ the cell is strongly spontaneous. Rust itself forms a step later: $F e^{2 +}$ ions wander away from the anode, meet dissolved oxygen and are oxidised to iron(III), which separates as hydrated iron(III) oxide, $F e_{2} O_{3} \cdot x H_{2} O$ , regenerating $H^{+}$ as it does. Because rust is porous and flakes away, it never seals the metal beneath. Salt makes everything faster by turning the water film into a much better ionic conductor, which is why vehicles rust sooner near the sea.

4/4Fe²⁺ meets oxygen and is oxidised further: a ring of rust forms.

Fig. 2.18Rusting is an electrochemical cell: iron dissolves at the anode, electrons flow through the metal, and oxygen is reduced at the cathode.NCERT Fig. 2.13 · p. 57

A cross-section shows an iron plate with a water drop resting on it. In the first step oxygen from the air dissolves mostly near the thin edge of the drop. In the second step a pit forms under the centre of the drop, labelled anode, where iron loses electrons and iron(II) ions enter the water; arrows show the electrons moving through the iron towards the edge. In the third step the edge of the drop is labelled cathode, where oxygen and hydrogen ions take up the electrons to form water. In the fourth step iron(II) ions meet oxygen in the water and a ring of rust, hydrated iron(III) oxide, forms near the edge.

Every method of protection breaks the corrosion cell somewhere: it keeps water and oxygen off the metal, or it makes some other metal play the anode.

Barrier coatings keep air and water away: paint, grease or oil, plastic films and chemical coatings such as epoxy paints. They protect only while unbroken.
Metal coatings. Tin plating, used for food cans, seals the iron off, but tin is less reactive than iron, so once it is scratched the exposed iron becomes the anode and rusts even faster. A zinc coating (galvanising) protects even when scratched, because zinc is more reactive and is oxidised in preference to the iron.
Sacrificial (cathodic) protection. A block of a more reactive metal such as magnesium or zinc is wired to the iron, or bolted to a ship's hull. It becomes the anode and corrodes instead, leaving the iron as the cathode, and is simply replaced when used up. Buried pipelines and water-heater tanks are protected this way.

Try this at home

Which nail rusts first?

You’ll need

5 iron nails
sandpaper
a strip of zinc or galvanised wire, and a piece of bare copper wire
salt, tap water, cooking oil
5 clean glass jars with labels

Predict first

Which nail do you expect to rust fastest?

nail 1, in tap waternail 3, under oilnail 4, wrapped in zincnail 5, wrapped in copper

Clean five iron nails with sandpaper and label them 1 to 5. Wrap a strip of zinc (from a hardware shop) or a piece of galvanised wire tightly round nail 4, and a piece of bare copper wire round nail 5.
Put each nail in its own labelled jar: nail 1 in tap water; nail 2 in salt water (one teaspoon of salt per cup); nail 3 in water that has been boiled and cooled, with a layer of cooking oil poured on top; nails 4 and 5 in salt water.
Leave them for a week, checking every day. Record when rust first appears and how much forms.

Table 2.8Rusting observations: note any rust on day 2 and how much by day 7

Nail	Observations
1: tap water
2: salt water
3: boiled, oil
4: salt, zinc
5: salt, copper

Your entries are saved automatically.

Nail 5, in contact with copper in salt water, rusts first and worst, because the iron is the anode of an iron–copper cell. Nail 2 rusts faster than nail 1, because salt makes the water a better ionic conductor. Nail 3 hardly rusts: boiling drove out the dissolved oxygen and the oil stops more from dissolving. Nail 4 stays bright while the zinc corrodes instead, since zinc is the anode: this is sacrificial protection in a jar.

In the real world

The hydrogen economy

In 2023 India launched its National Green Hydrogen Mission, aiming to produce about five million tonnes of 'green' hydrogen a year by 2030. Most hydrogen made today is 'grey': it comes from natural gas and releases carbon dioxide. Green hydrogen is made instead by electrolysing water with electricity from sunlight or wind, so no carbon dioxide is released in making it. It can then feed fertiliser plants, refineries and steelworks, or be turned back into electricity in a fuel cell, where its only product is water. The catch is cost: green hydrogen is still considerably dearer than grey, partly because electrolysers rely on scarce catalyst metals such as platinum and iridium. Cheaper, longer-lasting electrode materials, for electrolysers and fuel cells alike, are a major research goal.

Practice · 3 questions

Intext questions 2.13–2.15: questions with answers

Short answerMediumNCERT Intext Question 2.13
A car's alternator recharges its lead storage battery. Write the reaction at each plate during charging, name what is used up and what is formed, and say what happens to the acid.
Show solution
AnswerBoth plates' $P b S O_{4}$ is converted back to Pb and $P b O_{2}$ ; water is used up and sulphuric acid is regenerated.
At the plate joined to the charger's negative terminal (the discharge anode), lead sulphate is reduced: $P b S O_{4} (s) + 2 e^{-}$ $\overset{}{\to} P b (s) + S O_{4}^{2 -} (a q)$ .
At the other plate it is oxidised back to lead dioxide: $P b S O_{4} (s) + 2 H_{2} O (l)$ $\overset{}{\to} P b O_{2} (s)$ $+ S O_{4}^{2 -} (a q)$ $+ 4 H^{+} (a q) + 2 e^{-}$ .
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 and water are used up; lead, lead dioxide and sulphuric acid are formed, so the acid becomes stronger and denser again. Electrical energy from the alternator is stored as chemical energy.
Short answerMediumNCERT Intext Question 2.14
Name two fuels other than hydrogen that can be used in fuel cells, and write the anode half-reaction for one of them in an acidic electrolyte.
Show solution
AnswerMethanol and methane; e.g. $C H_{3} O H + H_{2} O$ $\overset{}{\to} C O_{2} + 6 H^{+} + 6 e^{-}$ .
Methanol and methane are common choices (ethanol and natural gas are others). For methanol, carbon goes from −2 to +4, so six electrons are released: $C H_{3} O H (l) + H_{2} O (l)$ $\overset{}{\to} C O_{2} (g)$ $+ 6 H^{+} (a q) + 6 e^{-}$ .
For methane, carbon goes from −4 to +4: $C H_{4} (g) + 2 H_{2} O (l)$ $\overset{}{\to} C O_{2} (g)$ $+ 8 H^{+} (a q) + 8 e^{-}$ .
Short answerMediumNCERT Intext Question 2.15
A drop of rainwater sits on a steel railing. Explain, as a galvanic cell, why a pit forms under the middle of the drop while rust appears as a ring near its edge.
Show solution
AnswerThe centre is the anode (little oxygen); the edge is the cathode (oxygen-rich); rust forms where Fe²⁺ meets oxygen.
Oxygen dissolves into the drop mainly through its thin edge, so the edge becomes the cathode, where $O_{2} + 4 H^{+} + 4 e^{-}$ $\overset{}{\to} 2 H_{2} O$ .
Under the oxygen-poor centre, iron is the anode: $F e \overset{}{\to} F e^{2 +} + 2 e^{-}$ , so the metal is eaten away there, forming a pit. The electrons travel through the steel to the edge, and the water carries ions to complete the circuit.
$F e^{2 +}$ ions drift outwards and are oxidised by dissolved oxygen to iron(III), which separates as $F e_{2} O_{3} \cdot x H_{2} O$ , so the rust appears near the edge, not over the pit.

Recap

Electrochemistry: key points and summary

An electrochemical cell has two electrodes in an electrolyte. A galvanic cell lets a spontaneous redox reaction do electrical work; an electrolytic cell uses electrical energy to drive a non-spontaneous reaction (Section 2.1).
Oxidation happens at the anode and reduction at the cathode. In a galvanic cell the anode is negative; electrons flow through the wire from anode to cathode, and a salt bridge keeps the solutions neutral (Section 2.2).
Standard electrode potentials are reduction potentials measured against the standard hydrogen electrode (0 V), and $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ . An oxidising agent reacts with any reducing agent below it in the table (Section 2.2.1).
The Nernst equation, $E_{cell} = E_{cell}^{\circ} - (0.0591 / n) \log Q$ at 298 K, gives the potential at any concentration (Section 2.3).
At equilibrium $E_{cell} = 0$ , so $E_{cell}^{\circ} = (0.0591 / n) \log K_{c}$ at 298 K (Section 2.3.1).
$Δ_{r} G = - n F E_{cell}$ and $Δ_{r} G^{\circ} = - n F E_{cell}^{\circ} = - R T \ln K$ . $E$ is intensive; $Δ_{r} G$ is extensive (Section 2.3.2).
$R = ρ l / A$ , $G = 1 / R = κ A / l$ and $Λ_{m} = κ / c$ . Metals conduct by electrons; solutions by ions, and better when warm (Section 2.4).
Conductivity is measured with AC in a conductivity cell of known cell constant $G^{*} = l / A = R κ$ , using a Wheatstone bridge or a meter (Section 2.4.1).
On dilution $κ$ falls but $Λ_{m}$ rises: gently for strong electrolytes ( $Λ_{m} = Λ_{m}^{\circ} - A \sqrt{c}$ ), steeply for weak ones (Section 2.4.2).
Kohlrausch's law: $Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}$ . It gives $Λ_{m}^{\circ}$ of weak electrolytes, then $α = Λ_{m} / Λ_{m}^{\circ}$ and $K_{a}$ (Law).
Faraday's laws: the amount deposited is proportional to the charge $Q = I t$ , and one faraday ( $96 485 C m o l^{- 1}$ ) deposits $1 / z$ mole of an ion of charge $z$ (Section 2.5).
The products of electrolysis depend on the electrolyte, on water competing at both electrodes, on the electrodes and on overpotential (Section 2.5.1).
Primary cells (dry cell, mercury cell) are used once; secondary cells (lead storage, Ni–Cd) are recharged; fuel cells run as long as fuel is supplied (Section 2.6, Section 2.7).
Corrosion is an electrochemical cell on the metal surface; coatings, galvanising and sacrificial anodes prevent it (Section 2.8).

Flash cards · 11

Formulas to remember

1 / 11

Practice · 23 questions

Practice questions with answers (NCERT “Exercises”)

2.1
Quick answerEasyNCERT Exercises 2.1
Rank aluminium, lead, nickel, silver and zinc from the strongest to the weakest displacer: each metal in your list should be able to push every later one out of solution.
1. zinc
2. nickel
3. lead
4. silver
5. aluminium
Show answer
Answer
aluminium
zinc
nickel
lead
silver
$E^{\circ}$ : Al −1.66, Zn −0.76, Ni −0.25, Pb −0.13, Ag +0.80 V. The lower the potential, the stronger the reducing agent, so each metal displaces the ones after it.
2.2
Quick answerEasyNCERT Exercises 2.2
Given $E^{\circ}$ : $L i^{+} / L i$ −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. Put the metals in order from the weakest reducing agent to the strongest.
1. Ag
2. Cu
3. Na
4. Li
5. Fe
Show answer
Answer
Ag
Cu
Fe
Na
Li
The highest $E^{\circ}$ (silver) is the weakest reducing agent; the lowest (lithium) the strongest.
2.3
Short answerMediumNCERT Exercises 2.3
A galvanic cell runs the reaction $2 A l (s) + 3 N i^{2 +} (a q)$ $\overset{}{\to} 2 A l^{3 +} (a q) + 3 N i (s)$ .
1. Write the cell in standard notation.
  Show solution
  Answer $A l (s) ∣ A l^{3 +} (a q)$ $∣ ∣ N i^{2 +} (a q) ∣ N i (s)$
  Aluminium is oxidised, so it is the anode and goes on the left: $A l (s) ∣ A l^{3 +} (a q)$ $∣ ∣ N i^{2 +} (a q) ∣ N i (s)$ .
2. Which electrode is the negative terminal?
  the aluminium electrodethe nickel electrode
  Show answer
  Answer (A) the aluminium electrode
  Aluminium is the anode: oxidation leaves electrons on it.
3. What carries the current in the wire, and what carries it in the cell?
  Show solution
  AnswerElectrons in the wire; ions in the solutions and salt bridge.
  Electrons flow through the wire from Al to Ni. Inside, $A l^{3 +}$ ions enter the left solution and $N i^{2 +}$ leave the right; anions from the salt bridge move towards the aluminium side and cations towards the nickel side.
4. Write the reaction at each electrode and find $E_{cell}^{\circ}$ .
  Show solution
  Answer $E_{cell}^{\circ} = 1.41 V$
  Anode: $A l \overset{}{\to} A l^{3 +} + 3 e^{-}$ . Cathode: $N i^{2 +} + 2 e^{-} \overset{}{\to} N i$ . Multiplying by 2 and 3 to balance six electrons gives the cell reaction.
  $E_{cell}^{\circ} = - 0.25 - (- 1.66) = 1.41 V$ .
2.4
NumericalMediumNCERT Exercises 2.4
Using Table 2.2, find $E_{cell}^{\circ}$ , $Δ_{r} G^{\circ}$ and $K_{c}$ at 298 K for each reaction.
1. $2 A l (s) + 3 S n^{2 +} (a q)$ $\overset{}{\to} 2 A l^{3 +} (a q)$ $+ 3 S n (s)$
  Show solution
  Answer1.52~V; −880~kJ~mol⁻¹; $K \approx 10^{154}$
  $E^{\circ} = - 0.14 - (- 1.66) = 1.52 V$ , and $n = 6$ .
  $Δ_{r} G^{\circ} = - 6 \times 96 485 \times 1.52 = - 8.80 \times 10^{5} J m o l^{- 1} = - 880 k J m o l^{- 1}$ .
  $\log K_{c} = 6 \times 1.52 / 0.0591 = 154.3$ , so $K_{c} \approx 2 \times 10^{154}$ .
2. $C l_{2} (g) + 2 B r^{-} (a q)$ $\overset{}{\to} 2 C l^{-} (a q) + B r_{2} (l)$
  Show solution
  Answer0.27~V; −52.1~kJ~mol⁻¹; $K \approx 1.4 \times 10^{9}$
  $E^{\circ} = 1.36 - 1.09 = 0.27 V$ ; $n = 2$ .
  $Δ_{r} G^{\circ} = - 2 \times 96 485 \times 0.27 = - 5.21 \times 10^{4} J m o l^{- 1} = - 52.1 k J m o l^{- 1}$ .
  $\log K_{c} = 9.14$ , so $K_{c} \approx 1.4 \times 10^{9}$ .
2.5
NumericalMediumNCERT Exercises 2.5
For each cell below, set up the Nernst equation and work out the emf at 298 K.
1. $N i (s) ∣ N i^{2 +} (0.0020 M)$ $∣ ∣ C u^{2 +} (0.050 M)$ $∣ C u (s)$
  V
  Show solution
  $E = E^{\circ} - \frac{0.0591}{2} \log \frac{[N i^{2 +}]}{[C u^{2 +}]}$ with $E^{\circ} = 0.59 V$ and $Q = 0.040$ .
  $E = 0.59 + 0.041 = 0.631 V$ .
2. $Z n (s) ∣ Z n^{2 +} (0.10 M)$ $∣ ∣ H^{+} (1 M)$ $∣ H_{2} (g, 1 b a r) ∣ P t (s)$
  V
  Show solution
  $E = 0.76 - \frac{0.0591}{2} \log \frac{0.10 \times 1}{1^{2}} = 0.76 + 0.030 = 0.790 V$ .
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
  Show solution
  $Q = [P b^{2 +}] p_{H_{2}} / [H^{+}]^{2} = 0.020 / (0.010)^{2} = 200$ .
  $E = 0.13 - 0.0296 \times \log 200 = 0.13 - 0.068 = 0.062 V$ .
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
  Show solution
  Cell reaction as written: $2 I^{-} + 2 H^{+}$ $\overset{}{\to} I_{2} + H_{2}$ , $E^{\circ} = 0.00 - 0.54 = - 0.54 V$ , $n = 2$ .
  $Q = 1 / ([I^{-}]^{2} [H^{+}]^{2})$ , and $[I^{-}]^{2} [H^{+}]^{2} = (0.020)^{2} (0.010)^{2} = 4.0 \times 10^{- 8}$ , so $Q = 2.5 \times 10^{7}$ .
  $E = - 0.54 - 0.0296 \times 7.40 = - 0.759 V$ . The negative sign means the cell runs the other way: hydrogen reduces iodine.
NumericalMediumNCERT Exercises 2.6
Zinc–air button cells in hearing aids run the reaction $2 Z n (s) + O_{2} (g)$ $\overset{}{\to} 2 Z n O (s)$ in an alkaline paste. The half-cell potentials are $Z n O (s) + H_{2} O (l) + 2 e^{-}$ $\overset{}{\to} Z n (s) + 2 O H^{-}$ , $E^{\circ} = - 1.26 V$ , and $O_{2} (g) + 2 H_{2} O (l) + 4 e^{-}$ $\overset{}{\to} 4 O H^{-}$ , $E^{\circ} = + 0.40 V$ . Find $E_{cell}^{\circ}$ and $Δ_{r} G^{\circ}$ for the equation as written.
Show solution
Answer $E^{\circ} = 1.66 V$ ; $Δ_{r} G^{\circ} = - 641 k J m o l^{- 1}$
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.40 - (- 1.26) = 1.66 V$ .
Two zinc atoms each lose two electrons, so $n = 4$ : $Δ_{r} G^{\circ} = - 4 \times 96 485 \times 1.66 = - 6.41 \times 10^{5} J m o l^{- 1} = - 641 k J m o l^{- 1}$ (or $- 320 k J$ per mole of zinc).
Long answerMediumNCERT Exercises 2.7
A friend says, 'Diluting a salt solution makes it a worse conductor and a better conductor at the same time.' Define conductivity and molar conductivity, and use them to explain what your friend means.
Show solution
Conductivity $κ$ is the conductance of a unit cube of the solution (a 1 cm or 1 m cube between opposite faces), in S cm⁻¹ or S m⁻¹. Molar conductivity $Λ_{m} = κ / c$ is the conductance of all the solution that contains one mole of electrolyte, placed between electrodes one unit apart, in S cm² mol⁻¹ or S m² mol⁻¹.
On dilution each unit cube holds fewer ions, so $κ$ falls: per cube, the solution is a worse conductor. But one mole now occupies a larger volume $V$ and its ions hinder each other less (a weak electrolyte also ionises more), so $Λ_{m} = κ V$ rises: per mole, it is a better conductor. For strong electrolytes the rise is gentle and linear in $\sqrt{c}$ ; for weak electrolytes it is steep at low concentration.
Quick answerEasyNCERT Exercises 2.8
A 0.150 mol L⁻¹ potassium nitrate solution has $κ = 0.0176 S c m^{- 1}$ at 298 K. What is its $Λ_{m}$ ?
S cm² mol⁻¹
Show answer
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
A conductivity cell filled with 0.0200 mol L⁻¹ KCl ( $κ = 2.77 \times 10^{- 3} S c m^{- 1}$ at 298 K) has a resistance of 420 Ω. Find the cell constant in cm⁻¹.
cm⁻¹
Show answer
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
The table gives $κ$ for potassium nitrate solutions at 298 K. Work out $Λ_{m}$ for each, draw the graph of $Λ_{m}$ against $\sqrt{c}$ , and use it to estimate $Λ_{m}^{\circ}$ .
Table 2.9KNO₃ solutions at 298 K
$c$ / (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
Show solution
Answer $Λ_{m}$ falls from 142.8 to 132.4 S~cm²~mol⁻¹; $Λ_{m}^{\circ} \approx 145 S c m^{2} m o l^{- 1}$
$Λ_{m} = 1000 κ / c$ with $κ$ in S cm⁻¹ gives 142.8, 141.8, 138.5, 135.8 and 132.4 S cm² mol⁻¹. The square roots of the concentrations are 0.0224, 0.0316, 0.0707, 0.100 and 0.141.
The points lie close to a straight line of slope about −87 S cm² mol⁻¹ (mol L⁻¹) $^{- 1 / 2}$ . Extending it to $\sqrt{c} = 0$ gives $Λ_{m}^{\circ} \approx 144.7$ , about 145 S cm² mol⁻¹ (compare $73.5 + 71.5 = 145.0$ from the ion table).
NumericalMediumNCERT Exercises 2.11
The conductivity of 0.0200 mol L⁻¹ formic acid ( $H C O O H$ ) is $7.30 \times 10^{- 4} S c m^{- 1}$ at 298 K. Using Table 2.7, find its molar conductivity, degree of dissociation and $K_{a}$ .
mol L⁻¹
Show solution
Answer $Λ_{m} = 36.5$ ; $α = 0.0903$ ; $K_{a} = 1.79 \times 10^{- 4}$
$Λ_{m} = 1000 \times 7.30 \times 10^{- 4} / 0.0200 = 36.5 S c m^{2} m o l^{- 1}$ .
$Λ_{m}^{\circ} = 349.6 + 54.6 = 404.2 S c m^{2} m o l^{- 1}$ , so $α = 36.5 / 404.2 = 0.0903$ .
$K_{a} = \frac{0.0200 \times (0.0903)^{2}}{1 - 0.0903} = 1.79 \times 10^{- 4} m o l L^{- 1}$ .
2.12
NumericalEasyNCERT Exercises 2.12
How much charge, in coulombs, is needed for each of these?
1. reducing 0.50 mol of $N i^{2 +}$ to Ni
  C
  Show answer
  Answer 96485 C
  $0.50 \times 2 = 1.0$ F $= 96 485 C$ .
2. reducing 2.0 mol of $A u^{3 +}$ to Au
  C
  Show answer
  Answer 579000 C
  $2.0 \times 3 = 6.0$ F $= 5.79 \times 10^{5} C$ .
3. reducing 1.0 mol of $N O_{3}^{-}$ to NO in acid
  C
  Show answer
  Answer 289000 C
  Nitrogen goes from +5 to +2: 3 F $= 2.89 \times 10^{5} C$ .
2.13
NumericalEasyNCERT Exercises 2.13
Find the charge, in faradays, that must pass to obtain:
1. 12.0 g of magnesium from molten $M g C l_{2}$ ( $M = 24.3 g m o l^{- 1}$ )?
  F
  Show answer
  Answer 0.988 F
  $12.0 / 24.3 = 0.494$ mol, × 2 = 0.988 F.
2. 13.9 g of lithium from molten LiCl ( $M = 6.94 g m o l^{- 1}$ )?
  F
  Show answer
  Answer 2 F
  $13.9 / 6.94 = 2.00$ mol, × 1 = 2.00 F.
2.14
NumericalMediumNCERT Exercises 2.14
Each of these oxidations is carried out at an anode. What charge, in coulombs, does each need?
1. 1.0 mol of $M n^{2 +}$ to $M n O_{4}^{-}$ ?
  C
  Show answer
  Answer 482000 C
  Mn goes from +2 to +7: 5 F $= 4.82 \times 10^{5} C$ .
2. 1.0 mol of $H_{2} O_{2}$ to $O_{2}$ ?
  C
  Show answer
  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
A steel bracket is zinc-plated in zinc sulphate solution at a steady 2.5 A. How much zinc does it gain in 40 minutes? ( $M = 65.4 g m o l^{- 1}$ .)
g
Show solution
$Q = 2.5 \times 2400 = 6000 C$ , which is $6000 / 96 485 = 0.0622$ mol of electrons.
$Z n^{2 +} + 2 e^{-} \overset{}{\to} Z n$ , so 0.0311 mol of zinc, or $0.0311 \times 65.4 = 2.03 g$ .
2.16
NumericalHardNCERT Exercises 2.16
A plating shop runs a nickel bath ( $N i S O_{4}$ ), a silver bath ( $A g N O_{3}$ ) and a tin bath ( $S n S O_{4}$ ) in series on one 2.0 A supply. The run stops when 0.880 g of nickel has been deposited ( $M$ : Ni 58.7, Ag 107.9, Sn 118.7 g mol⁻¹).
1. For how many minutes did the current flow?
  min
  Show solution
  $n (N i) = 0.880 / 58.7 = 0.0150$ mol, needing 0.0300 mol of electrons: $Q = 0.0300 \times 96 485 = 2.89 \times 10^{3} C$ .
  $t = Q / I = 1447 s = 24.1$ minutes.
2. What mass of silver was deposited?
  g
  Show solution
  The same 0.0300 mol of electrons deposits 0.0300 mol of silver: $0.0300 \times 107.9 = 3.24 g$ .
3. What mass of tin was deposited?
  g
  Show solution
  $S n^{2 +}$ needs two electrons, so 0.0150 mol of tin: $0.0150 \times 118.7 = 1.78 g$ .
2.17
Quick answerMediumNCERT Exercises 2.17
Which of these pairs will react when mixed under standard conditions? Use Table 2.2.
$Z n (s)$ with $F e^{2 +} (a q)$ $A g (s)$ with $C u^{2 +} (a q)$ $C l_{2} (g)$ with $F e^{2 +} (a q)$ $I_{2} (s)$ with $B r^{-} (a q)$ acidified $M n O_{4}^{-}$ with $C l^{-} (a q)$
Select all that apply
Show answer
Answer
(A) $Z n (s)$ with $F e^{2 +} (a q)$
(C) $C l_{2} (g)$ with $F e^{2 +} (a q)$
(E) acidified $M n O_{4}^{-}$ with $C l^{-} (a q)$
$E_{cell}^{\circ}$ : Zn/Fe²⁺ +0.32 V (yes); Ag/Cu²⁺ −0.46 V (no); Cl₂/Fe²⁺ +0.59 V (yes); I₂/Br⁻ −0.55 V (no); MnO₄⁻/Cl⁻ +0.15 V (yes, which is how chlorine is made in the laboratory).
2.18
Short answerMediumNCERT Exercises 2.18
What forms at each electrode in these electrolyses?
1. aqueous $N i S O_{4}$ with nickel electrodes
  Show solution
  AnswerNi deposits at the cathode; the nickel anode dissolves.
  Cathode: $N i^{2 +} + 2 e^{-} \overset{}{\to} N i$ (−0.25 V beats water at −0.83 V). Anode: nickel is oxidised far more easily than water, so $N i \overset{}{\to} N i^{2 +} + 2 e^{-}$ . Nickel is transferred from anode to cathode, as in electroplating.
2. aqueous KBr with platinum electrodes
  Show solution
  Answer $H_{2}$ at the cathode; $B r_{2}$ at the anode; KOH remains.
  Cathode: $K^{+}$ (−2.93 V) cannot compete with water, so $2 H_{2} O + 2 e^{-}$ $\overset{}{\to} H_{2} + 2 O H^{-}$ . Anode: $B r^{-}$ (1.09 V) is oxidised more easily than water (1.23 V), giving bromine. Potassium hydroxide is left in solution.
3. aqueous $N a_{2} S O_{4}$ with platinum electrodes
  Show solution
  Answer $H_{2}$ at the cathode; $O_{2}$ at the anode (water is split).
  Neither $N a^{+}$ nor $S O_{4}^{2 -}$ reacts. Water is reduced at the cathode and oxidised at the anode, giving hydrogen and oxygen in a 2 : 1 volume ratio; the salt just makes the water conduct.
4. molten $P b B r_{2}$ with graphite electrodes
  Show solution
  AnswerPb at the cathode; $B r_{2}$ at the anode.
  There is no water, so the only ions are reduced and oxidised: $P b^{2 +} + 2 e^{-} \overset{}{\to} P b$ (liquid lead collects at the bottom) and $2 B r^{-} \overset{}{\to} B r_{2} + 2 e^{-}$ (brown vapour).
Quick answerEasy
The main job of a salt bridge is to…
carry electrons between the two solutionskeep both solutions electrically neutral and complete the circuitsupply the ions that are reduced at the cathodestop the two metals from touching
Show answer
Answer (B) keep both solutions electrically neutral and complete the circuit
Ions move out of the bridge to balance the charge building up in each solution. Electrons never travel through the solutions or the bridge.
Quick answerMedium
A current of 5.0 A deposits copper from copper sulphate solution. How many minutes does it take to deposit 1.00 g? ( $M = 63.5 g m o l^{- 1}$ .)
min
Show answer
Answer 10.1 min
$1.00 / 63.5 = 0.01575$ mol Cu needs 0.0315 mol e⁻ = 3039 C; $t = 3039 / 5.0 = 608 s = 10.1$ min.
E3
Quick answerEasy
Match each device with what it does.
Tap an item on the left, then its match on the right.
Show answer
Answer
dry cell : primary cell, about 1.5 V
lead storage battery : rechargeable, about 2 V per cell
fuel cell : runs while reactants are fed in
conductivity cell : measures the resistance of a solution
Only the conductivity cell is not a source of energy: it is a measuring vessel used with AC.
Quick answerEasy
One faraday is the charge carried by one mole of .
Show answer
Answer electrons
$F = N_{A} e = 96 485 C m o l^{- 1}$ .
Quick answerMedium
In an electrolytic cell, the anode is joined to the positive terminal of the supply, and oxidation happens there.
TrueFalse
Show answer
Answer True
The supply pulls electrons out of the anode, so species there are oxidised. Oxidation is always at the anode; only the sign of the terminal changes from a galvanic cell.

$c$ / (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

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
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 on Electrochemistry

2.1 Electrochemical Cells3 questions
2.2 Galvanic Cells3 questions
2.2.1 Measurement of Electrode Potential3 questions
2.3 Nernst Equation3 questions
2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction3 questions
2.4 Conductance of Electrolytic Solutions3 questions
2.4.1 Measurement of the Conductivity of Ionic Solutions3 questions
2.4.2 Variation of Conductivity and Molar Conductivity with Concentration4 questions
2.5 Electrolytic Cells and Electrolysis3 questions
2.5.1 Products of Electrolysis3 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: The ratio $l / A$ for a conductivity cell, found from $G^{*} = R κ$ with a standard KCl solution.First seen in the chapter →
conductivityचालकताchalakta: The reciprocal of resistivity, $κ = 1 / ρ$ : the conductance of a unit cube of material, in S m⁻¹ or S cm⁻¹.First seen in the chapter →
corrosionसंक्षारणsankshaaran: The slow electrochemical oxidation of a metal surface by its surroundings, such as the rusting of iron.First seen in the chapter →

electrode potentialइलेक्ट्रोड विभवelectrode vibhav: The potential difference between an electrode and the electrolyte around it.First seen in the chapter →
electrolytic cellवैद्युत अपघटनी सेलvaidyut apghatani cell: A cell in which an external electrical supply drives a non-spontaneous reaction.First seen in the chapter →
emf of a cellविद्युत वाहक बलvidyut vahak bal: The potential difference between the electrodes of a cell when no current is drawn.First seen in the chapter →

Faraday constantफैराडे स्थिरांकFaraday sthirank: The charge on one mole of electrons, $F = 96 485$ C mol⁻¹.First seen in the chapter →
fuel cellईंधन सेलeendhan cell: A galvanic cell whose fuel and oxidant are supplied continuously, such as the hydrogen–oxygen cell.First seen in the chapter →

galvanic cellगैल्वेनी सेलgalvani cell: A cell in which a spontaneous redox reaction drives electrons round an external circuit, so chemical energy becomes electrical work.First seen in the chapter →

limiting molar conductivityसीमांत मोलर चालकताseemant molar chalakta: The molar conductivity as the concentration approaches zero, $Λ_{m}^{\circ}$ .First seen in the chapter →

molar conductivityमोलर चालकताmolar chalakta: Conductivity divided by molar concentration, $Λ_{m} = κ / c$ , in S m² mol⁻¹ or S cm² mol⁻¹.First seen in the chapter →

overpotentialअधिविभवadhivibhav: The extra voltage, beyond the value from electrode potentials, needed to make a slow electrode reaction run at a useful rate.First seen in the chapter →

salt bridgeलवण सेतुlavan setu: A tube of an inert electrolyte in jelly that joins two half-cells and keeps both solutions electrically neutral.First seen in the chapter →
standard electrode potentialमानक इलेक्ट्रोड विभवmanak electrode vibhav: The reduction potential of a half-cell with all species in their standard states, measured against the SHE; symbol $E^{\circ}$ .First seen in the chapter →
standard hydrogen electrodeमानक हाइड्रोजन इलेक्ट्रोडmanak hydrogen electrode: Platinised platinum in 1 M $H^{+}$ with hydrogen at 1 bar; its potential is defined as 0 V.First seen in the chapter →

2.1 Electrochemical Cells

2.2 Galvanic Cells

2.2.1 Measurement of Electrode Potential

2.3 Nernst Equation

2.3.1 Equilibrium Constant from Nernst Equation

2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction

2.4 Conductance of Electrolytic Solutions

2.4.1 Measurement of the Conductivity of Ionic Solutions

2.4.2 Variation of Conductivity and Molar Conductivity with Concentration

2.5 Electrolytic Cells and Electrolysis

2.5.1 Products of Electrolysis

2.6 Batteries

2.6.1 Primary Batteries

2.6.2 Secondary Batteries

2.7 Fuel Cells

2.8 Corrosion

Find your NCERT exercise question

Quick check: MCQs on Electrochemistry

You reached the end