Chapter 2 of 10

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CBSE Class 12 Chemistry★ Unit II · Electrochemistry, CBSE states no chapter-wise split

Electrochemistry

Galvanic and electrolytic cells, standard electrode potentials measured against the hydrogen electrode, the Nernst equation and its link to Gibbs energy and the equilibrium constant, conductivity and molar conductivity with Kohlrausch's law, Faraday's laws of electrolysis, and batteries, fuel cells and corrosion. CBSE Class 12 Chemistry Chapter 2.

⏱ 14 min readMedium difficulty✓ Free · NCERT-aligned

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Distinguish galvanic from electrolytic cells and describe how one becomes the other
2Represent a cell in standard notation and identify anode, cathode and current carriers
3Explain why electrode potentials are measured against the standard hydrogen electrode
4Apply the Nernst equation to find cell potential at any concentration
5Relate the standard cell potential to Gibbs energy and to the equilibrium constant
6Define conductivity and molar conductivity and account for their opposite behaviour on dilution
7Use Kohlrausch's law to find limiting molar conductivity, degree of dissociation and Ka
8Apply Faraday's laws to quantitative electrolysis problems
9Predict the products of electrolysis, including where overpotential overrides the potentials
10Describe primary and secondary batteries, fuel cells, and corrosion as an electrochemical process

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Why this chapter matters

Electrochemistry is where redox chemistry, thermodynamics and physical measurement meet. One voltmeter reading gives the Gibbs energy of a reaction and its equilibrium constant, and the same principles run every battery, every fuel cell and every case of rusting. It carries 18 exercises and 15 intext questions, and the numerical work is straightforward once the electron count is right.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Redox Reactions (Class 11 Ch 7): oxidation states, balancing half-equations and identifying oxidising and reducing agents are used on every page

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Thermodynamics (Class 11 Ch 5): Gibbs energy and the meaning of a spontaneous change underpin the whole link between emf and DrG

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Equilibrium (Class 11 Ch 6): the equilibrium constant, and Ka for a weak acid, are needed for the Nernst and Kohlrausch applications

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Some Basic Concepts of Chemistry (Class 11 Ch 1): the mole concept and molar mass drive every Faraday's law calculation

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Current Electricity (Class 12 Physics Ch 3): resistance, resistivity and the Wheatstone bridge are used directly in the conductance section

Electrochemistry

1. Check this before you revise anything

This was Unit 3 in the previous edition. Six chapters were removed from the Class 12 Chemistry book and the survivors renumbered, so Electrochemistry moved from 3 to 2.

The renumbering was not carried through the text. Five cross-references inside this chapter still point at Unit 3, and all five are wrong as printed:

Where	Printed as	Should be
Page 34, standard hydrogen electrode	Fig. 3.3	Fig. 2.3
Page 34, sum of the half-reactions	(3.5) and (3.6)	(2.5) and (2.6)
Page 48, the plot in Example 2.6	Fig. 3.7	Fig. 2.7
Page 50, data for Example 2.8	Table 3.4	Table 2.4
Page 60, Exercise 2.17	Table 3.1	Table 2.1

The last one matters most in an exam. Exercise 2.17 tells you to use "the standard electrode potentials given in Table 3.1", and Chapter 3 is Chemical Kinetics, which has no such table. The table you need is Table 2.1 on page 37 of this chapter.

This chapter has two separate question sets:

Intext Questions, 15 of them, in boxes on pages 36, 41, 51, 54 and 58.
Exercises, 18 of them, at the end.

Both sets are numbered 2.1, 2.2, 2.3 and so on, so an intext question and an exercise can share a label. Identify a question by its content, not its number.

The book answers almost nothing. The key on the last page covers intext 2.5, 2.6 and 2.9 only. The other twelve intext questions and all eighteen exercises have no printed answer. Every answer on this site is worked out independently, and for the three the book does give, ours agrees.

Textbook section	Topic
2.1	Electrochemical cells
2.2	Galvanic cells, electrode potential, the standard hydrogen electrode
2.3	Nernst equation, equilibrium constant, Gibbs energy
2.4	Conductance of electrolytic solutions, Kohlrausch's law
2.5	Electrolytic cells, Faraday's laws, products of electrolysis
2.6 to 2.8	Batteries, fuel cells, corrosion

2. Galvanic Cells and Electrode Potential (Textbook 2.1 to 2.2)

A galvanic cell converts the Gibbs energy of a spontaneous redox reaction into electrical work. An electrolytic cell does the reverse, using an external supply to drive a non-spontaneous reaction. The Daniell cell is the standard example of the first.

The chapter opens with a thought experiment worth remembering. Apply an opposing external voltage to a Daniell cell and raise it slowly. Below 1.1 V the cell runs normally. At exactly 1.1 V nothing flows at all. Above 1.1 V the whole thing reverses and becomes an electrolytic cell, with zinc plating out and copper dissolving.

Electrode potential is the potential difference that develops between an electrode and its electrolyte. When every species in the half-cell is at unit concentration, it is the standard electrode potential. By IUPAC convention these are always quoted as reduction potentials.

No half-cell potential can be measured on its own, since a voltmeter needs two terminals. The standard hydrogen electrode, Pt(s) | H2(g, 1 bar) | H+(aq, 1 M), is assigned exactly zero at all temperatures and everything else is measured against it.

Cell notation puts the anode on the left, a single bar for a phase boundary and a double bar for the salt bridge:

$Zn(s) ∣ Zn^{2 +} (aq) ∥ Cu^{2 +} (aq) ∣ Cu(s)$

$E_{cell}^{\circ} = E_{right}^{\circ} - E_{left}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

For the Daniell cell that is $0.34 - (- 0.76) = 1.10$ V.

Reading Table 2.1 is half of what this chapter asks. The species at the top left, fluorine, is the strongest oxidising agent; the species at the bottom right, lithium metal, is the strongest reducing agent. A metal displaces any metal below it in the table, and an oxidising agent attacks any reduced species below it.

3. The Nernst Equation, Gibbs Energy and K (Textbook 2.3)

Away from standard conditions the potential shifts with concentration. For the general reaction $a A + b B \to c C + d D$ transferring $n$ electrons:

$E_{cell} = E_{cell}^{\circ} - \frac{R T}{n F} ln Q = E_{cell}^{\circ} - \frac{0.0591}{n} lo g Q$

The constant $0.0591$ V is $2.303 R T / F$ at 298 K, so it is valid only at that temperature.

Three traps in applying it. First, solids and pure liquids never appear in $Q$ . Second, a stoichiometric coefficient becomes an exponent, so two Ag+ ions contribute $[Ag^{+}]^{2}$ . Third, $n$ is the number of electrons in the balanced overall reaction, not in one half-reaction as written in the table.

At equilibrium the cell is dead. Setting $E_{cell} = 0$ gives

$E_{cell}^{\circ} = \frac{0.0591}{n} lo g K_{c}$

The link to thermodynamics closes the loop:

$Δ_{r} G^{\circ} = - n F E_{cell}^{\circ} and Δ_{r} G^{\circ} = - R T ln K$

Note that $E_{cell}^{\circ}$ is intensive but $Δ_{r} G^{\circ}$ is extensive. Doubling every coefficient leaves the voltage unchanged and doubles both $n$ and $Δ_{r} G^{\circ}$ .

Because $lo g K$ is proportional to $n E_{cell}^{\circ}$ , small voltages hide enormous ranges of $K$ . Exercise 2.4 makes the point sharply: $0.34$ V with $n = 6$ gives $K \approx 3 \times 1 0^{34}$ , while $0.03$ V with $n = 1$ gives $K \approx 3$ .

4. Conductance of Electrolytic Solutions (Textbook 2.4)

Two quantities, and the whole section turns on the difference between them.

Conductivity $κ = 1/ ρ$ is the conductance of one unit volume of solution, held between electrodes of unit area at unit separation. Unit: S m-1 or S cm-1.

Molar conductivity $Λ_{m} = κ / c$ is the conductance of whatever volume contains one mole of electrolyte. Unit: S m2 mol-1 or S cm2 mol-1, related by $1$ S m2 mol-1 $= 1 0^{4}$ S cm2 mol-1.

With $κ$ in S cm-1 and $c$ in mol L-1 the working formula carries a factor of 1000, which converts litres to cubic centimetres:

$Λ_{m} = \frac{κ \times 1000}{c}$

Dropping that 1000 is the most common numerical error in the chapter.

Measurement uses a Wheatstone bridge fed by an a.c. source, since d.c. would electrolyse the solution. The cell constant $G^{*} = l / A$ is found by calibrating against a KCl solution of known conductivity: $G^{*} = κ R$ .

On dilution, $κ$ always falls and $Λ_{m}$ always rises. Both statements follow from the definitions. Dilution puts fewer ions in each unit volume, so $κ$ drops; but the volume holding one mole grows faster than $κ$ falls, so $Λ_{m} = κV$ climbs.

	Strong electrolyte	Weak electrolyte
Dissociation	Complete at all concentrations	Partial, rising towards 1 on dilution
Rise of $Λ_{m}$ on dilution	Gradual	Small, then very steep
Behaviour	$Λ_{m} = Λ_{m}^{\circ} - A c$	Not linear in $c$
Getting $Λ_{m}^{\circ}$	Extrapolate the straight line	Kohlrausch's law only

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity is the sum of independent ionic contributions:

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

It gives $Λ_{m}^{\circ}$ for weak electrolytes, which cannot be reached by extrapolation, and hence the degree of dissociation $α = Λ_{m} / Λ_{m}^{\circ}$ and the dissociation constant $K_{a} = c α^{2} / (1 - α)$ .

The extrapolation is a limiting law, not an exact one. Exercise 2.10 shows this in numbers. A straight line through all five NaCl points gives an intercept near 124.7 S cm2 mol-1, but Kohlrausch's law gives 126.4. Restrict the fit to the two most dilute points and the intercept climbs to 126.1. The most concentrated point is what bends the line.

5. Electrolysis and Faraday's Laws (Textbook 2.5)

Faraday's first law: the amount of chemical change at an electrode is proportional to the quantity of electricity passed.

Faraday's second law: the same quantity of electricity liberates masses of different substances in proportion to their chemical equivalents.

In practice everything reduces to three steps: $Q = I t$ , then $n (e^{-}) = Q / F$ with $F = 96487$ C mol-1, then divide by the electrons per formula unit.

Cells in series pass identical charge, which is what makes Exercise 2.16 solvable: the silver deposited in one cell fixes the copper and zinc in the other two.

Products of electrolysis (2.5.1) are decided by competition. At the cathode the species with the highest reduction potential wins; at the anode the species that is most easily oxidised wins. In aqueous solution H+ and OH- from water are always in the competition.

Overpotential breaks the rule, and the book relies on it twice. In brine, $E^{\circ}$ says water ( $1.23$ V) should be oxidised in preference to chloride ( $1.36$ V), yet chlorine is what comes off. The oxygen evolution reaction is kinetically slow and needs an extra voltage, so chloride wins in practice. The same reasoning gives Cl2 rather than O2 in Exercise 2.18(iv).

A reactive anode changes everything. With silver electrodes in AgNO3 the anode simply dissolves, so nothing in solution is oxidised at all. That is electroplating, and it is why Exercise 2.18 asks parts (i) and (ii) as a matched pair.

6. Batteries, Fuel Cells and Corrosion (Textbook 2.6 to 2.8)

Primary cells cannot be recharged. The dry (Leclanche) cell has a zinc container as anode and a graphite rod in MnO2 as cathode, giving about 1.5 V. The mercury cell uses a zinc amalgam anode and HgO cathode, giving about 1.35 V that stays constant, because no dissolved species changes concentration during its life.

Secondary cells can be recharged. In the lead storage battery:

$Pb + PbO_{2} + 2 H_{2} SO_{4} \to 2 PbSO_{4} + 2 H_{2} O$

Charging reverses this exactly. Since sulphuric acid is consumed on discharge and regenerated on charging, the specific gravity of the electrolyte reads the state of charge directly, which is how a garage tests a car battery.

Fuel cells feed reactants in continuously. The H2-O2 cell used in the Apollo programme runs in concentrated NaOH over porous carbon electrodes with Pt or Pd catalysts, at about 70% efficiency against 40% for a thermal plant. Methane and methanol are the alternative fuels named in the chapter.

Corrosion is a galvanic cell set up on the metal itself. One spot oxidises, another reduces oxygen, the moisture film is the electrolyte and the metal is its own external circuit:

$2 Fe + O_{2} + 4 H^{+} \to 2 Fe^{2 +} + 2 H_{2} O, E_{cell}^{\circ} = 1.67 V$

The Fe2+ is then oxidised further by air to hydrated ferric oxide, Fe2O3 . xH2O, which is rust. Prevention follows from the mechanism: paint or plate the surface, or attach a sacrificial block of Mg or Zn so that the iron is forced to act as cathode.

7. Where the printed chapter needs care

Example 2.3 has a printed arithmetic slip. It computes $Δ_{r} G^{\circ} = - 2 \times 1.1 \times 96487$ and prints the result as " $- 21227$ J mol-1", then as " $- 212.27$ kJ mol-1". The product is $212271$ J, so the joule figure has lost a digit. The kilojoule value is right.

Two exercises need data the chapter does not print. Exercise 2.4(i) requires $E^{\circ}$ for Cd2+/Cd and Exercise 2.6 requires $E^{\circ}$ for the Ag2O/Ag couple in alkali. Neither appears in Table 2.1. The standard values are $- 0.40$ V and $+ 0.344$ V, and both have to be brought in from outside the book.

Table 2.1 and Exercise 2.2 disagree by 0.01 V. The table gives Mg2+/Mg as $- 2.36$ V and the exercise supplies $- 2.37$ V. Nothing turns on it, but quote whichever the question gives you.

Equation 2.29 on page 51 prints the anode reaction of copper refining as Cu(s) gives Cu2+(s) + 2e-. The copper ion goes into solution, so it is Cu2+(aq).

Section 2.5.1 labels half-cell potentials as $E_{cell}^{\circ}$ , in each of the five competing electrode reactions used to decide the products of electrolysis. These are single-electrode values read from Table 2.1, not cell potentials. Write them as $E^{\circ}$ in an answer.

Table 2.4 has no entry for HCOO-, which is why intext 2.9 supplies $λ^{\circ} (HCOO^{-}) = 54.6$ inside the question rather than expecting you to look it up.

Summary

Electrochemistry runs in both directions: a spontaneous redox reaction can be made to deliver electrical work, and electrical work can be made to drive a non-spontaneous reaction. The Daniell cell under an increasing opposing voltage shows both, and the crossover happens at exactly 1.1 V.

Everything quantitative follows from three relations. The Nernst equation carries a standard potential to any concentration. Setting the cell potential to zero at equilibrium turns it into a route to $K_{c}$ . And $Δ_{r} G^{\circ} = - n F E_{cell}^{\circ}$ ties the whole subject back to thermodynamics, so that a voltmeter reading becomes a measurement of Gibbs energy.

The conductance half of the chapter turns on one distinction: conductivity is defined per unit volume and falls on dilution, while molar conductivity is defined per mole and rises. Strong electrolytes give up $Λ_{m}^{\circ}$ by extrapolation against the square root of concentration; weak ones will not, and need Kohlrausch's law instead.

Electrolysis reduces to counting electrons, with the products decided by competing potentials and, where kinetics intervenes, by overpotential. Batteries, fuel cells and corrosion are all the same chemistry applied: two of them useful, and the third the reason bridges are painted.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Cell potential from electrode potentials

E(cell) = E(cathode) - E(anode) = E(right) - E(left)

Anode on the left in the cell diagram; a positive result means a spontaneous cell

Standard hydrogen electrode

Pt(s) | H2(g, 1 bar) | H+(aq, 1 M), assigned E = 0 at all temperatures

Reference against which every other electrode potential is measured

Nernst equation, general form

E(cell) = E(cell, standard) - (RT/nF) ln Q

Q is the reaction quotient built from the balanced overall reaction

Nernst equation at 298 K

E(cell) = E(cell, standard) - (0.0591/n) log Q

The 0.0591 V is 2.303RT/F at 298 K only

Nernst equation for a single electrode

E = E(standard) - (0.0591/n) log (1 divided by the ion concentration)

Concentration of the solid metal is taken as unity

Hydrogen electrode at any pH

E = -0.0591 x pH

Follows from the Nernst equation with pH2 = 1 bar; the basis of the glass pH electrode

Equilibrium constant from cell potential

E(cell, standard) = (0.0591/n) log Kc

Obtained by setting E(cell) = 0, the condition of a dead cell

Gibbs energy from cell potential

DrG(standard) = -nF x E(cell, standard)

E is intensive but DrG is extensive, so doubling the equation doubles DrG

Gibbs energy and equilibrium constant

DrG(standard) = -RT ln K

Gives K to three significant figures where the 0.0591 shortcut does not

Faraday constant

F = NA x charge on the electron = 96487 C per mole of electrons

Approximated as 96500 C/mol for quick work

Conductivity

kappa = 1 divided by resistivity = cell constant divided by resistance

Conductance of unit volume; falls on dilution

Cell constant

G* = l/A = kappa x R

Found by calibrating against a KCl solution of known conductivity

Molar conductivity

Lambda(m) = kappa/c

Conductance of the volume holding one mole; rises on dilution

Molar conductivity in practical units

Lambda(m) in S cm2/mol = kappa in S/cm x 1000 divided by molarity

The factor of 1000 converts litres to cubic centimetres and is the most-dropped step

Debye-Huckel-Onsager form for strong electrolytes

Lambda(m) = Lambda(m, limiting) - A x square root of c

A plot against the square root of c is linear with intercept equal to the limiting value

Kohlrausch's law of independent migration

Lambda(m, limiting) = (number of cations x lambda of cation) + (number of anions x lambda of anion)

The only route to the limiting value for a weak electrolyte

Degree of dissociation from conductivity

alpha = Lambda(m) divided by Lambda(m, limiting)

Valid for weak electrolytes, where the rise on dilution comes from increasing dissociation

Dissociation constant from conductivity

Ka = c alpha squared divided by (1 - alpha)

Ostwald's dilution law; keep the (1 - alpha) unless alpha is very small

Charge passed

Q = I x t

I in amperes and t in seconds gives Q in coulombs

Mass deposited in electrolysis

mass = (molar mass x Q) divided by (n x F)

n is the number of electrons per formula unit of the deposited species

Corrosion of iron

2Fe + O2 + 4H+ gives 2Fe2+ + 2H2O, with E(cell, standard) = 1.67 V

Anodic and cathodic spots on the same metal, with the moisture film as electrolyte

Lead storage battery discharge

Pb + PbO2 + 2H2SO4 gives 2PbSO4 + 2H2O

Charging reverses this exactly, which is why specific gravity reads the state of charge

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Forgetting to square a concentration that carries a coefficient of 2

✓ A stoichiometric coefficient becomes an exponent in Q. In intext 2.5 the term is [Ag+] squared, and treating it as [Ag+] gives 0.98 V instead of 0.91 V.

WATCH OUT

✗ Taking n from a half-reaction as printed in Table 2.1

✓ n is the number of electrons transferred in the balanced overall reaction. In Exercise 2.4(i) that is 6, not 3, because two chromium atoms are oxidised.

WATCH OUT

✗ Dropping the factor of 1000 in the molar conductivity formula

✓ With kappa in S/cm and c in mol/L, Lambda(m) = kappa x 1000/c. Without the 1000 the answer is a thousand times too small and the units do not come out as S cm2/mol.

WATCH OUT

✗ Misreading a conductivity table headed with a power of ten

✓ Exercise 2.10 tabulates 10^2 x kappa in S/m, so the first entry means kappa = 1.237 x 10^-2 S/m, which is 1.237 x 10^-4 S/cm. Two conversions, not one.

WATCH OUT

✗ Saying molar conductivity falls on dilution because conductivity does

✓ They move in opposite directions. Conductivity is per unit volume and falls; molar conductivity is per mole and rises, because the volume holding one mole grows faster than kappa falls.

WATCH OUT

✗ Extrapolating molar conductivity to zero concentration for a weak electrolyte

✓ The curve rises too steeply near infinite dilution to extrapolate. Use Kohlrausch's law of independent migration of ions instead.

WATCH OUT

✗ Predicting electrolysis products from standard potentials alone

✓ Overpotential can override them. In brine and in aqueous CuCl2, chloride is oxidised in preference to water despite water having the lower potential.

WATCH OUT

✗ Forgetting that a reactive electrode dissolves

✓ With silver electrodes in AgNO3 the anode itself is oxidised, so nothing in solution reacts there. Compare Exercise 2.18 parts (i) and (ii).

WATCH OUT

✗ Counting electrons from the wrong formula in an oxidation question

✓ Exercise 2.14(ii) asks about 1 mol of FeO, not of Fe2O3. One mole of FeO holds one mole of iron, each losing one electron, so the answer is 1F not 2F.

WATCH OUT

✗ Treating DrG as intensive like the cell potential

✓ Doubling every coefficient leaves E unchanged but doubles n and therefore doubles DrG. Always state which balanced equation your DrG belongs to.

WATCH OUT

✗ Using the 0.059 shortcut when three significant figures are wanted

✓ Intext 2.6 gives Kc = 9.62 x 10^7 through DrG = -RT ln K, while the 0.059 shortcut gives about 1 x 10^8. Go through DrG when precision matters.

WATCH OUT

✗ Confusing the two question sets, which share the same numbering

✓ Intext questions and exercises are both numbered 2.1, 2.2 and so on. Identify a question by its content, not by its number alone.

WATCH OUT

✗ Looking for Table 3.1 because Exercise 2.17 says so

✓ The chapter was Unit 3 in the previous edition and five cross-references were never updated. The table of standard electrode potentials is Table 2.1 on page 37.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Intext Questions

Measuring a standard electrode potential against the SHE, storing copper sulphate in a zinc pot, oxidising agents for ferrous ion, the hydrogen electrode at pH 10, emf from the Nernst equation, Gibbs energy and equilibrium constant, conductivity on dilution, the limiting molar conductivity of water, degree of dissociation of methanoic acid, electron count from a current, metals extracted electrolytically, charge to reduce dichromate, recharging the lead storage battery, fuel cell fuels, and rusting as an electrochemical cell

Questions

Ex s

Exercises

Displacement order and reducing power from standard potentials, depicting a galvanic cell, standard cell potentials with Gibbs energy and equilibrium constant, four Nernst equation cells including one with negative emf, the silver oxide button cell, definitions and concentration dependence of conductivity and molar conductivity, cell constant, a full conductivity table for NaCl with extrapolation to the limiting value, dissociation constant of acetic acid, four Faraday's law problems, three cells in series, feasibility from standard potentials, and products of electrolysis for four systems

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Electrochemistry?

11 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

11 questions~8 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A galvanic cell converts chemical energy to electrical work; an electrolytic cell does the reverse
•A Daniell cell under an opposing voltage above 1.1 V runs backwards as an electrolytic cell
•Standard electrode potentials are reduction potentials, by IUPAC convention
•The standard hydrogen electrode is assigned exactly zero at all temperatures
•Anode on the left; single bar for a phase boundary, double bar for the salt bridge
•E(cell) = E(cathode) - E(anode), and a positive value means a spontaneous cell
•Fluorine is the strongest oxidising agent in Table 2.1 and lithium the strongest reducing agent
•A metal displaces any metal below it in the electrochemical series
•Nernst equation at 298 K carries the factor 0.0591/n, valid only at that temperature
•Solids and pure liquids never enter the reaction quotient
•A coefficient of 2 in the equation becomes an exponent of 2 in Q
•n is the electron count of the balanced overall reaction
•The hydrogen electrode gives E = -0.0591 x pH, the basis of the pH meter
•Setting E(cell) to zero at equilibrium gives Kc from the standard cell potential
•DrG = -nFE is intensive in E but extensive in DrG
•Conductivity is per unit volume and falls on dilution
•Molar conductivity is per mole and rises on dilution
•Lambda(m) = kappa x 1000/c when kappa is in S/cm and c in mol/L
•Cell constant G* = kappa x R, found by calibrating against standard KCl
•Strong electrolytes follow Lambda(m) = Lambda(limiting) - A root c and can be extrapolated
•Weak electrolytes cannot be extrapolated and need Kohlrausch's law
•alpha = Lambda(m)/Lambda(limiting), and Ka = c alpha squared over (1 - alpha)
•F = 96487 C per mole of electrons, approximated as 96500
•Cells in series pass identical charge
•At the cathode the highest reduction potential wins; overpotential can override it at the anode
•A reactive anode dissolves instead of oxidising anything in solution
•The lead storage battery consumes H2SO4 on discharge and regenerates it on charging
•Fuel cells run at about 70% efficiency against 40% for a thermal plant
•Rusting is a galvanic cell on the metal itself, with E(cell) = 1.67 V
•Sacrificial protection makes a more reactive metal the anode so the iron cannot dissolve
•This was Unit 3 in the previous edition, and five cross-references still say so
•Two question sets share the same numbering, and the book answers only intext 2.5, 2.6 and 2.9

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit II: Electrochemistry, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Galvanic Cells and Standard Electrode Potentials, Batteries, Fuel Cells and Corrosion	2-3	1	Cell notation, electrode identification, reading Table 2.1, and the descriptive work on batteries, fuel cells and rusting
Nernst Equation, Gibbs Energy and Equilibrium Constant	3-5	1	Emf at non-standard concentrations, DrG from emf, and Kc from the standard cell potential
Conductivity and Molar Conductivity, Faraday's Laws and Electrolysis	3-5	1	Cell constant, molar conductivity, Kohlrausch's law, degree of dissociation, and quantitative electrolysis

Prep strategy

Write the balanced overall reaction before touching the Nernst equation, because n and Q both come from it
Check whether each concentration carries a coefficient that must become an exponent
Convert kappa to S/cm and c to mol/L, then remember the factor of 1000
Read the heading of any conductivity table for a power of ten before substituting
For electrolysis, go Q = It, then moles of electrons, then divide by the electrons per formula unit
Ask whether the electrode is inert or reactive before predicting any anode product
Quote Table 2.1 by its correct number, not the Table 3.1 the exercise names

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Car batteries

The lead storage battery consumes sulphuric acid on discharge and regenerates it on charging, so a hydrometer reading of the electrolyte tells a mechanic the state of charge directly.

Electroplating and metal refining

A reactive anode dissolves and redeposits as pure metal at the cathode, which is how impure copper is refined to 99.9% purity and how chrome and silver plating are done.

The chlor-alkali industry

Electrolysis of brine produces chlorine, hydrogen and sodium hydroxide simultaneously, and depends entirely on the oxygen overpotential that lets chloride beat water at the anode.

Extraction of aluminium

Aluminium is too electropositive for carbon reduction, so it is obtained by electrolysis of alumina in molten cryolite, which is why smelters are built next to power stations.

Cathodic protection of pipelines and ships

Blocks of magnesium or zinc are bolted to steel structures so that the more reactive metal corrodes instead, keeping the steel as cathode.

pH meters

The linear -0.0591 volts per pH unit predicted by the Nernst equation for the hydrogen electrode is what a glass electrode converts into a digital pH reading.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the balanced overall reaction before anything else, since it fixes both n and Q

Check every concentration for a coefficient that must become an exponent

State whether each half-cell is anode or cathode explicitly before subtracting potentials

Convert conductivity to S/cm and concentration to mol/L, then apply the factor of 1000

Read table headings for powers of ten before substituting any conductivity value

For electrolysis, always go charge, then moles of electrons, then moles of substance

Ask whether the electrode is inert or reactive before naming an anode product

Use DrG = -RT ln K rather than the 0.059 shortcut when three significant figures are wanted

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

The Debye-Huckel-Onsager equation and the physical origin of the root-c dependence for strong electrolytes

STRETCH

Transport numbers and the Hittorf method for splitting limiting molar conductivity between cation and anion

STRETCH

The Butler-Volmer equation and the kinetic origin of overpotential

STRETCH

Concentration cells with and without transference, and liquid junction potentials

STRETCH

The temperature coefficient of emf as a route to the entropy and enthalpy of a cell reaction

STRETCH

Pourbaix diagrams, which map the stable species of a metal against potential and pH

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainSolubility product from a concentration cellNernst equation with an unknown ion concentration

The emf of the cell Ag(s) | AgCl(s) | Cl-(1 M) || Ag+(1 M) | Ag(s) is 0.58 V at 298 K. Calculate the solubility product of AgCl.

Stuck? Show the approach

Both electrodes are silver, so the cell potential comes entirely from the difference in Ag+ concentration on the two sides.

Show the full solution

The left electrode is silver in contact with saturated AgCl, so its Ag+ concentration is $[Ag^{+}]_{L} = K_{s p} / [Cl^{-}] = K_{s p}$ with $[Cl^{-}] = 1$ M. The right electrode has $[Ag^{+}]_{R} = 1$ M. For a concentration cell with $n = 1$ and $E_{cell}^{\circ} = 0$ : $E = \frac{0.0591}{1} lo g \frac{[ Ag ^{+} ] _{R}}{[ Ag ^{+} ] _{L}}$ . Substituting, $0.58 = 0.0591 lo g \frac{1}{K _{s p}}$ , so $lo g (1/ K_{s p}) = 9.81$ and $K_{s p} = 1.5 \times 1 0^{- 10}$ .

Answer: Ksp of AgCl = 1.5 x 10^-10

The trap

Trying to look up a standard potential for the AgCl electrode. In a concentration cell the standard potentials cancel and only the ratio matters.

JEE MainCharge shared between two electrolytic cellsFaraday's laws with a gas volume

The same quantity of electricity that deposits 0.108 g of silver from AgNO3 is passed through acidified water. What volume of oxygen, measured at STP, is liberated at the anode?

Stuck? Show the approach

Convert the silver mass to moles of electrons, then use the oxygen half-reaction to convert electrons to moles of gas.

Show the full solution

With $M (Ag) = 108$ g mol $^{- 1}$ and one electron per silver atom, $n (e^{-}) = 0.108/108 = 1.0 \times 1 0^{- 3}$ mol. At the anode $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ , so four electrons release one molecule of oxygen: $n (O_{2}) = 1.0 \times 1 0^{- 3} /4 = 2.5 \times 1 0^{- 4}$ mol. At STP one mole occupies 22.4 L, so $V = 2.5 \times 1 0^{- 4} \times 22400 = 5.6$ mL.

Answer: 5.6 mL of oxygen at STP

The trap

Using two electrons per oxygen molecule by analogy with hydrogen. The oxygen half-reaction needs four, because each of the two oxygen atoms changes by two.

JEE AdvancedTemperature dependence of cell potentialLinking emf to enthalpy and entropy of reaction

For a certain cell reaction with n = 2, the standard cell potential is 0.60 V at 298 K and 0.64 V at 318 K. Estimate the standard entropy and enthalpy of the reaction, assuming both are constant over this range.

Stuck? Show the approach

Differentiate DrG = -nFE with respect to temperature and identify the temperature coefficient of emf with the entropy change.

Show the full solution

Since $Δ_{r} G^{\circ} = - n F E^{\circ}$ and $(\partial Δ_{r} G^{\circ} / \partial T)_{p} = - Δ_{r} S^{\circ}$ , it follows that $Δ_{r} S^{\circ} = n F (\partial E^{\circ} / \partial T)_{p}$ . The temperature coefficient is $\frac{0.64 - 0.60}{318 - 298} = 2.0 \times 1 0^{- 3}$ V K $^{- 1}$ , so $Δ_{r} S^{\circ} = 2 \times 96487 \times 2.0 \times 1 0^{- 3} = 386$ J K $^{- 1}$ mol $^{- 1}$ . At 298 K, $Δ_{r} G^{\circ} = - 2 \times 96487 \times 0.60 = - 1.158 \times 1 0^{5}$ J mol $^{- 1}$ . Then $Δ_{r} H^{\circ} = Δ_{r} G^{\circ} + T Δ_{r} S^{\circ} = - 115.8 + 298 \times 0.386 = - 0.7$ kJ mol $^{- 1}$ , so the reaction is very nearly thermoneutral and is driven almost entirely by entropy.

Answer: DrS = +386 J/K/mol and DrH is approximately -0.7 kJ/mol, so the reaction is entropy-driven

The trap

Assuming a positive emf means an exothermic reaction. Here the enthalpy change is almost zero and the entropy term supplies the whole driving force.

Where else this chapter is tested

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because a potential difference needs two points. A voltmeter has two terminals, so any measurement you make is necessarily the difference between two half-cell potentials, never one of them alone. The way round this is to fix one half-cell by convention and measure everything against it. The standard hydrogen electrode, Pt(s) | H2(g, 1 bar) | H+(aq, 1 M), is assigned a potential of exactly zero at all temperatures. Every value in Table 2.1 is therefore a cell potential measured against the SHE, and the sign tells you whether the couple is a better or worse oxidising agent than H+/H2.

n is the number of electrons transferred in the balanced overall reaction, not the number appearing in a half-reaction as printed in Table 2.1. Balance the full equation first, cancel the electrons, and count how many were cancelled. In Exercise 2.4(i) two chromium atoms each lose three electrons, so n is 6 even though the Cr3+/Cr entry in the table shows 3. Getting n wrong changes both the Nernst correction and DrG, so it is worth writing the balanced equation out every time.

They are defined over different amounts of solution. Conductivity is the conductance of one unit volume, so what matters is how many ions are packed into that volume, and dilution spreads the same ions through more solvent. Molar conductivity is the conductance of whatever volume contains one mole of electrolyte, and on dilution that volume grows faster than the conductivity falls. Since Lambda(m) = kappa x V, the product rises. For a weak electrolyte there is a second effect on top: dilution actually increases the number of ions, because the degree of dissociation rises.

Only for a strong electrolyte, and only from dilute solutions. A strong electrolyte follows Lambda(m) = Lambda(limiting) - A times the square root of c, so a plot against root c is a straight line whose intercept is the limiting value. A weak electrolyte rises so steeply near infinite dilution that no straight line exists, and Kohlrausch's law of independent migration of ions must be used instead. Even for strong electrolytes the relation is a limiting law: in Exercise 2.10 fitting all five NaCl points gives an intercept of about 124.7, but using only the two most dilute gives 126.1, against the Kohlrausch value of 126.4.

Because of overpotential. On potentials alone, water at 1.23 V should be oxidised in preference to chloride at 1.36 V, and yet electrolysis of brine gives chlorine. The reason is kinetic rather than thermodynamic: evolving oxygen on a platinum or graphite anode is a slow, multi-step process that needs an extra voltage, called the overpotential, before it proceeds at any useful rate. Chloride oxidation has no such barrier, so it wins in practice. The same effect gives chlorine rather than oxygen in Exercise 2.18(iv), and it is the basis of the entire chlor-alkali industry.

A primary cell runs its reaction once and is then dead, because the reaction cannot be reversed usefully; the dry cell and the mercury cell are the examples in the chapter. A secondary cell can be recharged by driving the reaction backwards with an external supply, as in the lead storage battery and the nickel-cadmium cell. A fuel cell is different in kind: its reactants are not stored inside it at all but fed in continuously, with products removed continuously, so it runs as long as the supply lasts. The hydrogen-oxygen cell used in the Apollo programme is the standard example.

Because NCERT numbers both sets by chapter. The intext questions appear in boxes as each topic is introduced and run 2.1 to 2.15, while the end-of-chapter exercises run 2.1 to 2.18, so every label is used twice over. When a question is quoted without saying which set it belongs to, identify it by its content instead. On this site the two sets are kept as separate groups so there is no ambiguity.

Barely. The key on the last page of the chapter answers intext 2.5, 2.6 and 2.9 only. The other twelve intext questions and all eighteen end-of-chapter exercises have no printed answer at all. Every solution on this site has been worked out independently from the chapter's own methods, and for the three the book does answer, ours agrees with it exactly.

Because Electrochemistry used to be Unit 3. Six chapters were removed from the Class 12 Chemistry book and the survivors were renumbered, but the cross-references inside this chapter were never updated. Five of them still point at Unit 3: Fig. 3.3 for the hydrogen electrode, equations (3.5) and (3.6), Fig. 3.7, Table 3.4 in Example 2.8, and Table 3.1 in Exercise 2.17. In every case subtract one from the first number. The table of standard electrode potentials you need is Table 2.1 on page 37.

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