NCERT Solutions for Class 12 Chemistry Chapter 2: Electrochemistry

These Class 12 Chemistry Chapter 2 solutions cover Electrochemistry from the NCERT textbook (session 2026–27). Every Intext Question and every numbered Exercise is reproduced exactly as in the book and solved fully — with each EMF, Nernst-equation, conductivity and electrolysis numerical worked out step by step and the final answer cross-checked against the official NCERT answer key, complete with correct units.

Class: 12 Subject: Chemistry Chapter: 2 Title: Electrochemistry Exercises: 18 + 15 Intext Session: 2026–27
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Electrochemistry – Chapter Overview

Electrochemistry studies the interconversion of chemical energy and electrical energy. A galvanic (voltaic) cell converts the Gibbs energy of a spontaneous redox reaction into electrical work (e.g. the Daniell cell, 1.1 V), while an electrolytic cell uses external electrical energy to drive a non-spontaneous reaction. The chapter develops the idea of electrode potential measured against the standard hydrogen electrode, the Nernst equation for the effect of concentration on cell EMF, and the links between E°cell, ΔrG° and the equilibrium constant K. It then turns to the conductance of electrolytic solutions — resistivity, conductivity (κ), molar conductivity (Λm), their variation with concentration and Kohlrausch’s law — followed by Faraday’s laws of electrolysis, batteries and fuel cells, and the electrochemistry of corrosion.

Key Concepts & Definitions

Galvanic vs electrolytic cell: a galvanic cell produces electricity from a spontaneous reaction (anode −ve, cathode +ve); an electrolytic cell consumes electricity to force a non-spontaneous reaction (anode +ve, cathode −ve).

Standard electrode potential (E°): the reduction potential of a half-cell when all species are at unit activity (1 M, 1 bar), measured against the SHE which is assigned 0.00 V.

EMF of a cell: E°cell = E°cathode − E°anode (both as reduction potentials).

Conductivity (κ): conductance of a solution kept between electrodes 1 cm (or 1 m) apart of unit cross-section; SI unit S m−1.

Molar conductivity (Λm): conductance of all the ions from one mole of electrolyte; Λm = κ/c; units S cm2 mol−1 or S m2 mol−1.

Kohlrausch’s law: at infinite dilution Λ°m equals the sum of the independent contributions of the cation and anion: Λ°m = ν+λ°+ + ν−λ°−.

Faraday constant (F): charge on one mole of electrons = 96487 C mol−1 (≈ 96500 C mol−1).

Important Formulas

Nernst equation (298 K): Ecell = E°cell − (0.059/n) log Q

Electrode potential: E(Mn+/M) = E°(Mn+/M) − (0.059/n) log (1/[Mn+])

Gibbs energy: ΔrG° = −nFE°cell

Equilibrium constant: E°cell = (0.059/n) log Kc  ⇒  ΔrG° = −RT ln K

Conductivity: κ = (1/R) × (l/A) = G* / R, where cell constant G* = l/A = κ × R

Molar conductivity: Λm = κ × 1000 / c (with κ in S cm−1, c in mol L−1)

Strong electrolyte: Λm = Λ°m − A√c

Degree of dissociation: α = Λm / Λ°m;   Ka = cα2/(1 − α)

Faraday’s law: Q = It; mass deposited = (E × Q)/F, where E is equivalent mass

Intext Questions — Solutions

2.1 How would you determine the standard electrode potential of the system Mg2+|Mg?

ANSWER Set up a galvanic cell with the standard hydrogen electrode (SHE) as the reference and the magnesium half-cell as the other electrode: Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) || Mg2+(aq, 1 M) | Mg(s). Maintain all species at unit concentration (1 M) and 298 K, and measure the cell EMF with a potentiometer. Because the SHE is assigned 0.00 V, the measured EMF (with sign) equals the standard electrode potential of the Mg2+/Mg couple. In practice Mg is more easily oxidised than H2, so Mg acts as the anode and E°(Mg2+/Mg) comes out negative (−2.37 V).

2.2 Can you store copper sulphate solutions in a zinc pot?

ANSWER No. E°(Zn2+/Zn) = −0.76 V is lower than E°(Cu2+/Cu) = +0.34 V, so zinc is a stronger reducing agent than copper. Zinc will displace copper from the solution: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). The CuSO4 would be reduced to copper metal while the zinc pot dissolves, so the solution cannot be stored in a zinc vessel.

2.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

ANSWER Oxidation of Fe2+ to Fe3+ requires a species whose reduction potential is greater than E°(Fe3+/Fe2+) = +0.77 V. Any oxidising agent above this value can do it. Three suitable substances: F2 (E° = 2.87 V), Cl2 (E° = 1.36 V) and MnO4− in acid (E° = 1.51 V). Br2 (1.09 V) and Cr2O72− (1.33 V) would work too.

2.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

ANSWER For H+(aq) + e− → ½ H2(g),   E = E° − 0.059 log (1/[H+]) = 0 + 0.059 log [H+]. Since pH = −log [H+] = 10,   E = −0.059 × pH = −0.059 × 10 = −0.59 V.

2.5 Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+(0.002 M) → Ni2+(0.160 M) + 2Ag(s). Given that E°cell = 1.05 V.

ANSWER Here n = 2 and Q = [Ni2+]/[Ag+]2 = 0.160 / (0.002)2 = 0.160 / (4 × 10−6) = 4 × 104. Ecell = E°cell − (0.059/2) log Q = 1.05 − 0.0295 × log(4 × 104) = 1.05 − 0.0295 × 4.602 = 1.05 − 0.1358 = 0.91 V.

2.6 The cell in which the following reaction occurs: 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

ANSWER n = 2 electrons. ΔrG° = −nFE°cell = −2 × 96487 C mol−1 × 0.236 V = −45541 J mol−1 = −45.54 kJ mol−1. log Kc = nE°cell/0.059 = (2 × 0.236)/0.059 = 8.0 ⇒ Kc = 9.62 × 107.

2.7 Why does the conductivity of a solution decrease with dilution?

ANSWER Conductivity (κ) is the conductance of a unit volume of solution. On dilution, the number of current-carrying ions present per unit volume decreases. Fewer ions in a given volume means less charge is transported, so the conductivity falls as the solution is diluted (for both strong and weak electrolytes).

2.8 Suggest a way to determine the Λ°m value of water.

ANSWER Water is a very weak electrolyte, so Λ°m cannot be found by extrapolation. Use Kohlrausch’s law of independent migration of ions. Λ°m(H2O) = λ°(H+) + λ°(OH−). Equivalently, Λ°m(H2O) = Λ°m(HCl) + Λ°m(NaOH) − Λ°m(NaCl), each value being measurable for the strong electrolytes.

2.9 The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given λ°(H+) = 349.6 S cm2 mol−1 and λ°(HCOO−) = 54.6 S cm2 mol−1.

ANSWER Λ°m(HCOOH) = λ°(H+) + λ°(HCOO−) = 349.6 + 54.6 = 404.2 S cm2 mol−1. α = Λm/Λ°m = 46.1 / 404.2 = 0.114. Ka = cα2/(1 − α) = 0.025 × (0.114)2 / (1 − 0.114) = 0.025 × 0.013 / 0.886 = 3.67 × 10−4 mol L−1.

2.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

ANSWER Q = It = 0.5 A × (2 × 3600 s) = 0.5 × 7200 = 3600 C. Number of electrons = Q / (charge of one electron) = 3600 / (1.602 × 10−19) = 2.25 × 1022 electrons.

2.11 Suggest a list of metals that are extracted electrolytically.

ANSWER Highly reactive metals with no suitable chemical reducing agent are obtained by electrolysis of their fused salts/oxides: sodium (Na), potassium (K), calcium (Ca), magnesium (Mg), aluminium (Al) and lithium (Li). Copper and zinc are also refined/obtained electrolytically.

2.12 Consider the reaction: Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72−?

ANSWER The balanced equation shows 6 mol of electrons are needed to reduce 1 mol of Cr2O72−. Q = 6 × F = 6 × 96487 C = 5.79 × 105 C (i.e. 6 F).

2.13 Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

ANSWER On recharging, current is passed in the reverse direction and the discharge reaction is reversed. The overall recharging reaction is: 2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 2H2SO4(aq). PbSO4 on the anode is reduced back to spongy Pb, PbSO4 on the cathode is oxidised back to PbO2, and H2SO4 is regenerated, restoring the original cell.

2.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

ANSWER Methane (CH4) and methanol (CH3OH) can be used as fuels in fuel cells. Other examples include ethanol and hydrocarbons whose combustion energy is converted directly into electricity.

2.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

ANSWER In the presence of moisture and air, one spot on the iron surface acts as the anode, where iron is oxidised: 2Fe(s) → 2Fe2+ + 4e− (E° = −0.44 V). The released electrons travel through the metal to another spot acting as the cathode. At the cathode, oxygen is reduced in the presence of H+: O2(g) + 4H+ + 4e− → 2H2O (E° = 1.23 V). The water film acts as the electrolyte, completing the circuit with E°cell = 1.67 V. Fe2+ is then further oxidised by air to hydrated ferric oxide (Fe2O3·xH2O), which appears as rust — exactly like a tiny galvanic cell operating on the surface.

NCERT Exercises — Solutions

2.1 Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg and Zn.

ANSWER A metal displaces another from solution if it has a lower (more negative) standard reduction potential. Using E° values (Mg −2.37, Al −1.66, Zn −0.76, Fe −0.44, Cu +0.34 V), the more reactive metal lies first. Order of displacing power: Mg > Al > Zn > Fe > Cu. Each metal in this list can displace the metals that follow it from their salt solutions.

2.2 Given the standard electrode potentials, K+/K = −2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V. Arrange these metals in their increasing order of reducing power.

ANSWER The more negative the standard reduction potential, the stronger the reducing power. So reducing power increases as E° becomes more negative. Increasing order of reducing power: Ag < Hg < Cr < Mg < K (Ag, E° = +0.80 V, is the weakest reducing agent; K, E° = −2.93 V, is the strongest).

2.3 Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

ANSWER Cell representation: Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s). (i) The zinc electrode (anode) is negatively charged, since oxidation releases electrons there. (ii) Current is carried by ions within the solution and salt bridge, and by electrons through the external wire. (iii) Anode (oxidation): Zn(s) → Zn2+(aq) + 2e−.   Cathode (reduction): 2Ag+(aq) + 2e− → 2Ag(s).

2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) Calculate the ΔrG° and equilibrium constant of the reactions.

ANSWER (i) Cr is oxidised (anode), Cd2+ is reduced (cathode). E°cell = E°cathode − E°anode = E°(Cd2+/Cd) − E°(Cr3+/Cr) = (−0.40) − (−0.74) = 0.34 V. n = 6 electrons. ΔrG° = −nFE° = −6 × 96487 × 0.34 = −196834 J = −196.86 kJ mol−1. log K = nE°/0.059 = (6 × 0.34)/0.059 = 34.58 ⇒ K = 3.12 × 1034. (ii) Fe2+→Fe3+ (anode, E° = 0.77 V), Ag+→Ag (cathode, E° = 0.80 V). E°cell = 0.80 − 0.77 = 0.03 V. n = 1. ΔrG° = −1 × 96487 × 0.03 = −2895 J = −2.895 kJ mol−1. log K = (1 × 0.03)/0.059 = 0.508 ⇒ K = 3.2.

2.5 Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg2+(0.001 M)||Cu2+(0.0001 M)|Cu(s) (ii) Fe(s)|Fe2+(0.001 M)||H+(1 M)|H2(g)(1 bar)|Pt(s) (iii) Sn(s)|Sn2+(0.050 M)||H+(0.020 M)|H2(g)(1 bar)|Pt(s) (iv) Pt(s)|Br−(0.010 M)|Br2(l)||H+(0.030 M)|H2(g)(1 bar)|Pt(s)

ANSWER (i) Reaction: Mg + Cu2+ → Mg2+ + Cu; E°cell = 0.34 − (−2.37) = 2.71 V, n = 2. Ecell = E°cell − (0.059/2) log ([Mg2+]/[Cu2+]) = 2.71 − 0.0295 log(0.001/0.0001) = 2.71 − 0.0295 × log 10 = 2.71 − 0.0295 = 2.68 V. (ii) Reaction: Fe + 2H+ → Fe2+ + H2; E°cell = 0 − (−0.44) = 0.44 V, n = 2. Ecell = 0.44 − (0.059/2) log ([Fe2+]/[H+]2) = 0.44 − 0.0295 log(0.001/1) = 0.44 − 0.0295 × (−3) = 0.44 + 0.0885 = 0.53 V. (iii) Reaction: Sn + 2H+ → Sn2+ + H2; E°cell = 0 − (−0.14) = 0.14 V, n = 2. Ecell = 0.14 − 0.0295 log ([Sn2+]/[H+]2) = 0.14 − 0.0295 log(0.050/(0.020)2) = 0.14 − 0.0295 log(125) = 0.14 − 0.0295 × 2.097 = 0.14 − 0.0619 = 0.08 V. (iv) Reaction: 2Br− + 2H+ → Br2 + H2; E°cell = 0 − 1.09 = −1.09 V, n = 2. Ecell = −1.09 − (0.059/2) log (1/([Br−]2[H+]2)) = −1.09 − 0.0295 log(1/((0.010)2(0.030)2)) = −1.09 − 0.0295 log(1.11 × 107) = −1.09 − 0.0295 × 7.046 = −1.09 − 0.208 = −1.298 V.

2.6 In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH−(aq). Determine ΔrG° and E° for the reaction.

ANSWER The cell uses 2 electrons (Zn → Zn2+ + 2e−). Combining the half-cell potentials gives E° = 1.56 V. ΔrG° = −nFE° = −2 × 96487 C mol−1 × 1.56 V = −301040 J mol−1 = −301.04 kJ mol−1.

2.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

ANSWER Conductivity (κ): the conductance of a solution of unit length and unit cross-sectional area (i.e. conductance of one unit volume of solution). κ = 1/ρ (inverse of resistivity); units S m−1. Molar conductivity (Λm): the conducting power of all the ions produced by dissolving one mole of electrolyte in solution; Λm = κ/c; units S m2 mol−1. Variation: On dilution, conductivity decreases (fewer ions per unit volume), but molar conductivity increases (the volume containing one mole of ions increases). For strong electrolytes Λm rises slowly and follows Λm = Λ°m − A√c; for weak electrolytes Λm rises steeply near low concentration because the degree of dissociation increases.

2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm−1. Calculate its molar conductivity.

ANSWER Λm = (κ × 1000) / c = (0.0248 S cm−1 × 1000 cm3 L−1) / (0.20 mol L−1) = 24.8 / 0.20 = 124 S cm2 mol−1.

2.9 The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10−3 S cm−1.

ANSWER Cell constant G* = κ × R = (0.146 × 10−3 S cm−1) × (1500 Ω) = 0.219 cm−1 ⇒ G* = 0.219 cm−1.

2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below. Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of Λ°m.

Concentration / M0.0010.0100.0200.0500.100
102 × κ / S m−11.23711.8523.1555.53106.74
ANSWER Here κ values are given as 102 × κ in S m−1, so κ = (value)/100 S m−1. Using Λm = κ/c with c in mol m−3 (M × 1000) gives Λm in S m2 mol−1; multiply by 104 for S cm2 mol−1. For 0.001 M: κ = 1.237×10−2 S m−1, c = 1 mol m−3 ⇒ Λm = 1.237×10−2 S m2 mol−1 = 123.7 S cm2 mol−1. Computing the rest the same way:
c / M√c / M½Λm / S cm2 mol−1
0.0010.0316123.7
0.0100.1000118.5
0.0200.1414115.8
0.0500.2236111.1
0.1000.3162106.7
ANSWER (contd.) A plot of Λm (y-axis) versus √c (x-axis) is a straight line (NaCl is a strong electrolyte). Extrapolating the line to c = 0 (intercept on the y-axis) gives the limiting molar conductivity. From the intercept, Λ°m ≈ 126.0 S cm2 mol−1 (124–126 S cm2 mol−1), in agreement with the literature value for NaCl.

2.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity. If Λ°m for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

ANSWER Λm = (κ × 1000)/c = (7.896 × 10−5 × 1000)/0.00241 = 0.07896/0.00241 = 32.76 S cm2 mol−1. α = Λm/Λ°m = 32.76/390.5 = 0.0839. Ka = cα2/(1 − α) = (0.00241 × (0.0839)2)/(1 − 0.0839) = (0.00241 × 0.00704)/0.9161 = 1.85 × 10−5.

2.12 How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al ? (ii) 1 mol of Cu2+ to Cu ? (iii) 1 mol of MnO4− to Mn2+ ?

ANSWER (i) Al3+ + 3e− → Al needs 3 mol e− = 3F = 3 × 96487 = 2.89 × 105 C. (ii) Cu2+ + 2e− → Cu needs 2 mol e− = 2F = 1.93 × 105 C. (iii) MnO4− + 8H+ + 5e− → Mn2+ + 4H2O needs 5 mol e− = 5F = 4.82 × 105 C.

2.13 How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2? (ii) 40.0 g of Al from molten Al2O3?

ANSWER (i) Ca2+ + 2e− → Ca; moles of Ca = 20.0/40 = 0.5 mol. Charge = 0.5 × 2F = 1 F. (ii) Al3+ + 3e− → Al; moles of Al = 40.0/27 = 1.481 mol. Charge = 1.481 × 3F = 4.44 F.

2.14 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2? (ii) 1 mol of FeO to Fe2O3?

ANSWER (i) 2H2O → O2 + 4H+ + 4e−; for 1 mol H2O, electrons = 4/2 = 2 mol. Charge = 2F = 2 × 96487 = 1.93 × 105 C. (ii) In FeO, Fe is +2; in Fe2O3, Fe is +3 — each Fe loses 1 electron. For 1 mol FeO, electrons = 1 mol. Charge = 1F = 96487 C.

2.15 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

ANSWER Q = It = 5 A × (20 × 60 s) = 5 × 1200 = 6000 C. Ni2+ + 2e− → Ni: 2F (= 2 × 96487 C) deposit 1 mol Ni (58.7 g). Mass of Ni = (58.7 g mol−1 × 6000 C)/(2 × 96487 C mol−1) = 352200/192974 = 1.83 g.

2.16 Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

ANSWER Ag+ + e− → Ag: moles of Ag = 1.45/108 = 0.01343 mol ⇒ charge = 0.01343 × 96487 = 1296 C. Time: t = Q/I = 1296/1.5 = 864 s = 14.40 min. Copper (Cu2+ + 2e− → Cu): moles = 0.01343/2 = 0.006715; mass = 0.006715 × 63.5 = 0.427 g. Zinc (Zn2+ + 2e− → Zn): moles = 0.006715; mass = 0.006715 × 65.4 = 0.437 g.

2.17 Using the standard electrode potentials, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I−(aq) (ii) Ag+(aq) and Cu(s) (iii) Fe3+(aq) and Br−(aq) (iv) Ag(s) and Fe3+(aq) (v) Br2(aq) and Fe2+(aq)

ANSWER A reaction is feasible if E°cell = E°cathode − E°anode is positive (the oxidising agent must have the higher reduction potential). (i) Fe3+ (0.77 V) oxidises I− (I2/I− = 0.54 V): E° = 0.77 − 0.54 = +0.23 V → feasible. (ii) Ag+ (0.80 V) oxidises Cu (Cu2+/Cu = 0.34 V): E° = 0.80 − 0.34 = +0.46 V → feasible. (iii) Fe3+ (0.77 V) vs Br− (Br2/Br− = 1.09 V): E° = 0.77 − 1.09 = −0.32 V → not feasible. (iv) Fe3+ (0.77 V) oxidises Ag (Ag+/Ag = 0.80 V): E° = 0.77 − 0.80 = −0.03 V → not feasible. (v) Br2 (1.09 V) oxidises Fe2+ (Fe3+/Fe2+ = 0.77 V): E° = 1.09 − 0.77 = +0.32 V → feasible.

2.18 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes.

ANSWER (i) AgNO3, Ag electrodes: Cathode — Ag+ + e− → Ag (silver deposited). Anode — silver dissolves: Ag → Ag+ + e−. Net effect: silver transfers from anode to cathode (electro-refining). (ii) AgNO3, Pt electrodes: Cathode — Ag deposited. Anode (inert) — water oxidised: 2H2O → O2 + 4H+ + 4e−, so O2 is evolved. (iii) dilute H2SO4, Pt electrodes: Cathode — H2 gas (2H+ + 2e− → H2). Anode — O2 gas (2H2O → O2 + 4H+ + 4e−), since dilute acid favours water oxidation. (iv) CuCl2, Pt electrodes: Cathode — Cu deposited (Cu2+ + 2e− → Cu). Anode — Cl2 evolved (2Cl− → Cl2 + 2e−) due to overpotential of oxygen.

Extra Practice Questions

Short Answer Type Questions

Q1. Why is alternating current (AC) used while measuring the resistance of an electrolytic solution?

ANSWERPassing direct current would cause electrolysis and change the composition of the solution near the electrodes (polarisation). AC reverses direction rapidly, so no net electrolysis occurs and the true resistance is measured.

Q2. State Kohlrausch’s law of independent migration of ions and give one application.

ANSWERIt states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion: Λ°m = ν+λ°+ + ν−λ°−. Application: calculating Λ°m of weak electrolytes (e.g. acetic acid) which cannot be found by extrapolation.

Q3. Why does a dry cell become dead over time even if it has not been used?

ANSWERThe acidic NH4Cl paste slowly corrodes the zinc container even when the cell is idle, so the anode is consumed and the cell becomes dead with storage.

Q4. Differentiate between primary and secondary cells with one example each.

ANSWERA primary cell cannot be recharged (the reaction is irreversible), e.g. dry/Leclanché cell. A secondary cell can be recharged by reversing the current, e.g. lead storage battery.

Q5. Why is E°cell an intensive property while ΔrG° is extensive?

ANSWERE°cell does not depend on the amount of substance (multiplying the equation does not change it), so it is intensive. ΔrG° = −nFE° depends on n, so doubling the reaction doubles ΔrG° — making it extensive.

Long Answer Type Questions

Q1. Explain the construction and working of the standard hydrogen electrode (SHE).

ANSWERThe SHE consists of a platinum electrode coated with finely divided platinum black, dipped in 1 M H+ solution. Pure hydrogen gas at 1 bar is bubbled over it, and the electrode is represented as Pt(s) | H2(g, 1 bar) | H+(aq, 1 M). The equilibrium H+(aq) + e− &rightleftharpoons; ½H2(g) is established on the platinum surface. By international convention the SHE is assigned a potential of exactly 0.00 V at all temperatures. It serves as the reference: when coupled with any other half-cell at unit concentration, the measured EMF directly gives the standard electrode potential of that half-cell, including its sign.

Q2. Derive the relation between standard cell potential, Gibbs energy and the equilibrium constant.

ANSWERThe maximum (reversible) electrical work done by a cell equals the decrease in Gibbs energy: ΔrG = −nFEcell. Under standard conditions, ΔrG° = −nFE°cell. From thermodynamics, ΔrG° = −RT ln K. Equating the two expressions: −nFE°cell = −RT ln K, giving E°cell = (RT/nF) ln K = (2.303RT/nF) log K. At 298 K this becomes E°cell = (0.059/n) log K. Thus a measured E°cell yields both the standard Gibbs energy and the equilibrium constant of the cell reaction.

Q3. Describe the lead storage battery: electrode materials, electrolyte and the reactions during discharge.

ANSWERThe lead storage battery (used in automobiles) has a lead (Pb) anode and a grid packed with lead dioxide (PbO2) as cathode, with 38% sulphuric acid as electrolyte. During discharge: Anode — Pb(s) + SO42−(aq) → PbSO4(s) + 2e−; Cathode — PbO2(s) + SO42−(aq) + 4H+(aq) + 2e− → PbSO4(s) + 2H2O(l). Overall: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l). On recharging the reaction is reversed, regenerating Pb, PbO2 and H2SO4.

MCQs & Assertion–Reason

1. In a galvanic cell, oxidation occurs at the:

(a) cathode (+ve)    (b) anode (−ve)    (c) salt bridge    (d) voltmeter

2. The standard electrode potential of the standard hydrogen electrode is:

(a) +1.00 V    (b) −0.76 V    (c) 0.00 V    (d) +0.34 V

3. The Nernst equation at 298 K is E = E° − (0.059/n) log Q. The factor 0.059 comes from:

(a) F/RT    (b) 2.303RT/F    (c) nF    (d) RT/nF only

4. On dilution of an electrolytic solution, conductivity (κ) and molar conductivity (Λm):

(a) both increase    (b) both decrease    (c) κ decreases, Λm increases    (d) κ increases, Λm decreases

5. The SI unit of molar conductivity is:

(a) S m−1    (b) S m2 mol−1    (c) Ω cm    (d) S mol

6. One Faraday is equal to:

(a) 6.022 × 1023 C    (b) 1.6 × 10−19 C    (c) 96500 C mol−1    (d) 9650 C mol−1

7. The relation between ΔrG° and E°cell is:

(a) ΔrG° = nFE°    (b) ΔrG° = −nFE°    (c) ΔrG° = −E°/nF    (d) ΔrG° = RT/nF

8. During the electrolysis of molten NaCl, the product at the cathode is:

(a) Cl2    (b) Na metal    (c) H2    (d) O2

9. Kohlrausch’s law is most useful for finding Λ°m of:

(a) strong electrolytes    (b) weak electrolytes    (c) metals    (d) non-electrolytes

10. In the rusting of iron, the anodic reaction is:

(a) O2 + 4H+ + 4e− → 2H2O    (b) Fe → Fe2+ + 2e−    (c) Fe3+ + e− → Fe2+    (d) 2H+ + 2e− → H2

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(b), 6-(c), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Copper sulphate solution cannot be stored in a zinc vessel.

Reason: Zinc is a stronger reducing agent than copper and displaces copper from its salt solution.

A-R 2. Assertion: Conductivity of an electrolyte decreases on dilution.

Reason: The number of current-carrying ions per unit volume decreases on dilution.

A-R 3. Assertion: The standard hydrogen electrode has a fixed potential of 1.00 V.

Reason: Hydrogen gas is a powerful reducing agent.

A-R 4. Assertion: Λ°m of a weak electrolyte cannot be obtained by extrapolating the Λm vs √c plot.

Reason: Λm of a weak electrolyte rises steeply near low concentration and the plot is not linear.

A-R 5. Assertion: During electrolysis of aqueous NaCl, Cl2 is liberated at the anode rather than O2.

Reason: The oxidation of water has a high overpotential, so chloride is preferentially oxidised.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes & Exam Tips

Watch out for these

  • Forgetting that E°cell = E°cathode − E°anode uses reduction potentials for both electrodes — do not flip the sign of the cathode value.
  • Using the wrong value of n in the Nernst equation or in ΔrG° = −nFE° (it is the total electrons transferred in the balanced reaction).
  • Mixing up units: use κ in S cm−1 with c in mol L−1 and the factor 1000, or convert consistently to SI.
  • Writing the reaction quotient Q upside down — products over reactants, with pure solids/liquids taken as 1.
  • Confusing conductivity (decreases on dilution) with molar conductivity (increases on dilution).
  • Stating that water oxidises in preference to Cl− in aqueous NaCl — overpotential makes Cl2 the actual anode product.

How to score full marks in this chapter

Always show the balanced cell reaction, write the Nernst expression with the correct n and Q, substitute values with units, and box the final answer. For conductivity sums, state the formula Λm = κ × 1000/c first. Learn the key E° values (SHE = 0, Cu = +0.34, Zn = −0.76, Ag = +0.80, Fe3+/Fe2+ = +0.77 V) so feasibility questions take seconds. Remember 1 F = 96500 C deposits one equivalent of substance — this single idea solves every electrolysis numerical.

Frequently Asked Questions

What is Class 12 Chemistry Chapter 2 Electrochemistry about?

It covers galvanic and electrolytic cells, electrode potential and the standard hydrogen electrode, the Nernst equation, the link between E°cell, ΔrG° and the equilibrium constant, conductivity and molar conductivity of solutions, Kohlrausch’s law, Faraday’s laws of electrolysis, batteries, fuel cells and corrosion.

How many exercises are there in Class 12 Chemistry Chapter 2?

There are 18 end-of-chapter Exercise questions and 15 Intext questions. All of them are reproduced verbatim and solved step by step on this page, with every numerical verified against the NCERT answer key.

What is the Nernst equation used for?

The Nernst equation, Ecell = E°cell − (0.059/n) log Q at 298 K, gives the EMF of a cell when the ion concentrations are not 1 M. It is used to calculate cell potential, electrode potential at any concentration, pH and equilibrium constants.

Are these Class 12 Chemistry Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026–27.

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