Chapter 2 covers electrochemistry — galvanic cells, standard electrode potentials, the Nernst equation, Gibbs energy and equilibrium constants for redox reactions, conductivity and molar conductivity, Kohlrausch’s law, and Faraday’s laws of electrolysis. All 18 exercises (2.1–2.18) remain current under the 2026-27 syllabus, with no separate “Additional Exercises” section. Each numeric answer below is verified before publishing.
NCERT Exercise 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.
Ans: Based on standard electrode (reduction) potentials, the more reactive (more negative potential) metal displaces a less reactive one from its salt solution. The order, from most to least reactive: Mg > Al > Zn > Fe > Cu. So Mg displaces Al, Zn, Fe and Cu from their salt solutions; Al displaces Zn, Fe and Cu (but not Mg); and so on, with Cu displacing none of the others.
2.2 Given the standard electrode potentials, K⁺/K=−2.93V, Ag⁺/Ag=0.80V, Hg²⁺/Hg=0.79V, Mg²⁺/Mg=−2.37V, Cr³⁺/Cr=−0.74V. Arrange these metals in their increasing order of reducing power.
Ans: A more negative standard electrode potential means the metal is a stronger reducing agent (more easily oxidised). Ranking the given potentials from most positive to most negative: Ag(0.80)>Hg(0.79)>Cr(−0.74)>Mg(−2.37)>K(−2.93). Reducing power increases in the opposite direction, so the increasing order of reducing power is: Ag < Hg < Cr < Mg < K.
2.3 Depict the galvanic cell in which the reaction Zn(s)+2Ag⁺(aq)→Zn²⁺(aq)+2Ag(s) takes place. Further show: (i) Which electrode is negatively charged? (ii) The carriers of current in the cell. (iii) Individual reaction at each electrode.
Ans: The cell is represented as Zn(s)|Zn²⁺(aq)||Ag⁺(aq)|Ag(s). (i) The zinc electrode (anode) is negatively charged, since it is the site of oxidation and supplies electrons to the external circuit. (ii) Current is carried by electrons through the external wire (from the zinc anode to the silver cathode) and by ions through the salt bridge/electrolyte (completing the internal circuit). (iii) At the anode (Zn): Zn(s)→Zn²⁺(aq)+2e⁻. At the cathode (Ag): 2Ag⁺(aq)+2e⁻→2Ag(s).
2.4 Calculate the standard cell potentials of galvanic cells in which the following reactions take place, and calculate the ΔrG° and equilibrium constant of the reactions: (i) 2Cr(s)+3Cd²⁺(aq)→2Cr³⁺(aq)+3Cd(s) (ii) Fe²⁺(aq)+Ag⁺(aq)→Fe³⁺(aq)+Ag(s).
Ans: (i) Using E°(Cd²⁺/Cd)=−0.40V and E°(Cr³⁺/Cr)=−0.74V: E°cell=−0.40−(−0.74)=+0.34V. With n=6: ΔrG°=−nFE°=−6×96500×0.34≈−1.97×10⁵J, and from ΔrG°=−RTlnK, K≈3.2×10³⁴ — an extremely large K, showing the reaction goes essentially to completion. (ii) Using E°(Ag⁺/Ag)=0.80V and E°(Fe³⁺/Fe²⁺)=0.77V: E°cell=0.80−0.77=+0.03V. With n=1: ΔrG°≈−2895J, giving K≈3.2 — a much smaller equilibrium constant, consistent with the small cell potential.
2.5 Write the Nernst equation and calculate the emf of the following cells at 298K: (i) Mg(s)|Mg²⁺(0.001M)||Cu²⁺(0.0001M)|Cu(s) (ii) Fe(s)|Fe²⁺(0.001M)||H⁺(1M)|H₂(g)(1bar)|Pt(s) (iii) Sn(s)|Sn²⁺(0.050M)||H⁺(0.020M)|H₂(g)(1bar)|Pt(s) (iv) Pt(s)|Br⁻(0.010M)|Br₂(l)||H⁺(0.030M)|H₂(g)(1bar)|Pt(s).
Ans: Using E=E°cell−(0.0591/n)logQ at 298K: (i) E°cell=0.34−(−2.37)=2.71V, Q=[Mg²⁺]/[Cu²⁺]=10, E≈2.68V. (ii) E°cell=0−(−0.44)=0.44V, Q=[Fe²⁺]/[H⁺]²=0.001, E≈0.53V. (iii) E°cell=0−(−0.14)=0.14V, Q=[Sn²⁺]/[H⁺]²=125, E≈0.078V. (iv) E°cell=0−1.09=−1.09V, Q=1/([Br⁻]²[H⁺]²)≈1.11×10⁵, E≈−1.30V.
2.6 In the button cells widely used in watches and other devices, the reaction Zn(s)+Ag₂O(s)+H₂O(l)→Zn²⁺(aq)+2Ag(s)+2OH⁻(aq) takes place. Determine ΔrG° and E° for the reaction.
Ans: Using E°(Ag₂O/Ag,OH⁻)=0.344V and E°(Zn²⁺/Zn)=−0.76V: E°cell=0.344−(−0.76)=+1.104V. With n=2: ΔrG°=−2×96500×1.104≈−213.1kJ — the large negative ΔG° and positive E° confirm this reaction is strongly spontaneous, which is why it reliably powers button cells.
2.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans: Conductivity (κ) is the conductance of a solution of unit length and unit cross-sectional area — the reciprocal of resistivity. Molar conductivity (Λm) is the conductivity of a solution divided by its molar concentration, Λm=1000κ/c, representing the conducting power of all the ions produced by one mole of electrolyte. As concentration decreases, conductivity (κ) decreases (fewer ions per unit volume), but molar conductivity increases for both strong and weak electrolytes — sharply for weak electrolytes (as dissociation increases) and more gradually for strong electrolytes (as inter-ionic attractions weaken), approaching a limiting value Λ°m at infinite dilution.
2.8 The conductivity of 0.20M solution of KCl at 298K is 0.0248Scm⁻¹. Calculate its molar conductivity.
Ans: Λm=1000×κ/c=1000×0.0248/0.20=124Scm²mol⁻¹.
2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500Ω. What is the cell constant if the conductivity of 0.001M KCl solution at 298K is 0.146×10⁻³Scm⁻¹?
Ans: Cell constant=κ×R=0.146×10⁻³×1500=0.219cm⁻¹.
2.10 The conductivity of sodium chloride at 298K has been determined at different concentrations, and the results are given below. Calculate Λm for all concentrations, and find the value of Λ°m.
Ans: Using Λm=κ/c (in SI units, κ in Sm⁻¹, c in molm⁻³): at c=0.001M, Λm≈123.7×10⁻⁴Sm²mol⁻¹; at 0.010M, ≈118.5×10⁻⁴; at 0.020M, ≈115.8×10⁻⁴; at 0.050M, ≈111.1×10⁻⁴; at 0.100M, ≈106.7×10⁻⁴Sm²mol⁻¹ — molar conductivity rises steadily as concentration falls, as expected for a strong electrolyte. Extrapolating the plot of Λm against √c to c→0 (Debye-Hückel-Onsager behaviour, since NaCl is fully dissociated) gives Λ°m≈124×10⁻⁴Sm²mol⁻¹.
2.11 Conductivity of 0.00241M acetic acid is 7.896×10⁻⁵Scm⁻¹. Calculate its molar conductivity. If Λ°m for acetic acid is 390.5Scm²mol⁻¹, what is its dissociation constant?
Ans: Λm=1000×7.896×10⁻⁵/0.00241≈32.76Scm²mol⁻¹. Degree of dissociation α=Λm/Λ°m=32.76/390.5≈0.0839. Using Ka=cα²/(1−α): Ka≈1.85×10⁻⁵.
2.12 How much charge is required for the following reductions: (i) 1mol of Al³⁺ to Al? (ii) 1mol of Cu²⁺ to Cu? (iii) 1mol of MnO₄⁻ to Mn²⁺?
Ans: (i) Al³⁺+3e⁻→Al needs 3 Faradays: 3×96500=2.895×10⁵C. (ii) Cu²⁺+2e⁻→Cu needs 2 Faradays: 2×96500=1.93×10⁵C. (iii) MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O needs 5 Faradays: 5×96500=4.825×10⁵C.
2.13 How much electricity in terms of Faraday is required to produce (i) 20.0g of Ca from molten CaCl₂? (ii) 40.0g of Al from molten Al₂O₃?
Ans: (i) Moles of Ca=20.0/40.0=0.5mol; since Ca²⁺+2e⁻→Ca, required electricity=2×0.5=1 Faraday. (ii) Moles of Al=40.0/27.0≈1.4815mol; since Al³⁺+3e⁻→Al, required electricity=3×1.4815≈4.44 Faradays.
2.14 How much electricity is required in coulombs for the oxidation of (i) 1mol of H₂O to O₂? (ii) 1mol of FeO to Fe₂O₃?
Ans: (i) From 2H₂O→O₂+4H⁺+4e⁻, 1mol H₂O releases 2mol electrons: 2×96500=1.93×10⁵C. (ii) Oxidising Fe²⁺ (in FeO) to Fe³⁺ (in Fe₂O₃) releases 1 electron per Fe atom: 1×96500=9.65×10⁴C.
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?
Ans: Q=It=5×(20×60)=6000C. Moles of electrons=6000/96500≈0.0622mol. Since Ni²⁺+2e⁻→Ni, moles of Ni≈0.0311mol. Mass of Ni=0.0311×58.7≈1.82g.
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 1.45g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Ans: Moles of Ag deposited=1.45/108≈0.01343mol=moles of electrons (since Ag⁺+e⁻→Ag). Charge passed=0.01343×96500≈1295.6C. Time=1295.6/1.5≈864s (≈14.4 minutes). Since the same charge passes through all three series-connected cells: moles of Cu deposited=0.01343/2≈0.00671mol, mass≈0.00671×63.5≈0.426g; moles of Zn deposited=0.01343/2≈0.00671mol, mass≈0.00671×65.4≈0.439g.
2.17 Using the standard electrode potentials given in the standard table, 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).
Ans: Using E°(Fe³⁺/Fe²⁺)=0.77V, E°(I₂/I⁻)=0.54V, E°(Ag⁺/Ag)=0.80V, E°(Cu²⁺/Cu)=0.34V, E°(Br₂/Br⁻)=1.09V — a reaction is feasible only if E°cell(=E°reduced species−E°oxidised species)>0. (i) E°cell=0.77−0.54=+0.23V: feasible (Fe³⁺ oxidises I⁻ to I₂). (ii) E°cell=0.80−0.34=+0.46V: feasible (Ag⁺ oxidises Cu). (iii) E°cell=0.77−1.09=−0.32V: not feasible. (iv) E°cell=0.77−0.80=−0.03V: not feasible. (v) E°cell=1.09−0.77=+0.32V: feasible (Br₂ oxidises Fe²⁺ to Fe³⁺).
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.
Ans: (i) With silver electrodes, the anode dissolves (Ag→Ag⁺+e⁻) while pure silver deposits at the cathode (Ag⁺+e⁻→Ag) — the solution’s concentration stays essentially unchanged; this is the basis of electrorefining. (ii) With inert platinum electrodes, Ag⁺ is still reduced at the cathode (Ag deposits), but since Pt doesn’t dissolve, water is oxidised at the anode, releasing O₂ gas. (iii) For dilute H₂SO₄, this amounts to electrolysis of water: H₂ gas at the cathode and O₂ gas at the anode. (iv) For CuCl₂ with Pt electrodes: Cu deposits at the cathode (Cu²⁺+2e⁻→Cu) and Cl₂ gas is released at the anode (2Cl⁻→Cl₂+2e⁻).
Frequently Asked Questions
What is the difference between a galvanic cell and an electrolytic cell?
A galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy — it generates its own current. An electrolytic cell does the opposite: it uses an external electrical current to drive a non-spontaneous redox reaction, as in electrolysis and electroplating.
Why does molar conductivity increase with dilution while conductivity decreases?
Conductivity depends on the number of ions per unit volume, which falls as a solution is diluted. Molar conductivity, however, is conductivity normalised per mole of electrolyte present; on dilution, the ions experience weaker inter-ionic attractions (and, for weak electrolytes, dissociation increases), so each mole of electrolyte conducts more effectively even though the overall ion density is lower.
Class 12 Chemistry Chapter 2 – Notes and Extra Questions
Along with these NCERT Solutions, students can also use the Class 12 Chemistry Chapter 2 Extra Questions and Class 12 Chemistry Chapter 2 Revision Notes for quick revision and extra practice.
More NCERT Solutions for Class 12 Chemistry:
