Question 233 marksShort AnswerSection C

$(a) Answer the following: (3 \times 1) (i) Why is the Equilibrium Constant (Kc) related to E°cell and not to Ecell? (ii) Two metals 'A' and 'B' have standard electrode potential values of -0.24 V and +0.80 V respectively. Which of these will liberate hydrogen gas from dil. H₂SO₄? (iii) Write the cell reaction which occurs in lead storage battery when it is in charging. OR (b) What type of battery is Mercury cell? Why it is more advantageous than dry cell? Write overall reaction taking place in Mercury cell.$

Q: (a) Answer the following: (3 × 1) (i) Why is the Equilibrium Constant (Kc) related to E°cell and not to Ecell? (ii) Two metals 'A' and 'B' have standard electrode potential values of −0.24 V and +0.80 V respectively. Which of these will liberate hydrogen gas from dil. H₂SO₄? (iii) Write the cell reaction which occurs in lead storage battery when it is in charging. OR (b) What type of battery is Mercury cell? Why it is more advantageous than dry cell? Write overall reaction taking place in Mercury cell.

**Part (a)** 1. **Kc relates to E°cell, not Ecell:** at equilibrium the cell does no net work, so Ecell = 0 and Q = Kc. From the Nernst equation, E°cell = (0.059/n) log Kc. Since E°cell is a fixed constant for a given cell (standard states) and Kc is also a constant at a given temperature, Kc is tied to the constant E°cell — not to Ecell, which varies with concentration. 2. **Metal that liberates H₂:** a metal liberates H₂ from dil. H₂SO₄ only if its electrode potential is **more negative than that of hydrogen (0 V)**. Metal **A (−0.24 V)** is below hydrogen and will liberate H₂; B (+0.80 V) will not. 3. **Lead storage battery on charging (reverse of discharge):** 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq) **OR — Part (b) Mercury cell** - **Type:** it is a **primary cell** (non-rechargeable button cell). - **Advantage over dry cell:** its voltage stays **constant (~1.35 V)** throughout its life because the overall reaction does not involve any ion whose concentration changes; a dry cell's voltage falls steadily during use. - **Overall reaction:** Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

Electrochemistry

Electrochemical cells, electrode potentials and batteries

Official answer

Part (a) Kc relates to E°cell, not Ecell: at equilibrium the cell does no net work, so Ecell = 0 and Q = Kc. From the Nernst equation, E°cell = (0.059/n) log Kc. Since E°cell is a fixed constant for a given cell (standard states) and Kc is also a constant at a given temperature, Kc is tied to the constant E°cell — not to Ecell, which varies with concentration. Metal that liberates H₂: a metal liberates H₂ from dil. H₂SO₄ only if its electrode potential is more negative than that of hydrogen (0 V) . Metal A (-0.24 V) is below hydrogen and will liberate H₂; B (+0.80 V) will not. Lead storage battery on charging (reverse of discharge): 2PbSO₄(s) + 2H₂O(l) \to Pb(s) + PbO₂(s) + 2H₂SO₄(aq) OR — Part (b) Mercury cell Type: it is a primary cell (non-rechargeable button cell). Advantage over dry cell: its voltage stays constant (~1.35 V) throughout its life because the overall reaction does not involve any ion whose concentration changes; a dry cell's voltage falls steadily during use. Overall reaction: Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)

E°cell and Kc relation

Ecell = 0 at equilibrium

negative electrode potential liberates H₂

metal A liberates hydrogen

lead storage battery charging

2PbSO₄ + 2H₂O \to Pb + PbO₂ + 2H₂SO₄

mercury cell primary cell

constant voltage 1.35 V

Marking scheme

1. $1 mark: (a)(i) at equilibrium Ecell = 0, Q = Kc, so E°cell (a constant) relates to Kc while Ecell varies with concentration.$
2. $1 mark: (a)(ii) metal A (-0.24 V, more negative than H₂) liberates hydrogen.$
3. $1 mark: (a)(iii) correct charging reaction 2PbSO₄ + 2H₂O \to Pb + PbO₂ + 2H₂SO₄.$
4. $OR (b): 1 mark primary cell; 1 mark constant voltage (~1.35 V, no concentration change) vs falling dry-cell voltage; 1 mark overall reaction Zn(Hg) + HgO \to ZnO + Hg.$

Hint

$At equilibrium Ecell = 0 and Q = Kc, so only E°cell survives; a metal with negative E° (below H₂) liberates H₂; charging a lead battery reverses discharge; mercury cell is a constant-voltage primary cell.$

Quick oral answer

$At equilibrium the cell is dead so Ecell is zero and Q equals Kc, leaving E°cell = (0.059/n) log Kc; a metal like A with a negative potential lies below hydrogen and so liberates H₂ from acid, while charging a lead battery simply reverses its discharge reaction.$

Common mistakes

1. $Writing Ecell instead of E°cell in the equilibrium relation, ignoring that Ecell is zero at equilibrium.$
2. $Choosing metal B (positive E°) to liberate hydrogen — only metals below hydrogen (negative E°) can displace H₂.$
3. $Giving the discharge reaction for a lead battery that is being charged, instead of its reverse.$

Analysis and explanation

Concept — why E° carries the equilibrium information The Nernst equation Ecell = E°cell - (0.059/n) log Q shows Ecell depends on the instantaneous quotient Q, which changes as the cell runs. Only at equilibrium (Ecell = 0, Q = Kc) does the relation collapse to E°cell = (0.059/n) log Kc, linking the standard emf to the equilibrium constant. The activity/reactivity series idea Metals with negative reduction potentials are stronger reducing agents than H₂ and can displace it from acids. Here A at -0.24 V lies above hydrogen in the activity series; B at +0.80 V (like silver) is a noble, unreactive metal. Battery contrasts The lead-acid battery is a secondary cell: charging drives the discharge reaction in reverse, regenerating Pb and PbO₂ and consuming PbSO₄. The mercury cell keeps a steady voltage because its solid-solid reaction (Zn to ZnO, HgO to Hg) leaves the electrolyte composition unchanged. Exam trap Do not write Ecell in place of E°cell in the Kc relation. For charging, remember to reverse the discharge equation; do not repeat the discharge reaction.

Interesting facts

$A lead-acid battery, invented by Gaston Planté in 1859, is still the world's most widely used rechargeable battery, powering nearly every car's starter motor.$

$The mercury cell's rock-steady 1.35 V made it the standard reference for hearing aids and early cameras before environmental concerns over mercury phased it out.$

$The relation E°cell = (0.059/n) log Kc means even a modest cell emf of 0.6 V corresponds to a huge equilibrium constant of about 10²⁰ for a two-electron reaction.$

Question details

Q23 of 37 questions

Board: CBSE
Class: Class 12
Subject: Chemistry
Year: 2026
Section: Section C
Marks: 3
Type: Short Answer
Chapter: Electrochemistry
Topic: $Electrochemical cells, electrode potentials and batteries$
Bloom's level: understand
Syllabus topic: $Unit: Electrochemistry — E°cell and Kc, electrode potential and H₂ liberation, lead storage battery and mercury cell$
Author: CBSE
Reviewed by: Padhantu Academic Team

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Frequently asked questions

$Why is Kc linked to E°cell rather than Ecell?$

$As a cell discharges, Ecell falls and reaches zero at equilibrium, where the reaction quotient Q becomes Kc. Substituting Ecell = 0 and Q = Kc into the Nernst equation gives E°cell = (0.059/n) log Kc. Because E°cell and Kc are both fixed constants at a given temperature, they are directly related, whereas Ecell keeps changing with concentration.$

$Why is the mercury cell better than a dry cell?$

$The mercury cell delivers an almost constant 1.35 V throughout its life because its overall reaction (Zn to ZnO and HgO to Hg) involves only solids and does not change the electrolyte's ion concentration. In a dry cell, ion concentrations change as it discharges, so its voltage gradually drops.$