CBSE Class 12 Chemistry 2026: question 23(a)

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Step-by-step solution

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1. (i) At equilibrium the cell reaction stops, so $E_{cell}=0$. The Nernst equation then reduces to $E_{cell}^{\circ }=\frac{0.059}{n}\log K_c$ (at 298 K). Hence $K_c$ is related to $E_{cell}^{\circ }$, while $E_{cell}$ is zero.

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2. (ii) Metal A ($E^{\circ }=-0.24$ V). Its reduction potential is below that of the SHE (0.00 V), so it can reduce ${\mathrm{H}}^{+}$ to ${\mathrm{H}}_2$. Metal B (+0.80 V) cannot do this.

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3. (iii) On charging, the discharge reaction is reversed:
   $2\mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Pb}(\mathrm{s})+\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})$

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Answer: (i) At equilibrium E_cell = 0, so E°_cell = (0.059/n) log K_c; (ii) Metal A; (iii) 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq)

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
• Topic: Nernst Equation, Gibbs Energy and Equilibrium Constant