CBSE Class 12 Chemistry sample paper 2025-26: question 23

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Step-by-step solution

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1. I. Here $n=2$: $E_{cell}=E_{cell}^o-\frac{2.303RT}{2F}\log \frac{[{\text{Ni}}^{2+}]}{[{\text{Cu}}^{2+}]}$. At 298 K this is $E_{cell}^o-\frac{0.059}{2}\log \frac{[{\text{Ni}}^{2+}]}{[{\text{Cu}}^{2+}]}$.

   1 mark

2. II. Balanced ionic equation: $2\text{Al(s)}+3{\text{Fe}}^{2+}\text{(aq)}\to 2{\text{Al}}^{3+}\text{(aq)}+3\text{Fe(s)}$, so $n=6$.

   $E_{cell}=E_{cell}^o-\frac{2.303RT}{6F}\log \frac{[{\text{Al}}^{3+}{]}^2}{[{\text{Fe}}^{2+}{]}^3}$

   1 mark

3. III. Cell reaction: $\text{Mg(s)}+2{\text{Ag}}^{+}\text{(aq)}\to {\text{Mg}}^{2+}\text{(aq)}+2\text{Ag(s)}$, so $n=2$.

   $E_{cell}=E_{cell}^o-\frac{2.303RT}{2F}\log \frac{[{\text{Mg}}^{2+}]}{[{\text{Ag}}^{+}{]}^2}$

   1 mark

Answer: I. E = E° − (2.303RT/2F) log([Ni²⁺]/[Cu²⁺]); II. E = E° − (2.303RT/6F) log([Al³⁺]²/[Fe²⁺]³); III. E = E° − (2.303RT/2F) log([Mg²⁺]/[Ag⁺]²)

Revise the chapter

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