CBSE Class 12 Chemistry 2025: question 31(A)

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Step-by-step solution

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1. (a) Sn is the anode and the hydrogen electrode is the cathode. Cell reaction (n = 2):
   $\mathrm{Sn}(\mathrm{s})+2{\mathrm{H}}^{+}(\mathrm{aq})\to \mathrm{S}{\mathrm{n}}^{2+}(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

2. Nernst equation:
   $E_{cell}=(E_c^{\circ }-E_a^{\circ })-\frac{0.059}{2}\log \frac{[\mathrm{S}{\mathrm{n}}^{2+}]}{[{\mathrm{H}}^{+}{]}^2}$
   $=[0-(-0.14)]-\frac{0.059}{2}\log \frac{0.004}{(0.02{)}^2}$

   1 mark

3. $=0.14-0.0295\log 10=0.14-0.0295$
   $E_{cell}=0.1105$ V

   1 mark

4. (b)(i) Oxygen evolution at the anode has a high overpotential, so in practice $\mathrm{C}{\mathrm{l}}^{-}$ is oxidised in preference to water and $\mathrm{C}{\mathrm{l}}_2$ is liberated.

   1 mark

5. (b)(ii) Conductivity is the conductance of unit volume of solution. On dilution the number of ions carrying current per unit volume decreases, even though ionisation of $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ increases. So conductivity decreases.

   1 mark

Answer: Sn + 2H⁺ → Sn²⁺ + H₂; E_cell = 0.1105 V; (b)(i) overpotential of O₂; (ii) fewer ions per unit volume on dilution

Revise the chapter

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