CBSE Class 12 Chemistry 2025: question 23(b)

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Step-by-step solution

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1. At cathode: two reductions are possible.

   • $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Na}$, with $E^{\circ }=-2.71$ V
   • ${\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2+\mathrm{O}{\mathrm{H}}^{-}$, with $E^{\circ }=-0.83$ V

   The reaction with the higher $E^{\circ }$ is preferred, so water is reduced and **${\mathrm{H}}_2$** is evolved.

   1 mark

2. At anode: two oxidations are possible.

   • $\mathrm{C}{\mathrm{l}}^{-}\to \frac{1}{2}\mathrm{C}{\mathrm{l}}_2+{\mathrm{e}}^{-}$, with $E^{\circ }=1.36$ V
   • $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$, with $E^{\circ }=1.23$ V

   On $E^{\circ }$ values alone, water should be oxidised. However, oxygen evolution has a high overpotential, so oxidation of $\mathrm{C}{\mathrm{l}}^{-}$ is preferred and **$\mathrm{C}{\mathrm{l}}_2$** is liberated.

   1 mark

3. Overall reaction: $2NaCl(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2NaOH(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})$

   1 mark

Answer: H₂O (E° = −0.83 V) is reduced in preference to Na⁺ → H₂ at cathode; Cl₂ forms at anode because of the overpotential of O₂; 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)