0.5 F of charge is used for the electrolysis of above solution. Ethyl alcohol is oxidised
to ethanal at anode and water is reduced at cathode.
Information $\Rightarrow . (P^{K a_{1}}), (P^{K a_{2}})$ values of $H_{2} S$ are x, y respectively
) Volume of solution is constant during electrolysis.
) $N a_{2} S, N a H S$ do not participate in electrolytic reactions
*) Gram molar volume at STP is 22.7 litres.
Which of the following is/are correct?

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0.25 mole ethanal is produced at anode due to electrolysis

Volume of $H_{2}$ produced at cathode at STP is 5.675 lit.

pH of cathode after electrolysis is y + log 3

pH of anode after electrolysis is y – log3

answer is A, B, C, D.

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Detailed Solution

$\to C_{2} H_{6} O \to C_{2} H_{4} O + 2 H^{+} + 2 e a n o d e \begin{array}{l} 2 F \equiv 2 m o l e H^{+} \\ 0.5 F \equiv x \\ x = \frac{0.5 \times 2}{2} = 0.5 m o l e H^{+} \\ S_{\underset{0.5}{1}}^{- 2} + \underset{0.5}{H^{+}} \to \underset{\underset{1.5}{1}}{H} S^{-} \\ p^{H} = p^{K_{2}} + \log \frac{0.5}{1.5} \\ p^{H} = y - \log 3 \end{array} \begin{array}{l} \to 2 H_{2} O + 2 e \to H_{2} + 2 \bar{O H} \\ 0.5 F \equiv 0.5 m o l . \bar{O H} \\ \underset{\underset{0.5}{1}}{H \bar{S} + \bar{\underset{\underline{0.5}}{O}}} H \to {\underset{\underset{1.5}{1}}{S}}^{- 2} + H_{2} O \\ p^{H} = y + \log 3 \end{array} \begin{array}{l} \to 2 F \equiv 1 m o l e e t h a n o l \\ 0.5 F \equiv x \\ x = \frac{0.5 \times 1}{2} = 0.25 m o l \\ \to \underset{0.5}{2 F} = 22.7 l i t H_{2} \\ \frac{0.5 \times 22.7}{2} = \end{array}$