Percentage of aniline hydrochloride hydrolysed in its M/40 solution at 25°C is (Given: V, ) — Electrochemistry Chemistry Question

💡 Solution & Explanation

The half cell represents the hydrogen electrode dipping in the aniline hydrochloride solution: $H^+ + e^- \rightarrow 1/2 H_2$. Given $E = -0.18$ V, using Nernst equation $E = 0 - 0.06 \times pH \Rightarrow -0.18 = -0.06 \times pH \Rightarrow pH = 3$. Thus, $[H^+] = 10^{-3}$ M. Aniline hydrochloride hydrolyses as $C_6H_5NH_3^+ \rightleftharpoons C_6H_5NH_2 + H^+$. Degree of hydrolysis $h = [H^+] / C = 10^{-3} / (1/40) = 40 \times 10^{-3} = 0.04$. The percentage hydrolysis is $0.04 \times 100 = 4\%$. Therefore, correct answer is 4.