(c) The cell reaction is $H_{2}(g) + I_{2}(s) \rightleftharpoons 2H^{+}(aq) + 2I^{-}(aq)$ $0.7714 = 0.535 - \frac{0.0591}{2}\log\frac{\left\lbrack H^{+} \right\rbrack^{2}\left\lbrack I^{-} \right\rbrack^{2}}{p_{H_{2}}}$ $\therefore pH = 3$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.