Question

A cell contains two hydrogen electrode. The negative electrode is in contact with a solution of ${10}^{-6}M$ hydrogen ions. The emf of the cell is $0.118$ volt at ${25}^{\circ }C$. Calculate the concentration of hydrogen ions at the positive electrode.

Answer 1

The cell may be represented as
$Pt|H_1\left(1atm\right)|\underset{{10}^{-6}M}{H^{+}}\parallel \underset{CM}{H^{+}}|H_2\left(1atm\right)|Pt$
$underset\underset{H_2\to 2H^{+}+2e^{-}}{(-ve)}Anodeunderset\underset{2H^{+}+2e^{-}\to H_2}{(+ve)}Cathode$
$E_{cell}=\frac{0,0591}{2}\log \frac{[H^{+}{]}_{Cathode}^2}{[{10}^{-6}{]}^2}$
$0.118=\left(0.0591\right)\log \frac{[H^{+}{]}_{Cathode}}{{10}^{-6}}$
$\log \frac{[H^{+}{]}_{Cathode}}{{10}^{-6}}=\frac{0.118}{0.0591}=2$
$\frac{[H^{+}{]}_{Cathode}}{{10}^{-6}}={10}^2$
$[H^{+}{]}_{Cathode}={10}^{-6}\times 1-0^2={10}^{-4}M$.