Question

A cell contains two hydrogen electrode. The negative electrode is in contact with a solution of ${10}^{-6}\text{M}$ hydrogen ions. the EMF of the cell is $0.118\text{V}$ at ${25}^{o}C$. If the concentration of hydrogen ions at the positive electrode is $x\times {10}^{-5}\text{M}$, then the value of $x$ is

Answer 1

Here we are considering a concentration cell as both the electrodes are made of same element.
For a concentration cell,
$E_{\text{cell}}=\frac{0.059}{n}\log \frac{C_1}{C_2}$
$\text{At anode;}{H}_2\to 2H^{+}+2e^{-}\left[H^{+}\right]={10}^{-6}\text{M}$
$\text{At cathode;}2H^{+}+2e^{-}\to H_2$
So,
$E_{\text{cell}}=\frac{0.059}{1}log\left[\frac{C_{H^{+}}}{{10}^{-6}}\right]$
$\Rightarrow 0.118=\frac{0.059}{1}log\left[\frac{C_{H^{+}}}{{10}^{-6}}\right]$
$\Rightarrow log\left[\frac{C_{H^{+}}}{{10}^{-6}}\right]=\frac{0.118}{0.059}=2$
Therefore, $\left[C_{H^{+}}\right]={10}^{-4}\text{M}$
So the vale of $x$ will be $10$