Question

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

Answer 1

In electrochemical cell/galvanic cell:

Concentration in anode half cell $={10}^{-6}M$

${\epsilon }_{cell}=0.118V=-\frac{0.0592}{2}\log \frac{\left[H^{+}\right]anode}{\left[H^{+}\right]cathode}$

$\Rightarrow 0.18V=-\frac{0.0592}{2}\log \frac{{10}^{-6}}{\left[H^{+}\right]cathode}\Rightarrow -2\log \frac{{10}^{-6}}{\left[H^{+}\right]cathode}$

$\Rightarrow \log \frac{{10}^{-6}}{\left[H^{+}\right]cathode}={10}^{-2}$

$\left[H^{+}\right]cathode=\frac{{10}^{-6}}{{10}^{-2}}={10}^{-4}M$