Question

Calculate the potential of hydrogen electrode in contact with a solution whose $pH$ is $10$

Answer 1

Concentration of $H^{+}$ ions

For hydrogen electrode, $pH=10$
We know, $pH=-{\log }_{10}\left[H^{+}\right]$

So, $\left[H^{+}\right]={10}^{-pH}M$
$\Rightarrow \left[H^{+}\right]={10}^{-10}M$

potential of hydrogen electrode

Electrode reaction will be:

$H^{+}+e^{-}\to \frac{1}{2}{H}_2$
Using Nerst equation,

$E_{cell}=E_{H^{+}/\frac{1}{2}{H}_2}^{\circ }-\frac{0.0591}{1}\log \frac{P_{H_2}^{\frac{1}{2}}}{\left[H^{+}\right]}$

Now, $E_{H^{+}/\frac{1}{2}{H}_2}^{\circ }=P_{H_2}=1bar$

Add number of electron transfer (n)=1

Hence,
$E_{cell}=0-\frac{0.0591}{1}\log \frac{(1{)}^{\frac{1}{2}}}{{10}^{-10}}$

$\Rightarrow E_{cell}=-0.0591\times 10=-0.591V$

Final answer:
The potential of hydrogen electrode is $-0.591V$.