Question

# Reduction potential of a hydrogen electrode at $pH = 0$ is near about?A. 0 VB. -0.592 VC. 0.592 VD. Data insufficient

Hint- In order to deal with this question first by taking the help of the formula of $pH$ and its relation with the concentration of ${H^ + }$ion we will find the concentration of ${H^ + }$ ion. Then we will proceed further by writing the electrode reaction and the formula of potential of the reaction or reduction potential.

Formula used- $\left[ {{H^ + }} \right] = {10^{ - pH}},{E_{cell}} = {E_0} - \dfrac{{0.0591}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}$
For hydrogen electrode,
Given that $pH = 0$
As we know that the concentration of ${H^ + }$ ion is given as:
$\left[ {{H^ + }} \right] = {10^{ - pH}}$
So, the concentration of ${H^ + }$ ion is:
$\Rightarrow \left[ {{H^ + }} \right] = {10^{ - 0}} \\ \Rightarrow \left[ {{H^ + }} \right] = 1 \\$
Electrode reaction for hydrogen is given as:
${H^ + } + {e^ - } \to \dfrac{1}{2}{H_2}$
Now, in order to find the reduction potential we know that for the given reaction reduction potential is the overall potential of the reaction.
Potential of the reaction = reduction potential =
$\Rightarrow {E_{cell}} = {E^0} - \dfrac{{0.0591}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}$
For the given reaction:
${E_0} = 0 \\ \left[ {{H_2}} \right] = 1 \\$
Let us substitute these values and also the concentration of ${H^ + }$ ion.
So, reduction potential =
$\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1}\log \left( {\dfrac{1}{1}} \right) \\ \Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1}\log 1 \\ \Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1} \times 0{\text{ }}\left[ {\because \log 1 = 0} \right] \\ \Rightarrow {E_{cell}} = 0 - 0 \\ \Rightarrow {E_{cell}} = 0V \\$
Hence, reduction potential of the hydrogen electrode is 0V

So, the correct answer is option A.