Question

The oxidation potential of hydrogen electrode at pH = ${\text{10}}$ and ${\text{p}}{{\text{H}}_{\text{2}}}\,{\text{ = 1}}$atm is….
A. $0.51\,{\text{V}}$
B. ${\text{0}}{\text{.00}}\,{\text{V}}$
C. ${\text{ + }}\,{\text{0}}{\text{.59}}\,{\text{V}}$
D. $- \,{\text{0}}{\text{.059}}\,{\text{V}}$

Answer 1

First we will write the oxidation reaction of hydrogen to determine the number of electrons involved in oxidation. We can calculate the potential of the electrode by using Nernst equation. Nernst equation tells the relation in potential and concentration of oxidised and reduced species.

Formula used: ${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{oxidized}}} \right]}}{{\left[ {{\text{reduced}}} \right]}}$

Complete step-by-step answer:
The Nernst equation is used to determine the potential of a cell. The Nernst is represented as follows:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{reduced}}} \right]}}{{\left[ {{\text{oxidized}}} \right]}}$
${{\text{E}}_{{\text{cell}}\,}}$ is the potential of the cell.
${\text{E}}_{{\text{cell}}}^ \circ \,$is the standard reduction potential of the cell.
${\text{n}}$is the number of electrons involved in a redox reaction.
Oxidation potential of hydrogen is given it means oxidation is taking place at hydrogen electrode so, we can write the oxidation reaction for hydrogen as follows:
${{\text{H}}_2}\, \to \,{\text{2}}{{\text{H}}^ + } + \,2\,{{\text{e}}^ - }$
So, the Nernst equation for the hydrogen electrode is as follows:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^{\text{o}} - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}^2}}}{{\left[ {{\text{p}}{{\text{H}}_2}} \right]}}$
The standard reduction potential of the cell ${\text{E}}_{{\text{cell}}}^ \circ \,$ is zero for hydrogen electrode and the number of electrons involved in oxidation is $2$.
On substituting $0\,{\text{V}}$ for${\text{E}}_{{\text{cell}}}^ \circ \,$, $2$for number of electrons, $1$atm for the pressure of ${{\text{H}}_{\text{2}}}$.
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,0\,{\text{V}}\, - \dfrac{{0.0591}}{2}\log \,\dfrac{{{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}^2}}}{{\left[ 1 \right]}}$
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\, - \dfrac{{0.0591}}{2} \times \,2\log \,\left[ {{{\text{H}}^{\text{ + }}}} \right]$
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,\, - 0.0591\,\,\log \,\left[ {{{\text{H}}^{\text{ + }}}} \right]$…..$(1)$
Now, the formula of pH is as follows:
${\text{pH}}\, = \, - \log \,\left[ {{{\text{H}}^ + }} \right]$ ……$(2)$
On substituting pH from equation $(2)$ in equation $(1)$,
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,\, + \,0.0591\,\,{\text{pH}}\,$
On substituting $10$for pH in above equation,
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\, + \,\,0.0591\, \times \,10$
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\, + \,0.59\,{\text{V}}$
So, the oxidation potential of a hydrogen electrode is $+ \,0.59$V.

Therefore, option (C) $+ \,0.59$V, is correct.

Note: Standard reduction potential for hydrogen electrode is zero. The stoichiometry does not affect standard reduction potential. Standard reduction potential ${\text{E}}_{{\text{cell}}}^ \circ \,$is calculated by subtracting the reduction potential of anode from reduction potential of cathode. The negative logarithm of hydrogen ion concentration is known as pH. The reduction potential for hydrogen electrodes can be calculated in a similar manner. The formula comes for reduction potential for hydrogen electrode is ${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,\, - \,0.0591\,\,{\text{pH}}\,$.