Question

A hydrogen gas electrode is made by dipping platinum wire in a solution of ${{HCl}}$ of ${{pH = 10}}$ and by passing hydrogen gas around the platinum wire at ${{1atm}}$ pressure. The oxidation potential of electrodes would be ………
A. $0.59{{V}}$
B. ${{0}}{{.118V}}$
C. ${{1}}{{.18V}}$
D. ${{0}}{{.059V}}$

Answer 1

This problem is based on the Nernst equation. Nernst equation is used in Galvanic cells which are involved in thermodynamics. The two half-cell reactions are combined to give Nernst equation. The two half-cell reactions are reduction and oxidation half-cell reactions. Oxidation potential is the reverse of reduction potential.

Complete step by step answer:
When the electron is in contact with its own ions in solution, the ability of electrodes to transfer electrons is called electrode potential.
The standard reduction potential of hydrogen electrode is zero volt under ${{1atm}}$ pressure, ${{298K}}$ temperature.
It is given that the ${{pH}}$ of ${{HCl}}$ is $10$.
And the pressure, ${{P = 1atm}}$
The oxidation half-reaction which occurs in the hydrogen electrode is given below:
${{{H}}_2} \rightleftharpoons 2{{{H}}^ + } + 2{{{e}}^ - }$
Here, hydrogen gets oxidized to ${{{H}}^ + }$ ion.
We can calculate the concentration of ${{{H}}^ + }$ ion in molar concentration from the ${{pH}}$ value.
i.e. $\left[ {{{{H}}^ + }} \right] = {10^{ - {{pH}}}} \Leftrightarrow \left[ {{{{H}}^ + }} \right] = {10^{ - 10}}{{M}}$
The oxidation potential of the electrode can be calculated by the Nernst equation which is given below:
${{{E}}_{{{ox}}}} = {{{E}}^ \circ }_{{{ox}}} - \dfrac{{0.0591}}{{{n}}}\log \dfrac{{{{\left[ {{{{H}}^ + }} \right]}^2}}}{{{{{P}}_{{{{H}}_2}}}}}$, where ${{{E}}_{{{ox}}}}$ is the oxidation potential, ${{n}}$ is the number of electrons transferred, ${{{P}}_{{{H}}{ _2}}}$ is the pressure and ${{{E}}^ \circ }_{{{ox}}}$ is the oxidation electrode potential.
We know that ${{{E}}^ \circ }_{{{ox}}}$ of hydrogen electrodes is zero. On substitution we get
${{{E}}_{{{ox}}}} = 0 - \dfrac{{0.0591}}{2}\log \dfrac{{{{\left[ {{{10}^{ - 10}}} \right]}^2}}}{1}$
On simplification, we get
${{{E}}_{{{ox}}}} = - \dfrac{{0.0591}}{2}\log \left( {{{10}^{ - 20}}} \right) \Leftrightarrow {{{E}}_{{{ox}}}} = - \dfrac{{0.0591}}{2} \times - 20 = 0.591 \times 10 = 0.591{{V}}$
Thus the oxidation potential of the hydrogen electrode is $0.59{{V}}$

Hence the correct option is A.

Additional information:
Hydrogen gas is bubbled through a solution of hydrogen ion unit at ${{298K}}$ is fixed as zero. Standard oxidation and reduction potential are determined using standard hydrogen electrodes.

Note: Reduction potential can also be calculated using the given formula:
i.e. ${{{E}}_{{{red}}}} = {{{E}}^ \circ }_{{{red}}} - \dfrac{{0.0591}}{2}\log \dfrac{{{{{a}}_{{{red}}}}}}{{{{{a}}_{{{ox}}}}}}$, where ${{{E}}_{{{red}}}}$ is the reduction potential, ${{{E}}^ \circ }_{{{red}}}$ is the standard reduction potential, ${{{a}}_{{{red}}}}$ is the concentration of the species which is reduced and ${{{a}}_{{{ox}}}}$ is the concentration of species which is oxidized.