Question

The hydrogen electrode is dipped in a solution of $pH = 3$ at ${25^ \circ }C$. The reduction potential of the cell would be:
A. $0.177V$
B. $- 0.177V$
C. $0.087V$
D. $0.059V$

Answer 1

We can calculate the reduction potential using the Nernst equation. We have to calculate the reduction potential of the cell using $pH$ of the solution. We first have to calculate the concentration of hydrogen ion/hydronium ion using $pH$. The antilog of the $pH$ of the solution would give the concentration of the hydronium ion.

Complete step by step answer:
Given data contains,
The value of $pH$ of the solution is three.
The given temperature is ${25^ \circ }C$.
For the temperature at ${25^ \circ }C$ or $298K$, we can give the Nernst equation as,
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0592}}{n}{\log _{10}}Q$
Here,
${E_{cell}}$ represents the reduction potential of single electron
${E^ \circ }_{cell}$ represents the cell potential under fixed conditions
$n$ represents the number of electrons transferred in redox reaction
$Q$ represents the reaction quotient
For hydrogen electrode, we can write the electrochemical equation as,
$2{H^ + }_{\left( {aq} \right)} + 2{e^ - } \to {H_{2\left( g \right)}}$ ${E_{cell}}^ \circ = 0$
We know that $pH$ of the solution is three. So, let is now calculate the concentration of hydronium ion/hydrogen ion using the $pH$.
$\left[ {{H^ + }} \right] = {10^{ - pH}}$
Let us now substitute the value of $pH$ in the expression to calculate the concentration of hydrogen ions.
$\left[ {{H^ + }} \right] = {10^{ - pH}}$
$\left[ {{H^ + }} \right] = {10^{ - 3}}$
The concentration of hydrogen ions is ${10^{ - 3}}M$.
Let us now calculate the reduction potential with the help of Nernst equation.
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0592}}{n}{\log _{10}}Q$
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0592}}{n}\log \dfrac{{{P_{{H_2}}}}}{{{{\left[ {{H^ + }} \right]}^2}}}$
The reaction happens at pressure of one atmosphere.
The standard cell potential for the reaction is zero.
The pressure is one atmosphere.
The number of electrons involved in the redox reaction is two.
Let us now substitute the values in the expression of reduction potential.
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.0592}}{n}\log \dfrac{{{P_{{H_2}}}}}{{{{\left[ {{H^ + }} \right]}^2}}}$
Substituting the known values we get,
$\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0592}}{2}\log \dfrac{1}{{{{\left[ {{{10}^{ - 3}}} \right]}^2}}}$
On simplification we get,
$\Rightarrow {E_{cell}} = - 0.177V$

So, the correct answer is Option B.

Note: We have to remember that based on Nernst equation, the total potential of an electrochemical cell depends on the reaction quotient. An alternate way to calculate the reduction potential of the cell is,

Reduction potential of the hydrogen electrode=$- pH \times 0.059$
Reduction potential of the hydrogen electrode=$- 3 \times 0.059$
Reduction potential of the hydrogen electrode=$- 0.177V$
The reduction potential of the hydrogen electrode for the given pH is $- 0.177V$.