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

Hint: We know that the potential of the cell can be the oxidation potential or the reduction potential. The power of donation is the oxidation potential and the power of acceptance is the reduction potential.

Complete step-by-step answer:
We know that the ability of an element to accept electrons and get reduced is known as the reduction potential while the ability of an element to donate electrons and get oxidised is known as the oxidation potential.
We are given a hydrogen electrode. The reaction for hydrogen electrode is as follows:
${{\text{H}}^ + } + {{\text{e}}^ - } \to \dfrac{1}{2}{{\text{H}}_2}$
The expression for potential of the cell is as follows:
${E_{{\text{cell}}}} = E_{{\text{cell}}}^0 - \dfrac{{0.059}}{n}\log \dfrac{{{\text{[Products]}}}}{{{\text{[Reactants]}}}}$ …… (1)
Where ${E_{{\text{cell}}}}$ is the potential of the cell,
$E_{{\text{cell}}}^0$ is the standard potential of the cell,
$n$ is the number of moles of electrons involved in the reaction.
We are given a hydrogen electrode. For the hydrogen electrode, standard potential of the cell is zero. Thus, $E_{{\text{cell}}}^0 = 0$.
From the reaction, we can see that the number of moles of electrons involved is 1. Thus, $n = 1$. The product is ${{\text{H}}_2}$ and the reactant is ${{\text{H}}^ + }$.
Now we can calculate the potential of the cell using equation (1) as follows:
${E_{{\text{cell}}}} = 0 - \dfrac{{0.059}}{1}\log \dfrac{{{\text{[}}{{\text{H}}_2}{\text{]}}}}{{{\text{[}}{{\text{H}}^ + }{\text{]}}}}$
${E_{{\text{cell}}}} = \dfrac{{0.059}}{1}\log \dfrac{{{\text{[}}{{\text{H}}_2}{\text{]}}}}{{{\text{[}}{{\text{H}}^ + }{\text{]}}}}$ …… (2)
We know that the pH is the negative logarithm of the hydrogen ion $\left( {{{\text{H}}^ + }} \right)$ concentration. Thus,
${\text{pH}} = - \log {\text{ }}[{{\text{H}}^ + }]$
Thus,
${\text{[}}{{\text{H}}^ + }] = {10^{ - {\text{pH}}}}$
We are given that the hydrogen electrode is dipped in a solution of ${\text{pH}} = 3$. Thus,
${\text{[}}{{\text{H}}^ + }] = {10^{ - 3}}$
Thus, equation (2) becomes as follows:
${E_{{\text{cell}}}} = \dfrac{{0.059}}{1}\log \dfrac{1}{{{{10}^{ - 3}}}}$
${E_{{\text{cell}}}} = 0.059 \times 3$
${E_{{\text{cell}}}} = 0.177{\text{ V}}$
The potential of the cell is $0.177{\text{ V}}$.

Thus, the correct answer is option (A) $0.177{\text{ V}}$.

Note: Here we have written the reaction for reduction potential. The reaction of oxidation potential can also be written by reversing the reaction. Then we can calculate the potential of the cell which gives the same value but with an opposite sign.