Question

The potential of hydrogen electrode (${{P}_{{{H}_{2}}}}$= 1atms ; ${{C}_{H}}$= 0.1 M) at $25{}^\circ C$will be-
(A)- 0.00 V
(B)- -0.059 V
(C)- 0.118 V
(D)- 0.059 V

Answer 1

The power of donation is known as the oxidation potential while the power of acceptance is known as the reduction potential. The electrode potential can be calculated using the formula-
${{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.059}{1}\log \dfrac{\text{product}}{\text{reactant}}$
Where ${{E}_{cell}}$ = potential difference between two half cells
${{E}^{0}}_{cell}$ in this case, is taken as zero.

Complete solution:
-We know that potential word means ability. The potential can either be the oxidation potential of the hydrogen electrode or the reduction potential of the hydrogen electrode.
-The reduction potential is defined as the ability of the element to accept electrons thereby getting reduced while the oxidation potential is the ability of an element to get oxidized by donating the electrons.
-The reduction potential of an element is calculated using the formula-
${{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.059}{1}\log \dfrac{\text{product}}{\text{reactant}}$
We have ${{E}^{0}}_{cell}$ in this case is zero.
-The reaction for hydrogen electrode is given as-
$H+{{e}^{-}}\to \dfrac{1}{2}{{H}_{2}}$
$\begin{aligned} & {{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.059}{1}\log \left[ \dfrac{{{P}_{{{H}_{2}}}}}{({{H}^{+}})} \right] \\\ & {{E}_{cell}}=0-0.059\log \left[ \dfrac{1}{(0.1)} \right] \\\ & {{E}_{cell}}=0-0.059\log (10) \\\ & {{E}_{cell}}=-0.059V \\\ \end{aligned}$
This is the reduction potential of the hydrogen electrode.
-Let us now calculate the oxidation potential of hydrogen electrodes.
The reaction for oxidation potential is given as-
$\dfrac{1}{2}{{H}_{2}}\to {{H}^{+}}+{{e}^{-}}$

& {{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.059}{1}\log \dfrac{\text{product}}{\text{reactant}} \\\ & {{E}_{cell}}=0-0.059\log \left[ \dfrac{{{H}^{+}}}{{{H}_{2}}} \right] \\\ & {{E}_{cell}}=-0.059\log (-10) \\\ & {{E}_{cell}}=0.59V \\\ \end{aligned}$$ **Hence, the correct answers are option B and D, both are correct answers as one denotes the value of reduction potential and one denotes the value of oxidation potential.** **Note:** The oxidation potential of an element is equal to the reduction potential of the same element but with a negative sign. This implies that the oxidation potential and the reduction potential are equal but opposite signs. So, if we know the value of one, we can easily calculate the value of another potential.