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Question: The potential of a hydrogen electrode at pH = 10 is : a.) 0.59 V b.) 0.00 V c.) -0.59 V d.) ...

The potential of a hydrogen electrode at pH = 10 is :
a.) 0.59 V
b.) 0.00 V
c.) -0.59 V
d.) -0.059 V

Explanation

Solution

The oxidation potential is the power of donation while the reduction potential is the power of acceptance.
The electrode potential can be calculated as -
Ecell=Ecell0−0.0591log⁡productreactant{E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{1}\log \dfrac{{product}}{{reactant}}
Where Ecell{E_{cell}} is the potential difference between two half cells.
In this case, Ecell0E_{cell}^0 is taken as zero.

Complete answer:
We know that the word potential means the ability. The potential can be oxidation potential of hydrogen electrode or the reduction potential of hydrogen electrode.
The reduction potential can be defined as the ability of the element to accept electrons and get reduced while the oxidation potential is the ability of an element to get oxidized by donating the electrons.

The reduction potential can be calculated as -
We know the formula
Ecell=Ecell0−0.0591log⁡productreactan⁡t{E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{1}\log \dfrac{{product}}{{reac\tan t}}
We have, Ecell0E_{cell}^0 = 0
Thus, the formula is -
Ecell=−0.059log⁡productreactan⁡t{E_{cell}} = - 0.059\log \dfrac{{product}}{{reac\tan t}}
The reaction for hydrogen electrode can be written as -
H++e−→12H2{H^ + } + {e^ - } \to \dfrac{1}{2}{H_2}
Further, we have been given pH = 10
Thus, [H+]\left[ {{H^ + }} \right] = 10−10M{10^{ - 10}}M
Now, putting values in the formula; we have
Ecell=−0.059log⁡[H2][H+]{E_{cell}} = - 0.059\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}
Ecell=−0.059log⁡110−10{E_{cell}} = - 0.059\log \dfrac{1}{{{{10}^{ - 10}}}}
Ecell=−0.059×10{E_{cell}} = - 0.059 \times 10
Ecell{E_{cell}} = -0.59 V
Thus, this is the reduction potential of hydrogen electrodes.
So, option c.) is the correct answer.

Now, let us calculate the oxidation potential of hydrogen electrodes.
The reaction for oxidation potential may be written as -
12H2→H++e−\frac{1}{2}{H_2} \to {H^ + } + {e^ - }
So, now again using the formula; we can write
Ecell=Ecell0−0.0591log⁡productreactan⁡t{E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{1}\log \dfrac{{product}}{{reac\tan t}}
We have, Ecell0E_{cell}^0 = 0

Thus, the formula is -
Ecell=−0.059log⁡productreactan⁡t{E_{cell}} = - 0.059\log \dfrac{{product}}{{reac\tan t}}
We have pH = 10
Thus, [H+]\left[ {{H^ + }} \right] = 10−10M{10^{ - 10}}M
Now, putting values in the formula
Ecell=−0.059log⁡[H+][H2]{E_{cell}} = - 0.059\log \dfrac{{\left[ {{H^ + }} \right]}}{{\left[ {{H_2}} \right]}}
Ecell=−0.059log⁡10−101{E_{cell}} = - 0.059\log \dfrac{{{{10}^{ - 10}}}}{1}
Ecell=−0.059×(−10){E_{cell}} = - 0.059 \times ( - 10)
Ecell{E_{cell}} = 0.59 V
Thus, this is the oxidation potential of hydrogen electrodes.
So, option a.) is the correct answer.
Thus, we can overall write that the option a.) and option c.) both can be the correct answer depending on the oxidation electrode potential and reduction electrode potential.
So, the correct answer is “Option A and C”.

Note: The oxidation potential of an element is equal to reduction potential of the same element with the negative sign. This means they both are equal in values but opposite in signs. So, if we know one; we can easily calculate the other.
Further, mostly reduction electrode potential is used during calculations. So, we use (-0.591 V) as our value for electrode potential of hydrogen.