Question

The potential of hydrogen electrode at pH=10 is:
A) 0.59V
B) 0.00V
C) -0.59V
D) -0.059V

Answer 1

Hint: The answer is based on fact that hydrogen electrode which is also called as the standard hydrogen electrode or SHE that has the direct dependency on the activity or the concentration of hydronium ions and thereby on the pH value of the solution and the formula to calculate is given by,${{E}_{cell}}=E_{cell}^{0}-\dfrac{0.059}{1}\log \dfrac{product}{reac\tan t}$ Thus will lead you to the correct answer.

Complete step-by-step answer:In the classes of physical chemistry, we have studied about the practically used electrodes for the various reactions in the kinetics part of chemistry.
Let us now see the hydrogen potential and its variation in potential at different pH.
- Hydrogen electrode which is also called as the standard hydrogen electrode or SHE that has the direct dependency on the activity or the concentration of hydronium ions and thereby on the pH value of the solution.
- The reduction potential is the ability of element to accept electrons and get reduced whereas the oxidation potential is the ability of an element to get oxidised by the donation of electrons.
The reduction potential can be calculated using the formula,
${{E}_{cell}}=E_{cell}^{0}-\dfrac{0.059}{1}\log \dfrac{product}{reac\tan t}$
Here we have$E_{cell}^{0}$= 0
Therefore, above formula reduces to,
${{E}_{cell}}=-0.059\log \dfrac{product}{reac\tan t}$
The reaction of the hydrogen electrode is as shown below,
\[{{H}^{+}}+{{e}^{-}}\to \dfrac{1}{2}{{H}_{2}}\]
Now according to the data, pH = 10 and hence,$[{{H}^{+}}]={{10}^{-10}}M$
Now, substituting this in the above formula,
${{E}_{cell}}=-0.059\log \dfrac{[{{H}_{2}}]}{[{{H}^{+}}]}$
\[\Rightarrow {{E}_{cell}}=-0.059\log \dfrac{1}{{{10}^{-10}}}=-0.059\times 10=-0.059V\]
Therefore, the reduction potential of hydrogen electrode is – 0.059V
Thus, the correct answer is option C)
Also, since there is no mention of oxidation or reduction potential particularly, let us now calculate the oxidation potential of hydrogen electrode . The reaction will be,
\[\dfrac{1}{2}{{H}_{2}}\to {{H}^{+}}+{{e}^{-}}\]
Using the same formula and by substituting the values we have,
${{E}_{cell}}=-0.059\log \dfrac{[{{H}^{+}}]}{[{{H}_{2}}]}$
\[\Rightarrow {{E}_{cell}}=-0.059\log \dfrac{{{10}^{-10}}}{1}=-0.059\times -10=+0.59V\]
Therefore, the oxidation potential of hydrogen electrode is +0.59V
Thus, the correct answer will be option A)

Therefore, both option A) and C) are correct.

Note:Note that standard hydrogen potential electrode is the standard electrode used for the reference on all half cell potential reactions because the value of the standard electrode potential is zero which forms a basis for comparison with all the other electrode reactions at any temperatures.