Solveeit Logo
Login

Question

Question: For the half cell, at \({{pH}} = 2\), the electrode potential is: ![](https://www.vedantu.com/que...

For the half cell, at pH=2{{pH}} = 2, the electrode potential is:

A. 1.36V{{1}}{{.36V}}
B. 1.30V{{1}}{{.30V}}
C. 1.42V1.42{{V}}
D. 1.20V1.20{{V}}

Explanation

Solution

As we know, potential means ability. Electrode potential is the tendency of an electrode to get oxidized or reduced in its aqueous solution. It can be oxidation potential or reduction potential.
Also we know that pH{{pH}} is the negative logarithm of concentration of hydrogen ion. From the pH{{pH}} value, we can determine the hydrogen ion concentration.

Complete step by step answer:
In the half-cell reaction, hydroquinone is converted to quinone which is represented in the question.
It is given that the pH=2{{pH}} = 2 and the cell potential, E∘=1.30V{{{E}}^ \circ }{{ = 1}}{{.30V}}
We have to find the electrode potential.
pH=−log⁡[H+]=2{{pH}} = - \log \left[ {{{{H}}^ + }} \right] = 2
From the pH{{pH}} value, it is clear that the concentration of hydrogen ion, [H+]=10−2\left[ {{{{H}}^ + }} \right] = {10^{ - 2}}.
The reaction gives two moles of hydrogen ions. Let A be the reactant and B be the product formed. One mole of A reacts to form one mole of B and two moles of hydrogen ions.
So we can say that the concentration of A and B are equal.
i.e. [A]=[B]\left[ {{A}} \right] = \left[ {{B}} \right]
Electrode potential can be calculated using Nernst equation. It is expressed as given below:
E=E∘−0.0591Flog⁡[H+]2{{E}} = {{{E}}^ \circ } - \dfrac{{0.0591}}{{{F}}}\log {\left[ {{{{H}}^ + }} \right]^2}, E{{E}} is the electrode potential, E∘{{{E}}^ \circ } is the cell potential at standard conditions, F{{F}} is faraday which is the number of electrons transferred, [H+]\left[ {{{{H}}^ + }} \right] is the hydrogen ion concentration.
Substituting the values of cell potential and hydrogen ion concentration, we get
E=1.30−0.05912log⁡(10−2)2{{E}} = 1.30 - \dfrac{{0.0591}}{2}\log {\left( {{{10}^{ - 2}}} \right)^2}
Now on simplification, we get
E=1.30−0.05912log⁡(10−4){{E}} = 1.30 - \dfrac{{0.0591}}{2}\log \left( {{{10}^{ - 4}}} \right)
i.e. E=1.30−0.05912×−4=1.30+(0.0591×2)=1.30+0.1182=1.42V{{E}} = 1.30 - \dfrac{{0.0591}}{2} \times - 4 = 1.30 + \left( {0.0591 \times 2} \right) = 1.30 + 0.1182 = 1.42{{V}}
Thus the electrode potential is 1.42V1.42{{V}}.

Hence the correct option is C.

Note:
The product formed is called quinhydrone. Thus this electrode is called a quinhydrone electrode. It contains a solution of quinones and hydroquinones which is prepared from quinhydrone. The given reaction is a half-reaction taking place in a quinhydrone electrode.