Question

The EMF of the cell:
$H{{g}_{(l)}}|H{{g}_{2}}C{{l}_{2}}_{(s)},KCl.sol.(1.0N)|Quinohydrone|pt$
is 0.210V at 298K.What is the pH of the quinhydrone solution, the potential of the normal calomel electrode is 0.279V and ${E^0}$ for the quinhydrone electrode is 0.699V , both at the same temperature. [2.303 RT/F=0.06].
(A) 3.5
(B) 7.0
(C) 1.85
(D) -3.5

Answer 1

To solve this question we should know formula to calculate pH in terms of potential electrodes. ${E^0}$ is the standard potential of a cell. Understanding the determination of pH by using a quinhydrone electrode helps you to solve this question easily.

Complete answer:
Firstly, to understand the concept of determination of pH by using a quinhydrone electrode, we should know the quinone-hydroquinone system.
${{C}_{6}}{{H}_{4}}{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\overset{{}}{\leftrightarrows}{{C}_{6}}{{H}_{6}}{{O}_{2}}$
Quinone Hydroquinone
The above mentioned reaction is a reduction reaction. Here, the potential is developed on a platinum electrode immersed in that system is given by Nernst equation as:
${E_{el}} = E_{el}^0 - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{[left]}}{{[right]}}$
Where, ${E_{el}}$ = Emf of the electrode
n = number electron reduced or oxidized
$E_{el}^0$ = Standard potential of electrode
For above mentioned reaction,
${E_{el}} = E_{el}^0 - \dfrac{{2.303RT}}{{2F}}\log \dfrac{{[Q{H_2}]}}{{[Q]{{[{H^ + }]}^2}}}$ …………………equation 1
We know that, $- \log [{H^ + }] = pH$
Equation 1 can also be written as:

${E_{el}} = E_{el}^0 - [ - \dfrac{{2.303RT}}{{2F}}\log \dfrac{{[Q]}}{{[Q{H_2}]}} - \dfrac{{2.303RT}}{F}\log {H^ + }]$
${E_{el}} = E_{el}^0 + \dfrac{{2.303RT}}{{2F}}\log \dfrac{{[Q]}}{{[Q{H_2}]}} + \dfrac{{2.303RT}}{F}\log {H^ + }]$ ……………equation 2
The equation looks a bit complex, so instead of taking quinone and hydroquinone, a small amount of quinhydrone which is an equimolar mixture of quinone and hydroquinone. Reason behind this is that hydroquinone is a weak acid, so its ionization is very small particularly if the pH of the solution is less than 7.
Since the concentration of quinone and hydroquinone is the same.
Therefore, $\dfrac{{[Q]}}{{[Q{H_2}]}} = 1$
So, in equation 2 [$\dfrac{{2.303RT}}{{2F}}\log \dfrac{{[Q]}}{{[Q{H_2}]}}$] reduces to 0.
Hence, From equation 2 and above discussion,
${E_{el}} = E_{el}^0 + \dfrac{{2.303RT}}{{2F}}[\log {H^ + }]$
${E_{el}} = E_{el}^0 - 0.0591(pH)$……………..equation 3
$E_{el}^0$ is the standard potential electrode of quinhydrone electrode
Now, let solve the question:
Given, E = 0.210V
$E_{el}^0$ = 0.699V (Standard potential of quinhydrone electrode)
$E_{cal}^{}$ = 0.279V (Standard potential of calomel electrode)
$E = {E_{cal}} - {E_{el}}$………………….equation 4
${E_{el}} = E_{el}^0 - 0.0591(pH)$ [from equation 3]
= 0.699 – 0.0591(pH)
0.210 = 0.279 – [0.699 – 0.0591(pH)]
pH = 1.065
This is nearest to option c.

Thus, option C is the correct answer.

Note:
In exam point of view if known equation 4 and 3 will help us to solve this question within no time so remember the equation 4 and 3. There are other methods to determine the pH like determination of pH by using hydrogen electrode and glass electrode. The main limitation of quinhydrone electrodes is that this cannot be used for solutions of pH more than 8.