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Q. The cell reaction involving quinhydrone electrode is :-
Question
What will be the electrode potential at $pH=3$

NTA AbhyasNTA Abhyas 2022

Solution:

$E=E^\circ -\frac{0 . 0592}{n}log\left[H^{+}\right]^{2}$
$=1.30-2\times \frac{0 . 0592}{2}log\left[H^{+}\right]$
$=1.30+pH\times 0.0592$
$=1.48\,V$