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

Question

The cell reaction involving quinhydrone electrode is :- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-wbwphsf2j69ykn6l.jpg" /> What will be the electrode potential at pH=3 (A) 1.48 V (B) 1.20 V (C) 1.10 V (D) 1.30 V

Tardigrade Admin · Accepted Answer

E=E° -(0 . 0592/n)log[H+]2 =1.30-2× (0 . 0592/2)log[H+] =1.30+pH× 0.0592 =1.48 V

Q. The cell reaction involving quinhydrone electrode is :-

What will be the electrode potential at $p H = 3$

$1.48 V$

$1.20 V$

$1.10 V$

$1.30 V$

$E = E^{\circ} - \frac{0.0592}{n} l o g [H^{+}]^{2}$
$= 1.30 - 2 \times \frac{0.0592}{2} l o g [H^{+}]$
$= 1.30 + p H \times 0.0592$
$= 1.48 V$

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

1.48V

1.20V

1.10V

1.30V