Question

For the half-cell:

At $pH=2$, electrode potential is:

• A. $1.36\text{V}$
• B. $1.30\text{V}$
• C. $1.42\text{V}$
• D. $1.20\text{V}$

Answer 1

The correct option is C $1.42\text{V}$

This is quinhydrone electrode and here we take equimolar mixture of mixture quinone and hydroquinone.

So in this electrode, concentration of A = concentration of B

$\therefore$ From Nernst equation,
$E=E^{\circ }-\frac{0.059}{2}\text{log}[H^{+}{]}^2$
Given, $pH=2$
$\Rightarrow \left[H^{+}\right]={10}^{-2}$
Hence, $E=1.30-\frac{0.059}{2}log{10}^{-4}$
$\Rightarrow E=1.30+0.118=1.418\approx 1.42$