Question

$Pt\left(H_2\right)|H^{+}(pH=2)|H^{+}(pH=3)|Pt\left(H_2\right)$
1 atm 1 atm
The cell reaction for the given cell is:

• A. spontaneous
• B. non-spontaneous
• C. at equilibrium
• D. None of these

Answer 1

The correct option is B non-spontaneous

$E_{cell}=E_{cell}^0-\frac{0.0591}{n}{\log }_{10}\frac{[H^{+}{]}_{anode}}{[H^{+}{]}_{cathode}}$

This is a special type of cell in which we have $E^0=0$ because same type of species/ions are present in cathode and anode. It is called a concentration cell.
$=-\frac{0.0591}{1}{\log }_{10}\frac{{10}^{-2}}{{10}^{-3}}=-0.0591$
$=-\text{ve}$
Since $E_{cell}$ is negative, reaction is non-spontaneous as $△Gis+ve$