Chemistry - Electrochemistry Question with Solution | TestHub

The Nernst equation relates electrode potential to concentration. For the given half-cell reaction, a decrease in pH by one unit (increase in $[H^{+}]$ by a factor of 10) affects the electrode potential.

Using the Nernst equation:

$$E=E^0-\frac{0.05916}{n}\log \frac{[M{n}^{2+}]}{[Mn{O}_2][H^{+}{]}^4}$$

Since $[H^{+}]$ increases by 10, the term $\log [H^{+}{]}^4$ becomes $\log (10^4)=4$. Therefore, the change in potential is:

$$\Delta E=-\frac{0.05916}{2}\times (-4)=0.11832\approx 0.118V$$