Question

In acid medium, $Mn{O}_4^{\ominus }$ is an oxidizing agent.
$Mn{O}_4^{\ominus }+8H^{\oplus }+5e^{-}\to M{n}^{2+}+4H_{2}O$
If $H^{\oplus }$ ion concentration is doubled, electrode potential of the half cell $Mn{O}_4^{\ominus },M{n}^{2+}|Pt$ will :

• A. Increase by $28.36mV$
• B. Decrease by $28.36mV$
• C. Increase by $14.23mV$
• D. Decrease by $142.30mV$

Answer 1

The correct option is A Increase by $28.36mV$

Reactions:
$Mn{O}_4^{\ominus }+8H^{\oplus }+5e^{-}\to M{n}^{2+}+4H_{2}O$
$E_{Mn{O}_4^{\ominus }|M{n}^{2+}}$
$E_{Mn{O}_4^{\ominus }|M{n}^{2+}}=\frac{0.059}{5}log\left(\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{\ominus }\right][H^{\oplus }{]}^8}\right)$
$E_{Mn{O}_4^{\ominus }|M{n}^{2+}}=\frac{8}{5}\times 0.059pH-\frac{0.059}{5}log\left(\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{\ominus }\right]}\right)$
$\Rightarrow \left[H^{\oplus }\right]$ is doubled, i.e., pH is reduced by log 2 $equiv$ 0.3, then $E_{Mn{O}_4^{\ominus }|M{n}^{2+}}$ will be changed (increased) by $\frac{8}{5}\times 0.059\times 0.3V=28.36mV$