By how much is oxidizing power of $MnO_4^-/Mn^{+2}$ couple decreases if the $H^+$ concentration is decreased from 1M to $10^{-4}$M at $25^{\large\circ}C$ .Assume other species have no change in concentration.

Question

$\begin{array}{1 1}(a)\;0.37V\\(b)\;0.35V\\(c)\;0.4V\\(d)\;0.28V\end{array}$

Answer 1

In acidic medium $MnO_4^-$ acts as oxidizing agent as per the reaction.

$MnO_4^-+8H^++5e^-\rightarrow Mn^{+2}+4H_2O$

$E_1=E^0_{MnO_4^-/Mn^{+2}}-\large\frac{0.059}{5}$$\log \large\frac{[Mn^{+2}]}{[MnO_4^-][H^+]^8}$

$\Rightarrow E^0_{MnO_4^-/Mn^{+2}}-\large\frac{0.059}{5}$$\log \large\frac{[Mn^{+2}]}{[MnO_4^-][1]^8}$

$E_2=E^0_{MnO_4^-/Mn^{+2}}-\large\frac{0.059}{5}$$\log \large\frac{[Mn^{+2}]}{[MnO_4^-][10^{-4}]^8}$

$\therefore (E_1-E_2)=\large\frac{0.059}{5}$$[\log \large\frac{[MnO_4^-]}{[Mn^{+2}]}\large\frac{[Mn^{+2}]}{[MnO_4^-][10^{-4}]^8}]$

$\Rightarrow 0.377V$

Thus oxidising power of $MnO_4^-/Mn^{+2}$ couple decreases by 0.377 from its standard value.

Hence (a) is the correct answer.