Question

The extent of the oxidising power of (1 M) $Mn{O}_4^{-}/M{n}^{2+}$ (1 M) couple decreased if the $H^{+}$ concentration is decreased for 1 M to ${10}^{-4}$M at ${25}^o$C is:

• A. $E_{RP}^o-0.38V$
• B. $E_{RP}^o-0.36V$
• C. $E_{RP}^o-0.42V$
• D. None of these

Answer 1

Answer: (A)

$E_{RP}^o-0.38V$

The half cell reaction is,

$Mn{O}_4^{-}+8H^{+}+5e\to M{n}^{2+}+4H_{2}O$

$\therefore E_{Mn{O}_4^{-}/M{n}^{2+}}=E_{Mn{O}_4^{-}/M{n}^{2+}}^o+\frac{0.059}{5}log\frac{\left[Mn{O}_4^{-}\right][H^{+}{]}^8}{\left[M{n}^{2+}\right]}$

or $E_{RP}=E_{RP}^o+0.0118\times log\frac{1\times 1}{1}$

$\therefore E_{RP}=E_{RP}^o$

If $\left[H^{+}\right]={10}^{-4}M$

Then $E_{RP}=E_{RP}^o+0.0118log\frac{1\times ({10}^{-4}{)}^8}{1}$

$E_{RP}=E_{RP}^o-0.38V$