Question

If

$\left[H^{+}\right]$ changed from 1M to ${10}^{-4}$M
Find change in electrode potential$E_{Mn{O}_4^{-}/M{n}^{+2}}^o,(\frac{RT}{F}=0.059)$

$\left[Assume\right[Mn{O_4}^{-}]=[M{n}^{+2}]=1M]$

• A. 0.3767 V
• B. 0.3776 V
• C. 0.0376 V
• D. 0.3771 V

Answer 1

The correct option is C 0.0376 V

$5e^{-}+8H^{+}+M\underset{1M}{n}{O}_4^{-}\to M\underset{1M}{n}+4H_{2}O$

$E_1=E^o-\frac{0.59}{5}{\log }_{10}[\frac{1}{{\left[H^{+}\right]}^8}\times \frac{\left[M{n}^{+2}\right]}{[Mn{O}_4^{-}}]$

$=E^o-\frac{0.059}{5}{\log }_{10}\left[\frac{1}{{\left(1\right)}^8}\right]=E^o$

$=E_2=E^o-\frac{0.059}{5}{\log }_{10}[\frac{1}{{\left({10}^{-4}\right)}^8}\times \frac{\left[M{n}^{+2}\right]}{\left[Mn{O}_4^{-}\right]}]$

$=E^o-\frac{0.059}{5}{\log }_{10}\left[{10}^{32}\right]$

$=E^o-\frac{0.059}{5}\times 32$

$E_1-E_2=E^o+\frac{0.059}{5}\times 32$

=0.3776V