The change in cell potential of MnO4−,Mn2+,H+∣Pt when the H+ is changed to H+/100 is

# The change in cell potential of ${\mathrm{MnO}}_{4}^{-},{\mathrm{Mn}}^{2+},{\mathrm{H}}^{+}\mid \mathrm{Pt}$ when the $\left[{\mathrm{H}}^{+}\right]$ is changed to $\left[{\mathrm{H}}^{+}\right]/100$ is

1. A

0.19 V

2. B

-0.19 V

3. C

0.38 V

4. D

-0.38 V

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### Solution:

The reaction is ${\mathrm{MnO}}_{4}^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_{2}\mathrm{O}$. The cell potential is

$E={E}^{\circ }-\frac{RT}{5F}\mathrm{ln}\frac{\left[{\mathrm{Mn}}^{2+}\right]{\left[{\mathrm{H}}_{2}\mathrm{O}\right]}^{4}}{\left[{\mathrm{MnO}}_{4}^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^{8}}$

Hence

$\begin{array}{l}{E}_{1}={E}^{\circ }-\frac{RT}{5F}\mathrm{ln}\frac{\left[{\mathrm{Mn}}^{2+}\right]{\left[{\mathrm{H}}_{2}\mathrm{O}\right]}^{4}}{\left[{\mathrm{MnO}}_{4}^{-}\right]}-\frac{RT}{5F}\mathrm{ln}\frac{1}{{\left[{\mathrm{H}}^{+}\right]}^{8}}\\ {E}_{2}={E}^{\circ }-\frac{RT}{5F}\mathrm{ln}\frac{\left[{\mathrm{Mn}}^{2+}\right]{\left[{\mathrm{H}}_{2}\mathrm{O}\right]}^{4}}{\left[{\mathrm{MnO}}_{4}^{-}\right]}-\frac{RT}{5F}\mathrm{ln}\frac{1}{{\left(\left[{\mathrm{H}}^{+}\right]/100\right)}^{8}}\end{array}$

Thus

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