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

A
0.19 V
B
-0.19 V
C
0.38 V
D
-0.38 V

Solution:

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

$E = E^{\circ} - \frac{R T}{5 F} \ln \frac{[{Mn}^{2 +}] {[H_{2} O]}^{4}}{[{MnO}_{4}^{-}] {[H^{+}]}^{8}}$

Hence

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

Thus

$E_{2} - E_{1} = \frac{R T}{5 F} \ln \frac{{([H^{+}] / 100)}^{8}}{{[H^{+}]}^{8}} = - \frac{R T}{5 F} \ln (100)^{8} = \frac{- 8 (0.059 V)}{5} \times 2 = - 0.189 V$

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

Solution:

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