Q. ${MnO}_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow {Mn}^{2 +} + 4H_{2}O;\ \ E{^\circ} = 1.51\ V$ ${MnO}_{4} + 4H^{+} + 2e^{-} \rightarrow {Mn}^{2 + +}\ 2H_{2}O;\ \ E{^\circ} = 1.23\ V$ $E_{{MnO}_{4}^{-}|{MnO}_{2}}^{{^\circ}}$ is
A) 1.70 V
B) 0.91 V
C) 1.37 V
D) 0.548 V
Explanation
(a)
${MnO}_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow {Mn}^{2 +} + 4H_{2}O\ \ E^{{^\circ}} = 1.51\ V$
$\mathrm{\Delta}G_{1}{^\circ} = - 5(1.51)F = - 7.55F$
${MnO}_{2} + 4H^{+} + 2e^{-} \rightarrow Mn^{2 +} + 2H_{2}O\ \ \ E^{{^\circ}} = 1.23\ V$
$\mathrm{\Delta}G_{2}{^\circ} = - 2(1.23)F = - 2.46\ V$
On substrating
${MnO}_{4}^{-} + 4H^{+} + 3e^{-} \rightarrow {MnO}_{2} + 2H_{2}O\ \mathrm{\Delta}G_{3}{^\circ} = - 5.09\ F\$
$E_{{MnO}_{4}^{-}|\ {MnO}_{2}}^{{^\circ}} = \frac{\mathrm{\Delta}G_{3}{^\circ}}{- nF} = \frac{- 5.09\ F}{- 3F} = 1.70\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.