Consider the following reaction $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{~V}$. The quantity of electricity required in Faraday to reduce five moles of $\mathrm{MnO}_{4}^{-}$is ____ (Interger answer)

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Redox Reactions and Electrochemistry

Consider the following reaction

MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{+ 2} + 4 H_{2} O, E^{\circ} = 1.51 V

.
The quantity of electricity required in Faraday to reduce five moles of

MnO_{4}^{-}

is ____ (Interger answer)

Answer: 25

From the half-reaction, one mole of

MnO_{4}^{-}

requires 5 moles of electrons (5 Faradays) for complete reduction to

Mn^{2 +}

. For five moles of

MnO_{4}^{-}

, the quantity of electricity required is:

5 moles \times 5 F/mole = 25

Faradays. The answer is 25.

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer