Given are the reduction potentials:
$\begin{array}{ll} {MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O E_{1} \\ 2 {CO}_{2} + 2 H^{+} + 2 e^{-} \to HOOC - COOH E_{2} \end{array}$
The cell potential, having oxidation of oxalic acid with ${KMnO}_{4}$ as the cell reaction, will be

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$E_{cell} = E_{1} - E_{2}$

$E_{cell} = E_{2} - E_{1}$

$E_{cell} = 5 E_{1} - 2 E_{2}$

$E_{cell} = 5 E_{2} - 5 E_{1}$

answer is A.

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Detailed Solution

The half-cell reactions will be.

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O$ ; Reduction, right-hand electrode

$HOOC - COOH \to 2 {CO}_{2} + 2 H^{+} + 2 \bar{e}$ ; Oxidation, left-hand electrode

The cell potential, having oxidation of oxalic acid with ${KMnO}_{4}$ as the cell reaction, will be $E_{cell} = E_{R} - E_{L} = E_{1} - E_{2}$

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