Given are the reduction potentials:MnO4−+8H++5e−→Mn2++4H2O E12CO2+2H++2e−→HOOC−COOH E2The cell potential, having oxidation of oxalic acid with KMnO4 as the cell reaction, will be

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

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