Given are the reduction potentials:

$\begin{array}{l}{\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}_1\\ 2{\mathrm{CO}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to \mathrm{HOOC}-\mathrm{COOH}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}_2\end{array}$

The cell potential, having oxidation of oxalic acid with ${\mathrm{KMnO}}_4$ as the cell reaction, will be 

The half-cell reactions will be.

${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$;      Reduction, right-hand electrode

$\mathrm{HOOC}-\mathrm{COOH}\to 2{\mathrm{CO}}_2+2{\mathrm{H}}^{+}+2\overline{\mathrm{e}}$;       Oxidation, left-hand electrode

The cell potential, having oxidation of oxalic acid with ${\mathrm{KMnO}}_4$ as the cell reaction, will be $E_{\text{cell }}=E_{\mathrm{R}}-E_{\mathrm{L}}=E_1-E_2$