What is the standard electrode potential for the reduction of HCIO?
$$\begin{gathered} \mathrm{HClO}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \text{ Given : }{\mathrm{Cr}}^{2+}\left(\mathrm{aq}\right)⟶{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-},{\mathrm{E}}^{\circ }=0.41\mathrm{V} \\ \mathrm{HClO}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{Cr}}^{2+}\left(\mathrm{aq}\right)⟶2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right),{\mathrm{E}}^{\circ }=1.80 \end{gathered}$$

$$\begin{gathered} {\mathrm{Cr}}^{2+}\left(\mathrm{aq}\right)⟶{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-},{\mathrm{E}}^{\circ }=0.41\mathrm{V} ....\left(1\right) \\ \begin{array}{l}\mathrm{HClO}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{Cr}}^{2+}\left(\mathrm{aq}\right)⟶\\ 2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right),{\mathrm{E}}^{\circ }=1.8\end{array} \end{gathered}$$
[Reaction (1)]  x-2+reaction (2), we get.
$\begin{array}{l}\mathrm{HClO}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ {\mathrm{\Delta G}}^{\circ }=-2{\mathrm{FE}}^{\circ }-2\left(-1\times {\mathrm{FE}}^{\circ }\right)\\ -2{\mathrm{FE}}^{\circ }=-2\mathrm{F}(1.8)+2\mathrm{F}(0.41)\\ {\mathrm{E}}^{\circ }=+1.8-0.41=1.39\end{array}$