${{\mathrm{E}}^0}_{1/2{\mathrm{Cl}}_2|{\mathrm{Cl}}^{-}}=+1.36\mathrm{V}$ .The single electrode potential of Pt, ${\mathrm{Cl}}_2$ (1atm)|${\mathrm{Cl}}^{-}$ (0.1 M) is

Platinum is an electrode at which ${\mathrm{Cl}}_2\left(1\mathrm{atm}\right)|{\mathrm{Cl}}^{- }$at concentration of ${\mathrm{Cl}}^{-}=0.1\mathrm{M}$
Here $\frac{1}{2}{\mathrm{Cl}}_2\to {\mathrm{Cl}}^{-}$
Given:  ${\mathrm{E}}_{1/2{\mathrm{Cl}}_2|{\mathrm{Cl}}^{-}}^0=+1.36 \mathrm{V}$ ,n=1
By Nernst equation at ${25}^0$C

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{Cl}}_2|{\mathrm{Cl}}^{-} }={\mathrm{E}}_{{\mathrm{Cl}}_2|{\mathrm{Cl}}^{-}}^0-\frac{0.0591}{1}{\log }_{10}\left[{\mathrm{Cl}}^{-}\right] \\ =+1.36-0.0591{\log }_{10} {10}^{-} \\ =+1.36+0.0591 \\ =+1.42\mathrm{V} \end{gathered}$$

Therefore option C is correct.