The standard electrode potentials (E^o) for O | vedclass

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(C) Given half-reactions:$(I) \ OCl^{-} + H_2O + 2e^{-} \longrightarrow Cl^{-} + 2OH^{-}, \ E^o_1 = 0.94 \ V, \ n_1 = 2$$\Delta G^o_1 = -n_1 F E^o_1 =

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$+0.52$

Vedclass · Answer

(C) Given half-reactions:$(I) \ OCl^{-} + H_2O + 2e^{-} \longrightarrow Cl^{-} + 2OH^{-}, \ E^o_1 = 0.94 \ V, \ n_1 = 2$$\Delta G^o_1 = -n_1 F E^o_1 = -2 	imes F 	imes 0.94 = -1.88F$$(II) \ \frac{1}{2}Cl_2 + e^{-} \longrightarrow Cl^{-}, \ E^o_2 = 1.36 \ V, \ n_2 = 1$To get the reaction $OCl^{-} + H_2O + e^{-} \longrightarrow \frac{1}{2}Cl_2 + 2OH^{-}$,we perform $(I) - (II)$:$\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2 = -1.88F - (-1.36F) = -0.52F$For the target reaction,$n_3 = 1$:$E^o_3 = -\frac{\Delta G^o_3}{n_3 F} = -\frac{-0.52F}{1 	imes F} = +0.52 \ V$.

The standard electrode potentials $(E^o)$ for $OCl^{-}/Cl^{-}$ and $\frac{1}{2}Cl_2/Cl^{-}$ are $0.94 \ V$ and $+1.36 \ V$ respectively,the $E^o$ value for $OCl^{-}/\frac{1}{2}Cl_2$ will be ........... $V$.

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