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(C) To find the standard electrode potential for the reaction $OCl^{-} 	o \frac{1}{2}Cl_2$,we use the Gibbs free energy change relation $\Delta G^o =

Vedclass · Accepted Answer

$0.52$

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(C) To find the standard electrode potential for the reaction $OCl^{-} 	o \frac{1}{2}Cl_2$,we use the Gibbs free energy change relation $\Delta G^o = -nFE^o$.For reaction $1$: $OCl^{-} + H_2O + 2e^{-} 	o Cl^{-} + 2OH^{-}$,$E^o_1 = 0.94 \ V$,$n_1 = 2$.$\Delta G^o_1 = -2 	imes F 	imes 0.94 = -1.88F$.For reaction $2$: $Cl^{-} 	o \frac{1}{2}Cl_2 + e^{-}$,$E^o_2 = -1.36 \ V$,$n_2 = 1$.$\Delta G^o_2 = -1 	imes F 	imes (-1.36) = 1.36F$.Adding the two reactions: $OCl^{-} + H_2O + e^{-} 	o \frac{1}{2}Cl_2 + 2OH^{-}$.Total $\Delta G^o = \Delta G^o_1 + \Delta G^o_2 = -1.88F + 1.36F = -0.52F$.Since $\Delta G^o = -nFE^o_{total}$ and $n = 1$ for the net reaction:$-0.52F = -1 	imes F 	imes E^o_{total}$.$E^o_{total} = 0.52 \ V$.

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

Similar Questions

Using the data given below,find out the strongest reducing agent:
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$(iii)$ On left side or right side?

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