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(D) The efficiency of a cell is given by the ratio of Gibbs free energy change to the enthalpy change: $\eta = \frac{\Delta G^o}{\Delta H^o}$.Given $\

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(D) The efficiency of a cell is given by the ratio of Gibbs free energy change to the enthalpy change: $\eta = \frac{\Delta G^o}{\Delta H^o}$.Given $\eta = 0.70$ and $\Delta H^o = -551.5 \ kJ \ mol^{-1}$.$\Delta G^o = \eta 	imes \Delta H^o = 0.70 	imes (-551.5 \ kJ \ mol^{-1}) = -386.05 \ kJ \ mol^{-1} = -386050 \ J \ mol^{-1}$.The cell reaction is $A_{(s)} + B^{\oplus} \longrightarrow A^{\oplus} + B_{(s)}$,so $n = 1$.Using the relation $\Delta G^o = -nFE^o_{cell}$:$-386050 = -1 	imes 96500 	imes E^o_{cell}$.$E^o_{cell} = \frac{386050}{96500} = 4.0 \ V$.

Efficiency of a cell with cell reaction under standard conditions is $70\%$. Cell reaction:
$A_{(s)} + B^{\oplus} \longrightarrow A^{\oplus} + B_{(s)}$ ; $\Delta H^o = -551.5 \ kJ$. Calculate standard electrode potential of cell (in $volt$).

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Efficiency of a cell with cell reaction under standard conditions is $70\%$. Cell reaction:$A_{(s)} + B^{\oplus} \longrightarrow A^{\oplus} + B_{(s)}$ ; $\Delta H^o = -551.5 \ kJ$. Calculate standard electrode potential of cell (in $volt$).