Calculate Delta G^{circ} for the cell: Sn_{(s | vedclass

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Question

(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,the cell reaction is: $Sn_{(s)} + 2Ag

Vedclass · Accepted Answer

$-173.7$

Vedclass · Answer

(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,the cell reaction is: $Sn_{(s)} + 2Ag^{+}_{(aq)} ightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$.The number of electrons transferred,$n = 2$.The Faraday constant,$F = 96500 \ C \ mol^{-1}$.Given $E^{\circ}_{cell} = 0.90 \ V$.Substituting the values: $\Delta G^{\circ} = -2 	imes 96500 	imes 0.90 \ J$.$\Delta G^{\circ} = -173700 \ J = -173.7 \ kJ$.

Calculate $\Delta G^{\circ}$ for the cell: $Sn_{(s)} | Sn^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ at $25^{\circ} C$ given that $E^{\circ}_{cell} = 0.90 \ V$. (in $kJ$)

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