For the cell reaction,3 Sn^{4+} + 2 Cr longri | vedclass

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(A) The cell reaction is $3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$.Here,the number of electrons transferred $(n)$ is $6$ (since $Sn^{4+

Vedclass · Accepted Answer

$-515.31 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The cell reaction is $3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$.Here,the number of electrons transferred $(n)$ is $6$ (since $Sn^{4+} + 2e^- \longrightarrow Sn^{2+}$ occurs $3$ times and $Cr \longrightarrow Cr^{3+} + 3e^-$ occurs $2$ times).The standard Gibbs free energy change is given by the formula $\Delta G^{\circ} = -n F E^{\circ}_{cell}$.Substituting the values: $n = 6$,$F = 96500 \ C \ mol^{-1}$,and $E^{\circ}_{cell} = 0.89 \ V$.$\Delta G^{\circ} = -6 	imes 96500 	imes 0.89 \ J \ mol^{-1}$.$\Delta G^{\circ} = -515310 \ J \ mol^{-1} = -515.31 \ kJ \ mol^{-1}$.

For the cell reaction,$3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$,$E^{\circ}_{cell}$ is $0.89 \ V$. Then $\Delta G^{\circ}$ for the reaction is

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