For the given reactions:Sn^{2+} + 2e^{-} righ | vedclass

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(C) Given:$(1) \ Sn^{2+} + 2e^{-}$ $ightarrow Sn, \ E^{\circ}_{1} = -0.140 \ V, \ \Delta G^{\circ}_{1} = -nFE^{\circ}_{1} = -2 	imes F 	imes (-0.1

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$16$

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(C) Given:$(1) \ Sn^{2+} + 2e^{-}$ $ightarrow Sn, \ E^{\circ}_{1} = -0.140 \ V, \ \Delta G^{\circ}_{1} = -nFE^{\circ}_{1} = -2 	imes F 	imes (-0.140) = +0.280F$$(2) \ Sn^{4+} + 4e^{-}$ $ightarrow Sn, \ E^{\circ}_{2} = 0.010 \ V, \ \Delta G^{\circ}_{2} = -nFE^{\circ}_{2} = -4 	imes F 	imes (0.010) = -0.040F$We need the potential for $Sn^{4+} + 2e^{-} ightarrow Sn^{2+}$.This can be obtained by $(2) - (1)$:$(Sn^{4+} + 4e^{-}) - (Sn^{2+} + 2e^{-}) ightarrow Sn - Sn$$Sn^{4+} + 2e^{-} ightarrow Sn^{2+}$$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{2} - \Delta G^{\circ}_{1} = -0.040F - 0.280F = -0.320F$Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{Sn^{4+}/Sn^{2+}}$,where $n=2$:$-2FE^{\circ}_{Sn^{4+}/Sn^{2+}} = -0.320F$$E^{\circ}_{Sn^{4+}/Sn^{2+}} = 0.160 \ V = 16 	imes 10^{-2} \ V$The value is $16$.

Similar Questions

What is electrode potential?

Aluminium displaces hydrogen from dilute $HCl$ whereas silver does not. The e.m.f. of a cell prepared by combining $Al/Al^{3+}$ and $Ag/Ag^{+}$ is $2.46 \ V$. The reduction potential of silver electrode is $+0.80 \ V$. The reduction potential of aluminium electrode is $........... \ V$.

Assume the cell reaction,$A_{(s)} + B_{(aq)}^{+2} \rightarrow A_{(aq)}^{+2} + B_{(s)}$. If $\Delta G^{\circ} = -386 \ kJ$ at $298 \ K$,what is $E_{\text{cell}}^{\circ}$ (in $V$)? (Assume $n = 2$)

The standard Gibbs energy for the given cell reaction in $kJ \, mol^{-1}$ at $298 \, K$ is $Zn_{(s)} + Cu^{2+}_{(aq)} \to Zn^{2+}_{(aq)} + Cu_{(s)}$,given $E^o = 2 \, V$ at $298 \, K$ [Faraday's constant $F = 96500 \, C \, mol^{-1}$].

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