Given the standard electrode potentials E^{ci | vedclass

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(A) The given cell reaction is: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$In this reaction,$Sn$ is oxidized to $Sn^{2+}

Vedclass · Accepted Answer

$0.91$

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(A) The given cell reaction is: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$In this reaction,$Sn$ is oxidized to $Sn^{2+}$ (Anode) and $Fe^{3+}$ is reduced to $Fe^{2+}$ (Cathode).The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$Here,$E^{\circ}_{cathode} = E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \ V$ and $E^{\circ}_{anode} = E^{\circ}_{Sn^{2+}/Sn} = -0.14 \ V$.Substituting the values: $E^{\circ}_{cell} = 0.77 \ V - (-0.14 \ V) = 0.77 \ V + 0.14 \ V = 0.91 \ V$.

Given the standard electrode potentials $E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \ V$ and $E^{\circ}_{Sn^{2+}/Sn} = -0.14 \ V$,calculate the standard cell potential $E^{\circ}_{cell}$ for the reaction: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ (in $V$)

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