E^{circ}_{Fe^{2+}/Fe} = -0.441 V and E^{circ | vedclass

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(D) The given half-cell reactions are:$Fe^{2+} + 2e^{-} \rightarrow Fe, E^{\circ} = -0.441 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^{-}, E^{

Vedclass · Accepted Answer

$1.212$

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(D) The given half-cell reactions are:$Fe^{2+} + 2e^{-} \rightarrow Fe, E^{\circ} = -0.441 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^{-}, E^{\circ} = +0.441 \ V$)$Fe^{3+} + e^{-} \rightarrow Fe^{2+}, E^{\circ} = +0.771 \ V$ (Cathode reaction: $2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+}, E^{\circ} = +0.771 \ V$)For the overall reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$,the standard cell potential is calculated as:$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Fe^{2+}/Fe}$$E^{\circ}_{cell} = 0.771 \ V - (-0.441 \ V)$$E^{\circ}_{cell} = 1.212 \ V$

$E^{\circ}_{Fe^{2+}/Fe} = -0.441 \ V$ and $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771 \ V$,the standard $EMF$ of the reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$ will be $.......... \ V$.

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