Given the standard half-cell potentials (E^{c | vedclass

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(B) The given cell reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$. Oxidation half-reaction: $Zn \rightarrow Zn^{2+} + 2e^{-}$, $E^{\circ}_{ox} =

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$+0.35 \ V$

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(B) The given cell reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$. Oxidation half-reaction: $Zn \rightarrow Zn^{2+} + 2e^{-}$, $E^{\circ}_{ox} = +0.76 \ V$. Reduction half-reaction: $Fe^{2+} + 2e^{-} \rightarrow Fe$, $E^{\circ}_{red} = -0.41 \ V$ (since $E^{\circ}_{ox}$ for $Fe$ is $+0.41 \ V$, the reduction potential $E^{\circ}_{red} = -E^{\circ}_{ox}$). The standard e.m.f. of the cell is: $E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red}$ $E^{\circ}_{cell} = 0.76 \ V + (-0.41 \ V) = +0.35 \ V$.

Given the standard half-cell potentials $(E^{\circ})$ as: $Zn \rightarrow Zn^{2+} + 2e^{-}$; $E^{\circ} = +0.76 \ V$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$; $E^{\circ} = +0.41 \ V$. Then the standard e.m.f. of the cell with the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is:

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