For the half-reactions Zn rightarrow Zn^{2+} | vedclass

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(B) The given reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$. In this reaction,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to

Vedclass · Accepted Answer

$+0.35$

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(B) The given reaction is $Fe^{2+} + Zn ightarrow Zn^{2+} + Fe$. In this reaction,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to $Fe$ (Cathode). The standard cell potential is calculated using the formula: $E^{0}_{cell} = E^{0}_{Oxi}(	ext{anode}) + E^{0}_{Red}(	ext{cathode})$. Since $E^{0}_{Oxi}(	ext{Fe/Fe}^{2+}) = +0.41 \ V$,the reduction potential $E^{0}_{Red}(	ext{Fe}^{2+}	ext{/Fe}) = -0.41 \ V$. Thus,$E^{0}_{cell} = E^{0}_{Oxi}(	ext{Zn/Zn}^{2+}) + E^{0}_{Red}(	ext{Fe}^{2+}	ext{/Fe}) = 0.76 \ V + (-0.41 \ V) = +0.35 \ V$.

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