The standard reduction potential E^{circ} for | vedclass

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(B) The given reactions are oxidation half-reactions,so the given $E^{\circ}$ values are oxidation potentials $(E^{\circ}_{op})$:$Zn \rightarrow Zn^{2

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$0.35 \ V$

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(B) The given reactions are oxidation half-reactions,so the given $E^{\circ}$ values are oxidation potentials $(E^{\circ}_{op})$:$Zn ightarrow Zn^{2+} + 2e^-, E^{\circ}_{op} = +0.76 \ V$$Fe ightarrow Fe^{2+} + 2e^-, E^{\circ}_{op} = +0.41 \ V$For the cell reaction $Fe^{2+} + Zn ightarrow Zn^{2+} + Fe$,$Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).The cell potential is given by: $E^{\circ}_{cell} = E^{\circ}_{op}(	ext{anode}) + E^{\circ}_{rp}(	ext{cathode})$Since $E^{\circ}_{rp}(	ext{cathode}) = -E^{\circ}_{op}(	ext{cathode})$,we have:$E^{\circ}_{cell} = E^{\circ}_{op}(Zn) - E^{\circ}_{op}(Fe)$$E^{\circ}_{cell} = 0.76 \ V - 0.41 \ V = +0.35 \ V$.

$Zn \rightarrow Zn^{2+} + 2e^-$	$E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$	$E^{\circ} = +0.41 \ V$

The standard reduction potential $E^{\circ}$ for half reactions are
$Zn \rightarrow Zn^{2+} + 2e^-$ $E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$ $E^{\circ} = +0.41 \ V$

The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

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The standard reduction potential $E^{\circ}$ for half reactions are$Zn \rightarrow Zn^{2+} + 2e^-$$E^{\circ} = +0.76 \ V$$Fe \rightarrow Fe^{2+} + 2e^-$$E^{\circ} = +0.41 \ V$The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is