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A solution contains $Fe^{2+}$, $Fe^{3+}$ and $I^-$ ions. This solution was treated with iodine at $35^\circ C$. $E^\circ$ for $Fe^{3+}/Fe^{2+}$ is +0.77 V and $E^\circ$ for $I_2/2I^-$ = 0.536 V. The favourable redox reaction is
A
$I^-$ will be oxidised to $I_2$
B
$Fe^{2+}$ will be oxidised to $Fe^{3+}$
C
$I_2$ will be reduced to $I^-$
D
There will be no redox reaction
Detailed Solution
The couple with the higher standard reduction potential undergoes reduction and the other undergoes oxidation.
$E^\circ(Fe^{3+}/Fe^{2+}) = +0.77$ V is greater than $E^\circ(I_2/I^-) = +0.536$ V, so $Fe^{3+}$ is the stronger oxidising agent.
Reduction: $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$; oxidation: $2I^- \rightarrow I_2 + 2e^-$
Overall: $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.77 - 0.536 = +0.234$ V
$E^\circ_{cell}$ is positive, so this reaction is spontaneous; the reverse (oxidation of $Fe^{2+}$ by $I_2$) is not.
Hence $I^-$ will be oxidised to $I_2$.
$E^\circ(Fe^{3+}/Fe^{2+}) = +0.77$ V is greater than $E^\circ(I_2/I^-) = +0.536$ V, so $Fe^{3+}$ is the stronger oxidising agent.
Reduction: $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$; oxidation: $2I^- \rightarrow I_2 + 2e^-$
Overall: $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.77 - 0.536 = +0.234$ V
$E^\circ_{cell}$ is positive, so this reaction is spontaneous; the reverse (oxidation of $Fe^{2+}$ by $I_2$) is not.
Hence $I^-$ will be oxidised to $I_2$.