Q.Can $Fe^{3+}$ oxidise bromide to bromine under Standard Conditions ? Given : $E^{\circ}_{Fe^{3+}|Fe^{2+}} = 0.771\ V$ $E^{\circ}_{Br_2|Br^-} = 1.09\ V$

Comparing the two standard reduction potentials shows $Br_2/Br^-$ (1.09 V) is a stronger oxidising couple than $Fe^{3+}/Fe^{2+}$ (0.771 V), so $Fe^{3+}$ cannot spontaneously oxidise $Br^-$; the calculated $E^\circ_{cell}$ comes out negative.

For $Fe^{3+}$ to act as an oxidising agent toward $Br^-$, the desired overall reaction would be: $2Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2$ Here $Fe^{3+}$ is reduced (cathode half-reaction, $Fe^{3+}+e^-\rightarrow Fe^{2+}$, $E^\circ=0.771\ V$) and $Br^-$ is oxidised (anode half-reaction, the reverse of $Br_2+2e^-\rightarrow2Br^-$, $E^\circ=1.09\ V$). The overall cell potential is: $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=E^\circ(Fe^{3+}/Fe^{2+})-E^\circ(Br_2/Br^-)=0.771-1.09=-0.319\ V$ Since $E^\circ_{cell}$ is negative, $\Delta G^\circ=-nFE^\circ_{cell}$ is positive, meaning the reaction is non-spontaneous under standard conditions. Physically, this is because $Br_2/Br^-$ has a higher standard reduction potential than $Fe^{3+}/Fe^{2+}$, meaning $Br_2$ is the stronger oxidising agen …