A solution contains ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$ ,${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ and ${{\text{I}}^ - }$ ions. This solution was treated with iodine at ${\text{35}}{\,^{\text{o}}}{\text{C}}$ . ${{\text{E}}^{\text{o}}}$ for ${\text{F}}{{\text{e}}^{{\text{3 + }}}}/{\text{F}}{{\text{e}}^{{\text{2 + }}}}$ is $ + 0.77$ V and ${{\text{E}}^{\text{o}}}$ for ${{\text{I}}_2}{\text{/2}}{{\text{I}}^ - }\, = \,0.536$ V. The favourable redox reaction is:

Hint: Standard reduction potential of the half-cell is given and we have to determine the favourable redox reaction means the reduction and oxidation reaction. We will check the reduction potential. The species having high reduction potential will get reduced and the species having low reduction potential will get oxidized.

Complete step-by-step answer:The given half-cell reduction reactions are as follows:
\[{\text{F}}{{\text{e}}^{{\text{3 + }}}} + \,{{\text{e}}^ - }\, \to \,{\text{F}}{{\text{e}}^{{\text{2 + }}}}\]; ${{\text{E}}^{\text{o}}}\, = \, + 0.77\,{\text{volt}}$
\[{{\text{I}}_2}\, + \,{\text{2}}{{\text{e}}^ - }\, \to {\text{2}}{{\text{I}}^ - }\,\]; ${{\text{E}}^{\text{o}}}\, = \, + 0.536\,{\text{volt}}$
Now, we will decide the cathodic and anodic reaction as follows:
The species having high reduction potential get reduced at the cathode and the species having low reduction potential get oxidized at the anode.
The reduction potential of \[{\text{F}}{{\text{e}}^{{\text{3 + }}}}\] ($ + 0.77\,{\text{volt}}$) is greater than the reduction potential of \[{{\text{I}}_2}\]($ + 0.536\,{\text{volt}}$) so, \[{\text{F}}{{\text{e}}^{{\text{3 + }}}}\] will get reduce at the cathode and \[{{\text{I}}_2}\] will get oxidized at the anode.
So, the reduction reaction of \[{\text{F}}{{\text{e}}^{{\text{3 + }}}}\] will take place at cathode and oxidation reaction of \[{{\text{I}}_2}\] will take place at anode.
Reduction: \[{\text{F}}{{\text{e}}^{{\text{3 + }}}} + \,{{\text{e}}^ - }\, \to \,{\text{F}}{{\text{e}}^{{\text{2 + }}}}\]
Oxidation: \[{\text{2}}{{\text{I}}^ - }\, \to {{\text{I}}_2}\, + \,{\text{2}}{{\text{e}}^ - }\]
So, the complete redox reaction is,
\[{\text{2}}\,{\text{F}}{{\text{e}}^{{\text{3 + }}}} + \,{\text{2}}{{\text{I}}^ - }\,\, \to \,2\,{\text{F}}{{\text{e}}^{{\text{2 + }}}}\,{\text{ + }}\,\,{{\text{I}}_2}\]

Therefore, the favourable redox reaction is \[{\text{2}}\,{\text{F}}{{\text{e}}^{{\text{3 + }}}} + \,{\text{2}}{{\text{I}}^ - }\,\, \to \,2\,{\text{F}}{{\text{e}}^{{\text{2 + }}}}\,{\text{ + }}\,\,{{\text{I}}_2}\].

Note:According to the electrochemical series the element having high reduction work as oxidising agent and the element having low reduction work as reducing agent. By determining the standard reduction potential of the cell we can determine that the cell will work or not. The following expression is used for the determination of standard reduction potential of the cell;
\[{\text{E}}_{{\text{cell}}}^ \circ {\text{ = }}\,{\text{E}}_{{\text{cathode}}}^ \circ \, - {\text{E}}_{{\text{anode}}}^ \circ \]
Where,
\[{\text{E}}_{{\text{cell}}}^ \circ \] is the standard reduction potential of the cell
\[{\text{E}}_{{\text{cathode}}}^ \circ \] is the reduction potential of the cathode half-cell
\[{\text{E}}_{{\text{anode}}}^ \circ \] is the reduction potential of the anode half-cell.