Given that $E_{{\mathrm{Fe}}^{3+}\mid \mathrm{Fe}}^{\circ }\text{ and }{E}^{\circ }{\mathrm{Fe}}^{2+}\mid \mathrm{Fe}$  are -0.036 V and -0.439 V, respectively. The value of $E^{\circ }{\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}\mid \mathrm{Pt}$ would be

$\begin{array}{clc}{\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}& E_1^{\circ }=-0.036\mathrm{V}& \mathrm{\Delta }{G}_1^{\circ }=-3F{E}_1^{\circ }\\ {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Fe}& E_2^{\circ }=-0.439\mathrm{V}& \mathrm{\Delta }{G}_2^{\circ }=-2F{E}_2^{\circ }\\ {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}& E_3^{\circ }=?& \mathrm{\Delta }{G}_3^{\circ }=-F{E}_3^{\circ }\end{array}$

Third equation can be obtained by subtracting second from first. Hence

$\mathrm{\Delta }{G}_3^{\circ }=\mathrm{\Delta }{G}_1^{\circ }-\mathrm{\Delta }{G}_2^{\circ }\text{ i.e. }-F{E}_3^{\circ }=-3F{E}_1^{\circ }+2F{E}_2^{\circ }$

or $E_3^{\circ }=3E_1^{\circ }-2E_2^{\circ }=3(-0.036\mathrm{V})-2(-0.439\mathrm{V})$