$$\mathrm{FeO_4^{2 - }\buildrel { + 2.2V} \over \longrightarrow F{e^{3 + }}\buildrel { + 0.70V} \over \longrightarrow F{e^{2 + }}\buildrel { - 0.45V} \over \longrightarrow F{e^0}}$$

$E_{FeO_4^{2 - }/F{e^{2 + }}}^\theta$ is $x \times {10^{ - 3}}$ V. The value of $x$ is _________

$$ \begin{aligned} & \mathrm{FeO}_4^{2-} \stackrel{+2.2 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{3+} ~~~\Delta \mathrm{G}_1=-6.6 \mathrm{~F} \\\\ & \mathrm{Fe}^{3+} \stackrel{+0.70 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{2+} ~~~\Delta \mathrm{G}_2=-0.7 \mathrm{~F} \end{aligned} $$<br/><br/> Hence for<br/><br/> $$ \begin{aligned} & \mathrm{FeO}_4^{2-} \longrightarrow \mathrm{Fe}^{2+} \Delta \mathrm{G} =-7.3 \mathrm{~F} \\\\ & =-\mathrm{nEF} \\\\ & \mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{+2}}^0=\frac{-7.3 \mathrm{~F}}{-4 \mathrm{~F}}= 1.825, \mathrm{n}=4 \\\\ & =1825 \times 10^{-3} \mathrm{~V} \end{aligned} $$<br/><br/> $n=$ electron exchange of that half cell reaction.