QJEEM23Chabd1: JEE Main 2023 Chemistry - Class 12 JEEM23Chabd1 Solution: FeO 4 2 − ⟶ + 2.2 V F e 3 + ⟶ + 0.70 V F e 2 + ⟶ − 0.45 V F e 0 …

Question

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${FeO}_{4}^{2 -} \overset{+ 2.2 V}{⟶} F e^{3 +} \overset{+ 0.70 V}{⟶} F e^{2 +} \overset{- 0.45 V}{⟶} F e^{0}$

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

Accepted Answer

\begin{aligned}&amp; \mathrm{FeO}_4^{2-} \stackrel{+2.2 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{3+} ~~~\Delta \mathrm{G}_1=-6.6 \mathrm{~F} \\\\&amp; \mathrm{Fe}^{3+} \stackrel{+0.70 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{2+} ~~~\Delta \mathrm{G}_2=-0.7 \mathrm{~F}\end{aligned}

Hence for

\begin{aligned}&amp; \mathrm{FeO}_4^{2-} \longrightarrow \mathrm{Fe}^{2+} \Delta \mathrm{G} =-7.3 \mathrm{~F} \\\\&amp; =-\mathrm{nEF} \\\\&amp; \mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{+2}}^0=\frac{-7.3 \mathrm{~F}}{-4 \mathrm{~F}}= 1.825, \mathrm{n}=4 \\\\&amp; =1825 \times 10^{-3} \mathrm{~V}\end{aligned}

n=

electron exchange of that half cell reaction.

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