JEE MAIN - Chemistry (2023 - 10th April Morning Shift - No. 22)
$$\mathrm{FeO_4^{2 - }\buildrel { + 2.2V} \over \longrightarrow F{e^{3 + }}\buildrel { + 0.70V} \over \longrightarrow F{e^{2 + }}\buildrel { - 0.45V} \over \longrightarrow F{e^0}}$$
$$E_{FeO_4^{2 - }/F{e^{2 + }}}^\theta $$ is $$x \times {10^{ - 3}}$$ V. The value of $$x$$ is _________
Answer
1825
Explanation
$$
\begin{aligned}
& \mathrm{FeO}_4^{2-} \stackrel{+2.2 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{3+} ~~~\Delta \mathrm{G}_1=-6.6 \mathrm{~F} \\\\
& \mathrm{Fe}^{3+} \stackrel{+0.70 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{2+} ~~~\Delta \mathrm{G}_2=-0.7 \mathrm{~F}
\end{aligned}
$$
Hence for
$$ \begin{aligned} & \mathrm{FeO}_4^{2-} \longrightarrow \mathrm{Fe}^{2+} \Delta \mathrm{G} =-7.3 \mathrm{~F} \\\\ & =-\mathrm{nEF} \\\\ & \mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{+2}}^0=\frac{-7.3 \mathrm{~F}}{-4 \mathrm{~F}}= 1.825, \mathrm{n}=4 \\\\ & =1825 \times 10^{-3} \mathrm{~V} \end{aligned} $$
$n=$ electron exchange of that half cell reaction.
Hence for
$$ \begin{aligned} & \mathrm{FeO}_4^{2-} \longrightarrow \mathrm{Fe}^{2+} \Delta \mathrm{G} =-7.3 \mathrm{~F} \\\\ & =-\mathrm{nEF} \\\\ & \mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{+2}}^0=\frac{-7.3 \mathrm{~F}}{-4 \mathrm{~F}}= 1.825, \mathrm{n}=4 \\\\ & =1825 \times 10^{-3} \mathrm{~V} \end{aligned} $$
$n=$ electron exchange of that half cell reaction.
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