FeO_4^{2-} xrightarrow{2.2 V} Fe^{3+} xright | vedclass

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(B) The reaction $FeO_4^{2-} ightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so $n = 4$.Using the relation $\Delta G^{	he

Vedclass · Accepted Answer

$1825$

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(B) The reaction $FeO_4^{2-} ightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so $n = 4$.Using the relation $\Delta G^{	heta} = -nFE^{	heta}$,for the overall reaction:$n_{total} E_{total} = n_1 E_1 + n_2 E_2$Here,$n_1 = 3$ $(FeO_4^{2-} ightarrow Fe^{3+})$,$E_1 = 2.2 \ V$,$n_2 = 1$ $(Fe^{3+} ightarrow Fe^{2+})$,$E_2 = 0.70 \ V$,and $n_{total} = 4$.$4 	imes E_{FeO_4^{2-} / Fe^{2+}}^{	heta} = (3 	imes 2.2) + (1 	imes 0.70)$$4 	imes E_{FeO_4^{2-} / Fe^{2+}}^{	heta} = 6.6 + 0.70 = 7.3 \ V$$E_{FeO_4^{2-} / Fe^{2+}}^{	heta} = \frac{7.3}{4} = 1.825 \ V$$1.825 \ V = 1825 	imes 10^{-3} \ V$Therefore,$x = 1825$.

$FeO_4^{2-}$ $\xrightarrow{2.2 \ V} Fe^{3+}$ $\xrightarrow{0.70 \ V} Fe^{2+}$ $\xrightarrow{-0.45 \ V} Fe^0$
$E_{FeO_4^{2-} / Fe^{2+}}^{\theta}$ is $x \times 10^{-3} \ V$. The value of $x$ is $.........$.

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$FeO_4^{2-}$ $\xrightarrow{2.2 \ V} Fe^{3+}$ $\xrightarrow{0.70 \ V} Fe^{2+}$ $\xrightarrow{-0.45 \ V} Fe^0$$E_{FeO_4^{2-} / Fe^{2+}}^{\theta}$ is $x \times 10^{-3} \ V$. The value of $x$ is $.........$.