FeO_4^{2-} xrightarrow{2.0 V} Fe^{3+} xright | vedclass

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(A) The reaction $FeO_4^{2-} \rightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so the number of electrons involved is $n_4 =

Vedclass · Accepted Answer

$1.7$

Vedclass · Answer

(A) The reaction $FeO_4^{2-} ightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so the number of electrons involved is $n_4 = 4$.This process can be broken down into two steps:$1) FeO_4^{2-} + 3e^- ightarrow Fe^{3+} \quad (E_1^{\Theta} = 2.0 \ V, n_1 = 3)$$2) Fe^{3+} + 1e^- ightarrow Fe^{2+} \quad (E_2^{\Theta} = 0.8 \ V, n_2 = 1)$The total Gibbs free energy change is given by $\Delta G_4^{\Theta} = \Delta G_1^{\Theta} + \Delta G_2^{\Theta}$.Using $\Delta G^{\Theta} = -nFE^{\Theta}$,we get:$-n_4 F E_4^{\Theta} = -n_1 F E_1^{\Theta} - n_2 F E_2^{\Theta}$$n_4 E_4^{\Theta} = n_1 E_1^{\Theta} + n_2 E_2^{\Theta}$$4 	imes E_4^{\Theta} = (3 	imes 2.0) + (1 	imes 0.8)$$4 E_4^{\Theta} = 6.0 + 0.8 = 6.8$$E_4^{\Theta} = \frac{6.8}{4} = 1.7 \ V$.

$FeO_4^{2-}$ $\xrightarrow{2.0 \ V} Fe^{3+}$ $\xrightarrow{0.8 \ V} Fe^{2+}$ $\xrightarrow{-0.44 \ V} Fe^0$
In the above diagram,the standard electrode potentials are given in volts. The value of $E_{FeO_4^{2-}/Fe^{2+}}^{\Theta}$ is: (in $V$)

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$FeO_4^{2-}$ $\xrightarrow{2.0 \ V} Fe^{3+}$ $\xrightarrow{0.8 \ V} Fe^{2+}$ $\xrightarrow{-0.44 \ V} Fe^0$In the above diagram,the standard electrode potentials are given in volts. The value of $E_{FeO_4^{2-}/Fe^{2+}}^{\Theta}$ is: (in $V$)