For the reaction 2 Fe^{3+}_{(aq)} + 2 I^{-}_{ | vedclass

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(D) The reaction is $2 Fe^{3+} + 2 I^{-} \rightarrow 2 Fe^{2+} + I_{2}$.First,calculate $E^{\circ}_{Fe^{3+}/Fe^{2+}}$ using the given Latimer diagram

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) The reaction is $2 Fe^{3+} + 2 I^{-} ightarrow 2 Fe^{2+} + I_{2}$.First,calculate $E^{\circ}_{Fe^{3+}/Fe^{2+}}$ using the given Latimer diagram data:$n_1 E_1^{\circ} + n_2 E_2^{\circ} = n_3 E_3^{\circ}$$1 	imes E^{\circ}_{Fe^{3+}/Fe^{2+}} + 2 	imes E^{\circ}_{Fe^{2+}/Fe} = 3 	imes E^{\circ}_{Fe^{3+}/Fe}$$E^{\circ}_{Fe^{3+}/Fe^{2+}} + 2(-0.440) = 3(-0.036)$$E^{\circ}_{Fe^{3+}/Fe^{2+}} = -0.108 + 0.880 = 0.772 \ V$.Now,calculate the cell potential $E^{\circ}_{cell}$:$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{I_{2}/2I^{-}}$$E^{\circ}_{cell} = 0.772 - 0.539 = 0.233 \ V$.The standard Gibbs free energy change is $\Delta_{r} G^{\circ} = -n F E^{\circ}_{cell}$.Here,$n = 2$ (moles of electrons transferred).$\Delta_{r} G^{\circ} = -2 	imes 96500 	imes 0.233 = -44969 \ J \ mol^{-1} = -44.969 \ kJ \ mol^{-1}$.The magnitude is approximately $45 \ kJ \ mol^{-1}$.

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