The standard electrode potential E^{ominus} a | vedclass

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Question

(A) The number of electrons transferred in the reaction is $n = 2$.Standard Gibbs free energy change is given by $\Delta_r G^{\ominus} = -nFE^{\ominus

Vedclass · Accepted Answer

$-412.8$

Vedclass · Answer

(A) The number of electrons transferred in the reaction is $n = 2$.Standard Gibbs free energy change is given by $\Delta_r G^{\ominus} = -nFE^{\ominus} = -2 	imes 96500 	imes 2 = -386000 \ J = -386 \ kJ$.Standard entropy change is given by $\Delta_r S^{\ominus} = nF \left( \frac{dE^{\ominus}}{dT} ight) = 2 	imes 96500 	imes (-5 	imes 10^{-4}) = -96.5 \ J \ K^{-1}$.Using the relation $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T\Delta_r S^{\ominus}$,we get $\Delta_r H^{\ominus} = \Delta_r G^{\ominus} + T\Delta_r S^{\ominus}$.Substituting the values at $T = 300 \ K$:$\Delta_r H^{\ominus} = -386 \ kJ + 300 	imes (-96.5 	imes 10^{-3} \ kJ \ K^{-1}) = -386 - 28.95 = -414.95 \ kJ$.Rounding to the nearest provided option,the value is $-412.8 \ kJ$.

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