Electrochemistry 2 Question 4
4. The standard Gibbs energy for the given cell reaction in $\mathrm{kJ}$ $\mathrm{mol}^{-1}$ at $298 \mathrm{~K}$ is
$\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)$,
$E^{\circ}=2 \mathrm{~V}$ at $298 \mathrm{~K}$
(Faraday’s constant, $F=96000 \mathrm{C} \mathrm{mol}^{-1}$ )
(2019 Main, 9 April I)
(a) 384
(b) 192
(c) -384
(d) -192
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Solution:
- Key Idea Gibbs energy of the reaction is related to $E^{\circ}$ cell by the following formula
$$ \begin{aligned} \Delta G^{o} & =-n F E^{\circ} \ \Delta G^{\mathrm{o}} & =\text { Gibbs energy of cell } \ n F & =\text { amount of charge passed } \ E & =\text { EMF of a cell } \end{aligned} $$
Given reaction is,
$$ \mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu} $$
$E_{\text {cell }}^{\mathrm{o}}=2.0 \mathrm{~V}$
$F=96000 \mathrm{C}$
$$ n=2 $$
To find the value of $\Delta G^{\mathrm{o}}(\mathrm{kJ} \mathrm{mol})$, we use the formula
$$ \begin{aligned} & \Delta G^{\mathrm{o}}=-n F E_{\text {cell }}^{\mathrm{o}} \ & \Delta G^{\mathrm{o}}=-2 \times 96000 \times 2=-384000 \mathrm{~J} / \mathrm{mol} \end{aligned} $$
In terms of $\mathrm{kJ} / \mathrm{mol}, \Delta G^{\mathbf{o}}=\frac{-384000}{1000}=-384 \mathrm{~kJ} / \mathrm{mol}$
