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