Electrochemistry 1 Question 33
32. The following electrochemical cell has been set-up :
$$ \begin{aligned} & \operatorname{Pt}(1)\left|\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}(a=1)\right| \mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}(a=1) \mid \mathrm{Pt}(2) \ & E^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right)=0.77 \mathrm{~V} \ & \text { and } \quad E^{\circ}\left(\mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}\right)=1.61 \mathrm{~V} \end{aligned} $$
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current, will the current increases or decreases with time?
(2000, 2M)
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Solution:
- Since, activities of all the ions are unity, $E_{\text {cell }}=E_{\text {cell }}^{\circ}$. Also, left hand electrode is at lower reduction potential, it act as anode and
$$ E^{\circ}=E^{\circ}\left(\mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}\right)-E^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right)=0.84 $$
i.e. electrons will flow from left to right hand electrode and current from right hand electrode $[\mathrm{Pt}(2)]$ to left hand electrode $[\operatorname{Pt}(1)]$.
Also, $\quad E=E^{\circ}-0.0592 \log \frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{Ce}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{Ce}^{4+}\right]}$
As electrolysis proceeds, $E$ will decrease and therefore, current.
