Electrochemistry 1 Question 3
3. Calculate the standard cell potential (in $\mathrm{V}$ ) of the cell in which following reaction takes place
$\mathrm{Fe}^{2+}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ag}(s)$
Given that,
$$ \begin{aligned} & E^{\circ} \mathrm{Ag}^{+} / \mathrm{Ag}=x \mathrm{~V} \ & E^{\circ} \mathrm{Fe}^{2+} / \mathrm{Fe}=y \mathrm{~V} \ & E_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\circ}=z \mathrm{~V} \end{aligned} $$
(2019 Main, 8 April II)
(a) $x+2 y-3 z$
(b) $x-y$
(c) $x+y-z$
(d) $x-z$
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Answer:
Correct Answer: 3. (d)
Solution:
- $\mathrm{Fe}^{2+}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ag}(s)$
$\therefore \quad E_{\text {cell }}^{\circ}=E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=x \mathrm{~V}-E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}$
Now, for two half-cells
(i) $\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0}=E_{1}^{0}=y \mathrm{~V} \Delta G_{2}^{0}=-2 F E_{1}^{0}$
(ii) $\mathrm{Fe}^{3+}+3 e^{-} \longrightarrow \mathrm{Fe} ; E_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{0}=E_{2}^{\circ}=z \mathrm{~V} \Delta G_{2}^{\circ}=-3 F E_{2}^{\circ}$
So, $\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} ; E_{\mathrm{Fe}^{3+}}^{0} / \mathrm{Fe}^{2+}=E_{3}^{\circ}=$ ? $; \Delta G_{3}^{\circ}=-1 \times F E_{3}^{\circ}$
Again, $\Delta G_{3}^{\circ}=\Delta G_{2}^{\circ}-\Delta G_{1}^{\circ}$
$\Rightarrow \quad-F E_{3}^{\circ}=-3 F E_{2}^{\circ}-\left(-2 F E_{1}^{\circ}\right)$
$\Rightarrow \quad-E_{3}^{\circ}=2 E_{1}^{\circ}-3 E_{2}^{\circ} \Rightarrow E_{3}^{\circ}=3 E_{2}^{\circ}-2 E_{1}^{\circ}$
$\Rightarrow E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=(3 z-2 y) \mathrm{V}$
So, from equation (i)
$$ E_{\text {cell }}^{o}=x V-(3 z-2 y) \mathrm{V}=(x-3 z+2 y) \mathrm{V} $$
