Electrochemistry - Result Question 3
####3. Calculate the standard cell potential (in $V$ ) of the cell in which following reaction takes place
$Fe^{2+}(a q)+Ag^{+}(a q) \longrightarrow Fe^{3+}(a q)+Ag(s)$
Given that,
$$ \begin{aligned} & E^{\circ} Ag^{+} / Ag=x V \ & E^{\circ} Fe^{2+} / Fe=y V \ & E _{Fe^{3+} / Fe}^{\circ}=z V \end{aligned} $$
(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:
- $Fe^{2+}(a q)+Ag^{+}(a q) \longrightarrow Fe^{3+}(a q)+Ag(s)$
$\therefore \quad E _{\text {cell }}^{\circ}=E _{Ag^{+} / Ag}^{0}-E _{Fe^{3+} / Fe^{2+}}^{0}=x V-E _{Fe^{3+} / Fe^{2+}}^{0}$
Now, for two half-cells
(i) $Fe^{2+}+2 e^{-} \longrightarrow Fe ; E _{Fe^{2+} / Fe}^{0}=E _1^{0}=y V \Delta G _2^{0}=-2 F E _1^{0}$
(ii) $Fe^{3+}+3 e^{-} \longrightarrow Fe ; E _{Fe^{3+} / Fe}^{0}=E _2^{\circ}=z V \Delta G _2^{\circ}=-3 F E _2^{\circ}$
So, $Fe^{3+}+e^{-} \longrightarrow Fe^{2+} ; E _{Fe^{3+}}^{0} / Fe^{2+}=E _3^{\circ}=$ ? $; \Delta G _3^{\circ}=-1 \times F E _3^{\circ}$
Again, $\Delta G _3^{\circ}=\Delta G _2^{\circ}-\Delta G _1^{\circ}$
$\Rightarrow \quad-F E _3^{\circ}=-3 F E _2^{\circ}-\left(-2 F E _1^{\circ}\right)$
$\Rightarrow \quad-E _3^{\circ}=2 E _1^{\circ}-3 E _2^{\circ} \Rightarrow E _3^{\circ}=3 E _2^{\circ}-2 E _1^{\circ}$
$\Rightarrow E _{Fe^{3+} / Fe^{2+}}^{\circ}=(3 z-2 y) V$
So, from equation (i)
$$ E _{\text {cell }}^{o}=x V-(3 z-2 y) V=(x-3 z+2 y) V $$
