Electrochemistry Question 8
Question 8 - 30 January - Shift 1
Consider the cell
$Pt _{(s)}|H_2(g, 1 atm)| H^{+}(aq, 1 M)||Fe^{3+}(aq), Fe^{2+}(aq) \mid Pt(s)$
When the potential of the cell is $0.712 V$ at $298 K$, the ratio $\lfloor Fe^{2+}\rfloor / \lfloor Fe^{3+}\rfloor$ is
(Nearest integer)
Given: $Fe^{3+}+e^{-}=Fe^{2+}, E^{0} Fe^{3+}, Fe^{2+} \mid Pt=0.771$
$\frac{2.303 RT}{F}=0.06 V$
Show Answer
Answer: (10)
Solution:
$Pt _{(s)} \mid H_2(g$, latm $) \mid H^{+}(aq, 1 M) | Fe^{3+}(aq), Fe^{2+}(aq) \mid Pt(s)$
at anode $H_2 \longrightarrow 2 H^{+}+2 e^{-}$
At cathode $Fe_aq^{3+}+e^{– \text{thon }} Fe_aq^{2+}$
$E^{\circ}=E _{H_2 \mid H^{+}}^{\circ}+E _{Fe^{3+} \mid Fe^{2+}}^{\circ}=0.771 V$
$E=E^{\circ}-\frac{0 \cdot 06}{1} \log \frac{Fe^{2+}}{Fe^{3+}}$
$0 \cdot 712=(0+0 \cdot 771)-\frac{0 \cdot 06}{1} \log \frac{Fe^{2+}}{Fe^{3+}}$
$\log \frac{Fe^{2+}}{Fe^{3+}}=\frac{0 \cdot 059}{0 \cdot 06} \approx 1$
$\frac{Fe^{2+}}{Fe^{3+}}=10$
