Electrochemistry Question 3
Question 3 - 25 January - Shift 1
Consider the cell
$Pt(s)|H_2(s)(latm)| H^{+}(aq,[H^{+}]=1)|| Fe^{3+}(aq), Fe^{2+}(aq) \mid Pt(s)$
Given : $E _{Fe^{3+} / Fe^{2+}}^{\circ}=0.771 V$ and $E _{H^{+} / _2^{1} H_2}^{\circ}=0 V, T=298 K$
If the potential of the cell is $0.712 V$ the ratio of concentration of $Fe^{2+}$ to $Fe^{2+}$ is (Nearest integer)
Show Answer
Answer: (10)
Solution:
$\frac{1}{2} H_2(g)+Fe^{3+}($ aq. $) \xrightarrow{\longrightarrow} H^{+}($aq $)+Fe^{2+}($ aq. $)$
$E=E^{o}-\frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$
$\Rightarrow \quad 0.712=(0.771-0)-\frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$
$\Rightarrow \quad \log \frac{[Fe^{2+}]}{[Fe^{3+}]}=\frac{(0.771-0712)}{0.059}=1$
$\Rightarrow \quad \frac{[Fe^{2+}]}{[Fe^{3+}]}=10$
