Consider the cell
$$\mathrm{Pt(s)|{H_2}(g)\,(1\,atm)|{H^ + }\,(aq,[{H^ + }] = 1)||F{e^{3 + }}(aq),F{e^{2 + }}(aq)|Pt(s)}$$
Given $\mathrm{E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.771\,V}$ and $\mathrm{E_{{H^ + }/1/2\,{H_2}}^o = 0\,V,\,T = 298\,K}$
If the potential of the cell is 0.712 V, the ratio of concentration of Fe$^{2+}$ to Fe$^{3+}$ is _____________ (Nearest integer)
Answer (integer)
10
Solution
$$
\begin{aligned}
& \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{Fe}^{3+}(\mathrm{aq} .) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq} .) \\\\
& \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \\\\
& \Rightarrow \quad 0.712=(0.771-0)-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \\\\
& \Rightarrow \quad \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=\frac{(0.771-0712)}{0.059}=1 \\\\
& \Rightarrow \quad \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=10
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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