QJEEM23Chf191: JEE Main 2023 Chemistry - Class 12 JEEM23Chf191 Solution: Consider the cell Given and If the potential of the cell is…

Question

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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)

Accepted Answer

\begin{aligned}&amp; \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{Fe}^{3+}(\mathrm{aq} .) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq} .) \\\\&amp; \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \\\\&amp; \Rightarrow \quad 0.712=(0.771-0)-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \\\\&amp; \Rightarrow \quad \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=\frac{(0.771-0712)}{0.059}=1 \\\\&amp; \Rightarrow \quad \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=10\end{aligned}

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