JEE MAIN - Chemistry (2023 - 25th January Evening Shift - No. 17)
$$Pt(s)|{H_2}(g)(1\,bar)|{H^ + }(aq)(1\,M)||{M^{3 + }}(aq),{M^ + }(aq)|Pt(s)$$
The $$\mathrm{E_{cell}}$$ for the given cell is 0.1115 V at 298 K when $${{\left[ {{M^ + }(aq)} \right]} \over {\left[ {{M^{3 + }}(aq)} \right]}} = {10^a}$$
The value of $$a$$ is ____________
Given : $$\mathrm{E_{{M^{3 + }}/{M^ + }}^\theta = 0.2}$$ V
$${{2.303RT} \over F} = 0.059V$$
Answer
3
Explanation
Overall reaction :-
$$ \begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{M}_{\text {(aq) }}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\\\ & \mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \\\\ & 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & \therefore \mathrm{a}=3 \end{aligned} $$
$$ \begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{M}_{\text {(aq) }}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\\\ & \mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \\\\ & 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & \therefore \mathrm{a}=3 \end{aligned} $$
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