$Pt(s)|{H_2}(g)(1\,bar)|{H^ + }(aq)(1\,M)||{M^{3 + }}(aq),{M^ + }(aq)|Pt(s)$
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 (integer)
3
Solution
Overall reaction :-<br/><br/>
$$
\begin{aligned}
& \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{M}_{\text {(aq) }}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\\\
& \mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \\\\
& 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\
& 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\
& \therefore \mathrm{a}=3
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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