EC-M_JEE-Main_25-1-23{S2} - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2023

25-1-23 S2

Question

$\mathrm{Pt}(\mathrm{s}) \mathrm{H}_2(\mathrm{~g})(1 \mathrm{bar})\left|\mathrm{H}^{+}(\mathrm{aq})(1 \mathrm{M})\right|\left|\mathrm{M}^{3+}(\mathrm{aq}), \mathrm{M}^{+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$ $$ \text { The Ecell for the given cell is } 0.1115 \mathrm{~V} \text { at } 298 \mathrm{~K} \text { when } \frac{\left[\mathrm{M}^{+}(\mathrm{aq})\right]}{\left[\mathrm{M}^{3+}(\mathrm{aq})\right]}=10^{\mathrm{a}} \text {. The value of } \mathrm{a} \text { is } $$ $$ \text { Given: } \mathrm{E}_{\mathrm{M}^{3+} / \mathrm{M}^{+}}^{\ominus}=0.2 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V} $$

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Solution

Sol. Overall reaction: $$ \begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{sq})}^2 \longrightarrow \mathrm{M}_{(\mathrm{aq})}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cathode }}^s-\mathrm{E}_{\text {anode }}^s-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \end{aligned} $$ $\begin{aligned} & 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\ & 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\ & \therefore a=3\end{aligned}$

Question Tags

JEE Main

Chemistry

Medium

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