EC_JEE-Main_29-06-2022_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE Advanced 2025 - Mathematics

Numerical Type

Question

The cell potential for the given cell at $298 \mathrm{~K} \mathrm{Pt}\left|\mathrm{H}_2(\mathrm{~g}, 1 \mathrm{bar})\right| \mathrm{H}^{+}(\mathrm{ag}) \| \mathrm{Cu}^{2+}(\mathrm{aq}) \mid \mathrm{Cu}(\mathrm{s})$ is 0.31 V . The pH of the acidic solution is found to be 3 , whereas the concentration of $\mathrm{Cu}^{2+}$ is $10^{-x} \mathrm{M}$. The value of $x$ is $\_\_\_\_$ .
(Given: $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Ou}}^{\ominus}=0.34 \mathrm{~V}$ and $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$

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Solution

$\begin{aligned} & \mathrm{H}_2(\mathrm{~g})+\mathrm{Cu}^{2+}(\text { aq. }) \rightarrow 2 \mathrm{H}^{+}(\text {aq. })+\mathrm{Cu}(\mathrm{s}) \\ & 0.31=0.34-\frac{0.06}{2} \log \frac{\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{Cu}^{2+}\right]} \\ & {\left[\mathrm{Cu}^{2+}\right]=10^{-7} \mathrm{M}} \\ & \mathrm{X}=7\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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