At 298 K, standard reduction... | Electrochemistry JEE Main

JEE MAIN 2023

13-4-23 S2

Question

At 298 K , the standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ electrode is 0.34 V . Given: $\mathrm{K}_{\text {sa }} \mathrm{Cu}(\mathrm{OH})_2=1 \times 10^{-20}$ [JEE (Main) 13-4-23_S2] [EC_H] M0derate Take $\frac{2.303 R T}{F}=0.059 \mathrm{~V}$ The reduction potential at $\mathrm{pH}=14$ for the above couple is $(-) \mathrm{x} \times 10^{-2} \mathrm{~V}$. The value of x is $\_\_\_\_$

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Solution

$\begin{aligned} & \mathrm{Cu}(\mathrm{OH})_2(\mathrm{~s}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{ag})+2 \mathrm{OH}^{-}(\mathrm{ag}) \\ & \mathrm{K}_{\mathrm{seP}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & \mathrm{pH}=14 ; \mathrm{pOH}=0 ;\left[\mathrm{OH}^{-}\right]=1 \mathrm{M} \\ & \therefore\left[\mathrm{Cu}^{2+}\right]=\frac{\mathrm{K}_{\mathrm{sp}}}{[1]^2}=10^{-20} \mathrm{M} \\ & \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \\ & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log _{10} \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \\ & =0.34-\frac{0.059}{2} \log _{10} \frac{1}{10^{-20}} \\ & =-0.25=-25 \times 10^{-2}\end{aligned}$

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JEE Main

Chemistry

Medium

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