Given below are two statements Assertion (A): All Cu (II) halides are known except the iodide Reason (R):…
Given below are two statements
Assertion (A): All Cu (II) halides are known except the iodide
Reason (R): $\mathrm{Cu}^{2+}$ oxidizes $\mathrm{I}^{-}$to $\mathrm{I}_2$
The correct answer is
Both (A) and (R) are correct and (R) is the correct explanation of $(\mathrm{A})$
Both (A) and (R) are correct and (R) is the correct but $(\mathrm{R})$ is not the correct explanation of $(\mathrm{A})$
(A) is correct but (R) is not correct
(A) is not correct but ( $\mathrm{R})$ is correct
Solution
$\begin{aligned} & \text { } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{I}_2 / \mathrm{I}^{-}}^{\circ}=0.54 \\ & 2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow 2 \mathrm{CuI}+\mathrm{I}_2\end{aligned}$
Despite the above values of the standard reduction potentials, $\mathrm{Cu}^{2+}$ is able to oxidize iodide to $\mathrm{I}_2$ due to all the species like $\mathrm{Cu}^{2+}, \mathrm{Cu}^{+}$, and $\mathrm{CuI}$ being present together in the solution.
The cumulative effect of this results in the $\mathrm{Cu}^{2+} \rightarrow \mathrm{CuI}$ reduction potential being $+0.88 \mathrm{~V}$ that makes the reaction feasible.
Therefore, $\mathrm{CuI}$ or $\mathrm{Cu}_2 \mathrm{I}_2$ exists instead of $\mathrm{CuI}_2$ as $\mathrm{Cu}^{2+}$ oxidises $\mathrm{I}^{-}$to $\mathrm{I}_2$ and so (A) and (R) are true with (R) explaining (A).