Given below are two statements:Assertion (A): | vedclass

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(A) The standard reduction potential for $Cu^{2+} + e^{-} \rightarrow Cu^{+}$ is $+0.15 \ V$ and for $I_2 + 2e^{-} \rightarrow 2I^{-}$ is $+0.54 \ V$.

Vedclass · Accepted Answer

Both $(A)$ and $(R)$ are correct and $(R)$ is the correct explanation of $(A)$

Vedclass · Answer

(A) The standard reduction potential for $Cu^{2+} + e^{-} \rightarrow Cu^{+}$ is $+0.15 \ V$ and for $I_2 + 2e^{-} \rightarrow 2I^{-}$ is $+0.54 \ V$.Since the reduction potential of $I_2$ is higher,$Cu^{2+}$ is a strong enough oxidizing agent to oxidize $I^{-}$ to $I_2$.The reaction is: $2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2$.Because $Cu^{2+}$ oxidizes $I^{-}$ to $I_2$,$CuI_2$ is unstable and does not exist; instead,$CuI$ is formed.Thus,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

Given below are two statements:
Assertion $(A)$: All $Cu$ $(II)$ halides are known except the iodide.
Reason $(R)$: $Cu^{2+}$ oxidizes $I^{-}$ to $I_2$.
The correct answer is:

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Given below are two statements:Assertion $(A)$: All $Cu$ $(II)$ halides are known except the iodide.Reason $(R)$: $Cu^{2+}$ oxidizes $I^{-}$ to $I_2$.The correct answer is: