Cu^{2+} does not form CuI_2. This is because | vedclass

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(C) The standard reduction potential of $Cu^{2+} / Cu^{+}$ is $+0.15 \ V$,while the oxidation potential of $I^{-} / I_2$ is $-0.54 \ V$. Because the $

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oxidizes $I^{-}$ to $I_2$

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(C) The standard reduction potential of $Cu^{2+} / Cu^{+}$ is $+0.15 \ V$,while the oxidation potential of $I^{-} / I_2$ is $-0.54 \ V$. Because the $Cu^{2+}$ ion has a high reduction potential,it acts as a strong oxidizing agent. It readily oxidizes the iodide ion $(I^{-})$ to iodine $(I_2)$ and itself gets reduced to $Cu^{+}$. Therefore,$2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2$ occurs,and $CuI_2$ cannot be formed.

$Cu^{2+}$ does not form $CuI_2$. This is because $Cu^{2+}$

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