Cu^{2+} salt reacts with potassium iodide to | vedclass

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Question

(B) When $Cu^{2+}$ salts react with potassium iodide $(KI)$, the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by the iodide ions $(I^{-})$.The chemical

Vedclass · Accepted Answer

$CuI$

Vedclass · Answer

(B) When $Cu^{2+}$ salts react with potassium iodide $(KI)$, the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by the iodide ions $(I^{-})$.The chemical reaction is as follows:$2Cu^{2+} + 4I^{-} \longrightarrow 2CuI(s) + I_{2}$Here, $CuI$ is formed as a white precipitate, and $I_{2}$ is liberated.Since $CuI$ is the standard representation for copper$(I)$ iodide in this context, the correct option is $B$.

$Cu^{2+}$ salt reacts with potassium iodide to give:

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