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(C) In the reaction $Cu^{2+} (aq.) + 2I^{-} \longrightarrow CuI \downarrow + \frac{1}{2}I_2$,the oxidation state of $Cu$ changes from $+2$ to $+1$ (re

Vedclass · Accepted Answer

for either intermolecular redox reaction or displacement reaction.

Vedclass · Answer

(C) In the reaction $Cu^{2+} (aq.) + 2I^{-} \longrightarrow CuI \downarrow + \frac{1}{2}I_2$,the oxidation state of $Cu$ changes from $+2$ to $+1$ (reduction).The oxidation state of $I$ changes from $-1$ to $0$ (oxidation).Since two different species are being oxidized and reduced in an intermolecular process,this is an intermolecular redox reaction.Also,it can be classified as a displacement reaction where $I^-$ displaces $Cu^{2+}$ to form $CuI$.Therefore,the correct classification is $C$.

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$Cu^{2+} (aq.) + 2I^{-} \longrightarrow CuI \downarrow + \frac{1}{2}I_2$

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Assign $A$,$B$,$C$,$D$ from the given type of reaction.$Cu^{2+} (aq.) + 2I^{-} \longrightarrow CuI \downarrow + \frac{1}{2}I_2$