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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 $C$.

List-$I$	List-$II$
$P. \ PbO_2 + H_2SO_4 \xrightarrow{?} PbSO_4 + O_2 + \text{other product}$	$1. \ NO$
$Q. \ Na_2S_2O_3 + H_2O \xrightarrow{?} NaHSO_4 + \text{other product}$	$2. \ I_2$
$R. \ N_2H_4 \xrightarrow{?} N_2 + \text{other product}$	$3. \ \text{Warm}$
$S. \ XeF_2 \xrightarrow{?} Xe + \text{other product}$	$4. \ Cl_2$

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.$Cu^{2+} (aq.) + 2I^{-} \longrightarrow CuI \downarrow + \frac{1}{2}I_2$