
Why is \[Cu{I_2}\] unstable?
Answer
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Hint: Any chemical compound is stable in nature if it is present in its most common oxidation number. Different compounds show reducing power and oxidizing power depending upon their position in electrochemical series.
Complete answer:
Electrochemical series contain large numbers of elements which are organized in a sequence of their oxidizing power and reducing power. When we look into the series, we find that the conversion of ${I_2} \to {I^ - }$ lies above the conversion of $C{u^{ + 2}} \to Cu$.
But when we reverse the conversion of ${I_2} \to {I^ - }$ into ${I^ - } \to {I_2}$, iodide ion acts as a strong reducing agent and has tendency to convert $C{u^{ + 2}} \to C{u^ + }$.
Elements which lie atop of the electrochemical series have high value of electrode potential and therefore they easily undergo reduction and act as a good excellent oxidizing agent.
When $C{u^{ + 2}}$ ions reacts with ${I^ - }$ ions to form the compound \[Cu{I_2}\].
Redox reaction between ions is expressed as:
$C{u^{ + 2}}$$ + $${I^ - }$$ \to $$Cu{I_2}$
In the above reaction we see that iodide ions get converted into iodine and hence, a process of oxidation takes place.
But from the above discussion we already see that conversion of ${I_2} \to {I^ - }$ is more stable according to the electrochemical series. Therefore, formation of $Cu{I_2}$ is not preferred instead of which another halide of copper is formed which is stable in nature.
The redox reaction to from stable compound is expressed as-
${I_{2\left( {aq} \right)}} + 2C{u^ + }_{\left( {aq} \right)} \to 2Cu{I_{\left( S \right)}}$
$ \Rightarrow $ \[Cu{I_2}\] is unstable in nature because iodide ion act as very good reducing agent and it promote conversion of cupric ion $C{u^{ + 2}}$ into cuprous ion $C{u^ + }$ which further form stable compound $CuI$.
Note:
Electrochemical series also help us to determine the chances of feasibility of the redox reaction. According to the series the chemical reaction is feasible only if compounds with higher reduction potential get reduced during the reaction.
Complete answer:
Electrochemical series contain large numbers of elements which are organized in a sequence of their oxidizing power and reducing power. When we look into the series, we find that the conversion of ${I_2} \to {I^ - }$ lies above the conversion of $C{u^{ + 2}} \to Cu$.
But when we reverse the conversion of ${I_2} \to {I^ - }$ into ${I^ - } \to {I_2}$, iodide ion acts as a strong reducing agent and has tendency to convert $C{u^{ + 2}} \to C{u^ + }$.
Elements which lie atop of the electrochemical series have high value of electrode potential and therefore they easily undergo reduction and act as a good excellent oxidizing agent.
When $C{u^{ + 2}}$ ions reacts with ${I^ - }$ ions to form the compound \[Cu{I_2}\].
Redox reaction between ions is expressed as:
$C{u^{ + 2}}$$ + $${I^ - }$$ \to $$Cu{I_2}$
In the above reaction we see that iodide ions get converted into iodine and hence, a process of oxidation takes place.
But from the above discussion we already see that conversion of ${I_2} \to {I^ - }$ is more stable according to the electrochemical series. Therefore, formation of $Cu{I_2}$ is not preferred instead of which another halide of copper is formed which is stable in nature.
The redox reaction to from stable compound is expressed as-
${I_{2\left( {aq} \right)}} + 2C{u^ + }_{\left( {aq} \right)} \to 2Cu{I_{\left( S \right)}}$
$ \Rightarrow $ \[Cu{I_2}\] is unstable in nature because iodide ion act as very good reducing agent and it promote conversion of cupric ion $C{u^{ + 2}}$ into cuprous ion $C{u^ + }$ which further form stable compound $CuI$.
Note:
Electrochemical series also help us to determine the chances of feasibility of the redox reaction. According to the series the chemical reaction is feasible only if compounds with higher reduction potential get reduced during the reaction.
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