Question

${Cu}^{2+}\left(aq\right)+I^{-}\left(aq\right)\to CuI\left(s\right)+I_2\left(s\right)$
After balancing the above equation to lowest whole number, coefficient for $I^{-}\left(aq\right)$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. $5$

Answer 1

The correct option is D $4$

The unbalanced redox equation is:
$C{u}^{2+}\left(aq\right)+I^{-}\left(aq\right)\to CuI\left(s\right)+I_2\left(s\right)$.
Cu atoms are already balanced. Balance iodine atoms.
$C{u}^{2+}\left(aq\right)+3I^{-}\left(aq\right)\to CuI\left(s\right)+I_2\left(s\right)$
The oxidation number of $Cu$ changes from $+2$ to $+1$.
Total decrease in the oxidation number of $Cu=1$.
The oxidation number of iodine increases from $-1$ to $0$.
Increase in oxidation number of one iodine atom = $1$.
Total increase in oxidation number of two iodine atoms = $2$.
Balance increase in oxidation number with decrease in oxidation number by multiplying copper species with coefficent $2$.
$2C{u}^{2+}\left(aq\right)+3I^{-}\left(aq\right)\to 2CuI\left(s\right)+I_2\left(s\right)$
Balance iodine atoms.
$2C{u}^{2+}\left(aq\right)+4I^{-}\left(aq\right)\to 2CuI\left(s\right)+I_2\left(s\right)$
This is the balanced chemical equation. The coefficient for iodide ion $I^{-}$ is 4.