Question

$aC{u}^{2+}\left(aq\right)+b{I}^{-}\left(aq\right)\to cCuI\left(s\right)+d{I}_2\left(s\right)$

When the equation above is balanced and all coefficients are reduced to lowest whole-number terms, what will be the coefficient for $I^{-}$?

• 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 $+1$.
The 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 $=1$.
The 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 coefficient 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.