# How do you balance redox equations using the oxidation number change method?

## For example: $C l {O}_{3}^{-} \left(a q\right) + {I}^{-} \left(a q\right) \to C {l}^{-} \left(a q\right) + {I}_{2} \left(a q\right)$ [acidic solution

Feb 4, 2018

You introduce electrons as virtual particles and assign oxidation states...according to given rules.

#### Explanation:

Chlorate is REDUCED to chloride....$C l \left(+ V\right) \rightarrow C l \left(- I\right)$..and we account for the difference in oxidation number $+ V - \left(- I\right) = 6$ by the addition of electrons....

$C l {O}_{3}^{-} + 6 {H}^{+} + {\underbrace{6 {e}^{-}}}_{\text{i.e. +V/-I}} \rightarrow C {l}^{-} + 3 {H}_{2} O \left(l\right)$ $\left(i\right)$

Iodide is oxidized to iodine...$I \left(- I\right) \rightarrow {I}_{2} \left(0\right)$

${I}^{-} \rightarrow \frac{1}{2} {I}_{2} + {\underbrace{{e}^{-}}}_{\text{-I/0}}$ $\left(i i\right)$

In each case we account for the difference in oxidation number by the addition (reduction) or subtraction (oxidation) of electrons....

We take $\left(i\right) + 6 \times \left(i i\right) :$

$C l {O}_{3}^{-} + 6 H I \left(a q\right) + 6 {e}^{-} \rightarrow 3 {I}_{2} \left(s\right) + 6 {e}^{-} + C {l}^{-} + 3 {H}_{2} O$

....to give finally after cancellation...

$C l {O}_{3}^{-} + 6 H I \left(a q\right) \rightarrow 3 {I}_{2} \left(s\right) + C {l}^{-} + 3 {H}_{2} O$

The which is balanced with respect to mass and charge....as indeed it must be if we purport to represent chemical reality....

Feb 4, 2018

WARNING! Long answer! The balanced equation is

$\text{ClO"_3^"-" + "6I"^"-" + "6H"^"+" → "Cl"^"-" + "3I"_2 + "3H"_2"O}$

#### Explanation:

The unbalanced equation is

${\text{ClO"_3^"-" + "I"^"-" → "Cl"^"-" + "I}}_{2}$

Step 1. Identify the atoms that change oxidation number

Start by determining the oxidation numbers of every atom in the equation.

Ignore $\text{O}$. It comes in automatically during the balancing procedure.

$\stackrel{\textcolor{b l u e}{\text{+5")("Cl")"O"_3^"-" + "I"^"-" → "Cl"^"-" + stackrelcolor(blue)(0)("I}}}{_} 2$

We see that the oxidation number of $\text{Cl}$ decreases from +5 to -1 and the oxidation number of $\text{I}$ increases from -1 to 0.

The changes in oxidation number are:

$\text{Cl: +5 → -1"; "Change ="color(white)(m) "-6 (reducttion)}$
$\text{I: -1 → 0"; color(white)(mll)"Change ="color(white)(ll) "+1 (oxidation)}$

Step 2. Equalize the changes in oxidation number

We need 6 atoms of $\text{I}$ for every 1 atom of $\text{Cl}$.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} {\text{ClO"_3^"-" + color(red)(6)"I"^"-" → color(red)(1)"Cl"^"-" + color(red)(3)"I}}_{2}$

Step 4. Balance $\text{O}$

Add $\text{H"_2"O}$ molecules to the deficient side.

$\textcolor{red}{1} \text{ClO"_3^"-" + color(red)(6)"I"^"-" → color(red)(1)"Cl"^"-" + color(red)(3)"I"_2 + color(blue)(3)"H"_2"O}$

Step 5. Balance $\text{H}$

Add $\text{H"^"+}$ ions to the deficient side.

$\textcolor{red}{1} \text{ClO"_3^"-" + color(red)(6)"I"^"-" + color(brown)(6)"H"^"+" → color(red)(1)"Cl"^"-" + color(red)(3)"I"_2 + color(blue)(3)"H"_2"O}$

Every formula now has a coefficient. The equation should be balanced.

Step 6. Check that all atoms are balanced.

$\underline{\boldsymbol{\text{Atom"color(white)(m)"On the left"color(white)(m) "On the right}}}$
$\textcolor{w h i t e}{m} \text{Cl} \textcolor{w h i t e}{m m m m m} 1 \textcolor{w h i t e}{m m m m m m l l} 1$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m m l} 3 \textcolor{w h i t e}{m m m m m m l l} 3$
$\textcolor{w h i t e}{m} \text{I} \textcolor{w h i t e}{m m m m m l l} 6 \textcolor{w h i t e}{m m m m m m l l} 6$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m l} 6 \textcolor{w h i t e}{m m m m m m l l} 6$

Step 7. Check that charge is balanced.

$\underline{\boldsymbol{\text{On the left"color(white)(m) "On the right}}}$
$\textcolor{w h i t e}{m m m} \text{1-"color(white)(mmmmmm)"1-}$

The balanced equation is

$\textcolor{b l u e}{\text{ClO"_3^"-" + "6I"^"-" + "6H"^"+" → "Cl"^"-" + "3I"_2 + "3H"_2"O}}$