# How do you balance this redox reaction using the oxydation number method ? KClO_3 + H_2SO_4 -> ClO_4^(-) + ClO_2 + SO_4^(2-) + K^(+) + H_2O

May 12, 2015

Warning! This is a long answer. The balanced equation is

$\text{3KClO"_3 + "H"_2"SO"_4 → "ClO"_4^(-) + 2"ClO"_2 + "SO"_4^(2-) + 3"K"^+ + "H"_2"O}$

$\text{KClO"_3 + "H"_2"SO"_4 → "ClO"_4^(-) + "ClO"_2 + "SO"_4^(2-) + "K"^+ + "H"_2"O}$

Step 1. Identify the atoms that change oxidation number

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{K") stackrel(color(blue)(+5))("Cl") stackrel(color(blue)(-2))("O")_3 + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_4 → (stackrel(color(blue)(+7))("Cl") stackrel(color(blue)(-2))("O")_4)^(-) + stackrel(color(blue)(+4))("Cl") stackrel(color(blue)(-2))("O")_2 + (stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_4)^(2-) + (stackrel(color(blue)(+1))("K"))^+ + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(-2))("O}}$

Left hand side: $\text{K} = + 1$; $\text{Cl} = + 5$; $\text{O} = - 2$; $\text{H} = + 1$; $\text{S} = + 6$

Right hand side: ${\text{Cl in ClO}}_{4}^{-} = + 7$; $\text{O} = - 2$; ${\text{Cl in ClO}}_{2} = + 4$; $\text{S} = + 6$; $\text{K} = + 1$; $\text{H} = + 1$

The changes in oxidation number are:

$\text{Cl}$: +5 → +7 in ${\text{ClO}}_{4}^{-}$; Change =+2
$\text{Cl}$: +5 → +4 in ${\text{ClO}}_{2}$; Change = -1

Step 2. Equalize the changes in oxidation number

We need 2 atoms of $\text{Cl}$ in ${\text{ClO}}_{2}$ for every 1 atom of $\text{Cl}$ in ${\text{ClO}}_{4}^{-}$. This gives us total changes of -2 and +2.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{3} \text{KClO"_3 + "H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + "SO"_4^(2-) + "K"^+ + "H"_2"O}$

Step 4. Balance $\text{K}$ by adding ${\text{K}}^{+}$ ions to the appropriate side

$\textcolor{red}{3} \text{KClO"_3 + "H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + "SO"_4^(2-) + color(red)(3)"K"^+ + "H"_2"O}$

Step 5. Balance charge

We have no net charge on the left or on the right, so there must be only 1 ${\text{SO}}_{4}^{2 -}$

$\textcolor{red}{3} \text{KClO"_3 + "H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + color(green)(1)"SO"_4^(2-) + color(red)(3)"K"^+ + "H"_2"O}$

Step 6. Balance $\text{S}$

We have 1 $\text{S}$ atom on the right, so we must have 1 $\text{S}$ atom on the left

$\textcolor{red}{3} \text{KClO"_3 + color(purple)(1)"H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + color(green)(1)"SO"_4^(2-) + color(red)(3)"K"^+ + "H"_2"O}$

Step 7. Balance $\text{H}$

We have 2 $\text{H}$ on the left, so we need 2 $\text{H}$ on the right.

$\textcolor{red}{3} \text{KClO"_3 + color(teal)(1)"H"_2"SO"_4 → color(red)(1)"ClO"4^-) + color(red)(2)"ClO"_2 + color(green)(1)"SO"_4^(2-) + color(red)(3)"K"^+ + color(orange)(2)"H"_2"O}$

The balanced equation is

$\textcolor{red}{\text{3KClO"_3 + "H"_2"SO"_4 → "ClO"_4^(-) + 2"ClO"_2 + "SO"_4^(2-) + 3"K"^+ + "H"_2"O}}$