# How do you balance NaClO_3 -> NaCl + O_2?

Feb 16, 2016

$2 N a C l {O}_{3} \left(s\right) \to 2 N a C l \left(s\right) + 3 {O}_{2} \left(g\right)$
or
$N a C l {O}_{3} \left(a q\right) + N {a}_{2} S {O}_{3} \left(a q\right) \to N a C l \left(a q\right) + {O}_{2} \left(g\right) + N {a}_{2} S {O}_{4} \left(a q\right)$

#### Explanation:

To balance the reaction, if it is happening at the solid state (decomposition):

$2 N a C l {O}_{3} \left(s\right) \to 2 N a C l \left(s\right) + 3 {O}_{2} \left(g\right)$

However, to balance the reaction, if it is happening in aqueous medium:

$N a C l {O}_{3} \left(a q\right) \to N a C l \left(a q\right) + {O}_{2} \left(g\right)$

We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from $\left(- 2\right)$ in $N a C l {O}_{3}$ to $\left(0\right)$ in ${O}_{2}$.

The oxidation numbers for $N a$ and $C l$ are not changing and they are $\left(+ 1\right)$ and $\left(- 1\right)$ respectively.

To balance this reaction then:
$N a C l {O}_{3} \left(a q\right) + 2 {H}^{+} \left(a q\right) + 2 {e}^{-} \to N a C l \left(a q\right) + {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$

Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent $N a C l {O}_{3}$ such as for example $N {a}_{2} S {O}_{3}$.

The oxidation half equation will then be:

$S {O}_{3}^{2 -} \left(a q\right) + {H}_{2} O \left(l\right) \to S {O}_{4}^{2 -} \left(a q\right) + 2 {H}^{+} \left(a q\right) + 2 {e}^{-}$

The redox reaction will thus be:

Oxidation: $S {O}_{3}^{2 -} \left(a q\right) + \cancel{\textcolor{g r e e n}{{H}_{2} O \left(l\right)}} \to S {O}_{4}^{2 -} \left(a q\right) + \cancel{\textcolor{b l u e}{2 {H}^{+} \left(a q\right)}} + \cancel{\textcolor{red}{2 {e}^{-}}}$

Reduction: $N a C l {O}_{3} \left(a q\right) + \cancel{\textcolor{b l u e}{2 {H}^{+} \left(a q\right)}} + \cancel{\textcolor{red}{2 {e}^{-}}} \to N a C l \left(a q\right) + {O}_{2} \left(g\right) + \cancel{\textcolor{g r e e n}{{H}_{2} O \left(l\right)}}$

RedOx:
$N a C l {O}_{3} \left(a q\right) + S {O}_{3}^{2 -} \left(a q\right) \to N a C l \left(a q\right) + {O}_{2} \left(g\right) + S {O}_{4}^{2 -} \left(a q\right)$
or
$N a C l {O}_{3} \left(a q\right) + N {a}_{2} S {O}_{3} \left(a q\right) \to N a C l \left(a q\right) + {O}_{2} \left(g\right) + N {a}_{2} S {O}_{4} \left(a q\right)$

Here is a video that explains how to balance a redox reaction in acidic medium:
Balancing Redox Reactions | Acidic Medium.

And here is another one on how to balance a redox reaction in basic medium:
Balancing Redox Reactions | Basic Medium.