# Question def93

Aug 21, 2016

${\text{NaOCl"_ ((aq)) + "Na"_ 2"SO"_ (3(aq)) -> "NaCl"_ ((aq)) + "Na"_ 2"SO}}_{4 \left(a q\right)}$

#### Explanation:

Yes, you are dealing with a redox reaction, but you got the products wrong here.

For starters, the sodium cations are spectator ion, so you can discard them from the reaction altogether.

The idea here is that the hypochlorite anions, ${\text{ClO}}^{-}$, which are delivered to the reaction by sodium hypochlorite, $\text{NaOCl}$, can be reduced using sulfite anions, ${\text{SO}}_{3}^{2 -}$, which are delivered to the reaction by sodium sulfite, ${\text{Na"_2"SO}}_{3}$, to chloride anions, ${\text{Cl}}^{-}$.

The sulfite anions get oxidized to the sulfate anions, ${\text{SO}}_{4}^{2 -}$.

The balanced net ionic equation looks like this -- remember, the sodium cations are omitted here!

${\text{ClO"_ ((aq))^(-) + "SO"_ (3(aq))^(2-) -> "Cl"_ ((aq))^(-) + "SO}}_{4 \left(a q\right)}^{2 -}$

To better visualize that this is indeed a redox reaction, assign oxidation numbers to the atoms that take part in the reaction

stackrel(color(blue)(+1))("Cl") stackrel(color(blue)(-2))("O")""_ ((aq))^(-) + stackrel(color(blue)(+4))("S") stackrel(color(blue)(-2))("O")""_ (3(aq))^(2-) -> stackrel(color(blue)(-1))("Cl")""_ ((aq))^(-) + stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")""_ (4(aq))^(2-)#

As you can see, the oxidation number of chlorine goes from $\textcolor{b l u e}{+ 1}$ on the reactants' side to $\textcolor{b l u e}{- 1}$ on the products' side, which means that chlorine is being reduced.

On the other hand, the oxidation number of sulfur goes from $\textcolor{b l u e}{+ 4}$ on the reactants' side to $\textcolor{b l u e}{+ 6}$ on the products' side, which means that sulfur is being oxidized.

This confirms that the sulfite anions are acting as reducing agents because they are reducing the hypochlorite anions to chloride anions.

Similarly, the hypochlorite anions are acting as oxidizing agents because they are oxidizing the sulfite anions to sulfate anions.

The complete chemical equation is

${\text{NaOCl"_ ((aq)) + "Na"_ 2"SO"_ (3(aq)) -> "NaCl"_ ((aq)) + "Na"_ 2"SO}}_{4 \left(a q\right)}$