Question #def93
1 Answer
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,
The sulfite anions get oxidized to the sulfate anions,
The balanced net ionic equation looks like this -- remember, the sodium cations are omitted here!
#"ClO"_ ((aq))^(-) + "SO"_ (3(aq))^(2-) -> "Cl"_ ((aq))^(-) + "SO"_ (4(aq))^(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
On the other hand, the oxidation number of sulfur goes from
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
#"NaOCl"_ ((aq)) + "Na"_ 2"SO"_ (3(aq)) -> "NaCl"_ ((aq)) + "Na"_ 2"SO"_ (4(aq))#