Question #ae919

You're dealing with a disproportionation reaction, which is basically a redox reaction in which the same chemical species is both oxidized and reduced.

You can see this by assigning oxidation numbers to the atoms that take part in the reaction

#8stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O")_ (2(aq)) + 16stackrel(color(blue)(+1))("H")_ 2stackrel(color(blue)(-2))("S")_ ((aq)) -> 3stackrel(color(blue)(0))("S")_ (8(s)) darr + 16stackrel(color(blue)(+1))("H")_ 2stackrel(color(blue)(-2))("O")_((l))#

Notice that the sulfur atoms that are present in sulfur dioxide are being reduced, since their oxidation state decreases from #color(blue)(+4)# to #color(blue)(0)# in solid sulfur.

On the other hand, the sulfur atoms that are present in the hydrogen sulfide are being oxidized, since their oxidation state is increasing from #color(blue)(-2)# to #color(blue)(0)#.

In this case, you're dealing with a disproportionation reaction because sulfur is being oxidized and reduced at the same time.

So, sulfur dioxide will act as an oxidizing agent because it oxidizes the sulfur atoms in hydrogen sulfide to solid sulfur. Likewise, hydrogen sulfide will act as a reducing agent because it reduces the sulfur atoms in sulfur dioxide to solid sulfur.