The reaction between sodium metabisulfite (Na2S2O5) and bromine (Br2) is a redox reaction, where sodium metabisulfite acts as the reducing agent and bromine acts as the oxidizing agent. The exact reaction can be represented by the following chemical equation:

Na2S2O5 + Br2 + H2O → Na2SO4 + 2HBr + SO2

In this reaction, sodium metabisulfite undergoes oxidation, where the sulfur atom in the metabisulfite ion (S2O5^2-) is oxidized from a +4 oxidation state to a +6 oxidation state, forming sulfate ions (SO4^2-). On the other hand, bromine undergoes reduction, where the bromine molecules (Br2) are reduced from a 0 oxidation state to a -1 oxidation state, forming bromide ions (Br-).

Additionally, water (H2O) participates in the reaction, and sulfuric acid (H2SO3) is formed as an intermediate product. However, H2SO3 is unstable and decomposes into sulfur dioxide (SO2) and water, which are the gaseous products of the reaction.

Overall, the reaction between sodium metabisulfite and bromine results in the formation of sodium sulfate, hydrogen bromide, sulfur dioxide, and water. This reaction has several applications, including the use of sodium metabisulfite as a reducing agent in various industrial processes, such as in photography, textile dyeing, and water treatment.