In the first step, manganese(II) sulphate (at 48% of the total volume) is added to an environmental water sample.  Next, potassium iodide (15% in potassium hydroxide 70%) is added to create a pinkish-brown precipitate.  In the alkaline solution, dissolved oxygen will oxidize manganese(II) ions to the tetravalent state.
: 2 Mn + O + 2 HO  → 2 MnO(OH)
Mn has been oxidised to 4+, and MnO(OH) appears as a brown precipitate.  There is some uncertainty about whether the oxidised manganese is tetravalent or trivalent. Some sources claim that Mn(OH) is the brown precipitate, but hydrated MnO may also give the brown colour.
: 4 Mn(OH) + O + 2 HO → 4 Mn(OH)
The second part of the Winkler test reduces (acidifies) the solution.  The precipitate will dissolve back into solution as the H reacts with the O and OH to form water.
: MnO(OH) + 4 H → Mn + 3 HO
The acid facilitates the conversion by the brown, Manganese-containing precipitate of the Iodide ion into elemental Iodine.
The Mn(SO) formed by the acid converts the iodide ions into iodine, itself being reduced back to manganese(II) ions in an acidic medium.
: Mn(SO) + 2 I →  Mn + I + 2 
Thiosulfate is used, with a starch indicator, to titrate the iodine.
: 2  + I →  + 2 I