Writing the ON of all the atoms above their symbols, we have

The ON of I_{2} decreases from zero in I_{2} to -1 in HI, therefore, I_{2} is reduced and hence it acts as on oxidant.

The O.N. of S increases from -2 in H_{2}S to zero in S, therefore, H_{2}S is oxidized and hence it acts as the reductant.