Identify the oxidant and reductant in the following reactions:
`I_2(g)+H_2S(g)to2IH(g) +S(s)`.

Explanation

Writing the oxidation number of all the atoms above their symbols, we have
`overset0(I_2)(g)+overset(+1)(H_2)overset(-2)(S)(g)tooverset(+1)(2H)overset(-1)I(g) +overset(0)S(s)`
The oxidation number of `I_2` decreases from zero in `I_2` to -1 in HI, therefore, `I_2` is reduced and hence it acts as an oxidant.
THe O.N of S increase from -2 in `H_2S` to zero in S, therefore , `H_2 S` is oxidised and hence it acts as the reductant.