Since we do not consider Fluorine and Astatine in our reactions, we'll exclude them from this explanation. Hence, the remainder is chlorine, bromine and iodine. Down the group, the bond energy between the halogen atoms increases.
Cl2<Br2<I2
Also, the electron affinity decreases down the group:
Cl2>Br2>I2
Hence, Chlorine is the most electronegative whilst Iodine is the least. As a rule, the most electronegative element has the most oxidising power as it will readily accept an electron from a reductant.
When Chlorine dissolves in water, a disproportionation reaction occurs to form HCl and HOCl. HOCl is an unstable species and it decomposes to form HCl and O (reactive oxygen atoms), which is why it is important as bleach and in germicides in swimming pools. Hence, the actual halide in this reaction is only HCl.
However, in hydrogen halides there is a similar problem of oxidising ability reducing down the group. Hence, HCl is a weak reducing agent and strong oxidising agent. When it reacts as a metal halide with H2SO4, only HCl is formed (as an acid breaks up the salt) but it will not be oxidised by H2SO4, whilst that is not true for Br2 and I2.
When a metal bromide reacts with H2SO4, it also forms HBr initially. However, as it is not as strong as HCl when it comes to oxidising ability, it is oxidised by H2SO4 to Br2 and hence the following reactions take place:
NaBr + H2SO4 ----> HBr + NaHSO4
HBr + H2SO4 ----> Br2 + H2O + SO2 (not balanced)
In case of metal iodide, the same mechanism occurs but instead of just stopping at SO2, it is able to reduce H2SO4 to H2S, with the lowest oxidation state of Sulphur:
NaI + H2SO4 ----> HI + NaHSO4
HI + H2SO4 ----> I2 + H2S + H2O (not balanced)
This trend in the halides is due to the decreasing bond enthalpies of the H-X bond, which means each progressive halide is more unstable and hence is a better reducing agent, more willing to 'get rid' of the hydrogen.
