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    Hi guys,
    "When either solid KBr or solid KI reacts with concentrated sulphuric acid, a mixture of gases is evolved. In each case, one of the gases is produced by oxidation.
    Give the formula of the oxidation product formed in each reaction and state how this product would be observed"

    Im a bit confused as to which product to choose, so eg for KBr with H2SO4, the products are KHSO4+HBr,the HBr is then strong enough reductant to further reduce H2SO4 to SO2, giving other products of water and BR2 gas.
    I just dont know which ones are the oxidation products, and which the reduction products
    Thanks
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    Reduction products are probably what the sulphuric acid gets reduced to;
    so from the reduction reaction, SO2 is the reduction product with the KBr one and with KI, I think the Iodine reduces Sulphuric acid right down to H2S (Hydrogen sulphide gas) which is the reduction product, ie the product of the reaction equation

    I think, but I'm not sure, that the oxidation product would be Bromine gas, as the oxidation state goes from -1 to 0 in HBr to Br2

    I dont think the oxidation state changes in hydrogen or in oxygen from H2SO4 to H2O, as from KBr to HBr, the bromine doesn't change oxidation state, neither does the sulphur in H2SO4 to KHSO4. I'm being cautious here but what I said earlier is what I think would be the answers. What were you thinking?
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    In the case of Br- ions, Br- is oxidized by SO42- into Br2 and converts the sulphate to SO2 gas. Since the sulphate is being reduced (check the S oxidation numbers if you want - +6 in the sulphate, +4 in the sulphur dioxide) it is not the gas produced by oxidation. Instead, Br2 is (you observe an orange vapour for bromine gas). Similarly, in the case of I- ions, you don't even need to know how far the reduction of sulphate by iodide goes, because you're asked for the oxidation product, which will simply be I2 gas (purple vapour observed).

    For interest's sake and because you probably do have to know it, you'll get a mixture of SO2 gas, S (cyclic S8 really, you never get S on its own) solid and H2S gas from reduction of the sulphate ion by iodide ions. With sufficient excess of iodide you will get complete conversion to H2S gas.
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    Ah thank you so much guys i get it now
 
 
 
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