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    as above.. not in google.
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    It's a redox reaction.

    Do your half equations, but note that iodide is strong enough to reduce sulphuric acid all the way to hydrogen sulphide.
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    (Original post by JamesBlood)
    It's a redox reaction.

    Do your half equations, but note that iodide is strong enough to reduce sulphuric acid all the way to hydrogen sulphide.
    No wonder I couldn't find the equation on google LOl. Thanks alot.


    And we get sulfur on it's own as well from the reduced reaction ryt. :P
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    You'd probably get bits of sulphur dioxide and sulphur (oxidation states of +4 and 0 respectively), but like I said the I- can reduce it all the way to H2S (-2), so that would be the major product. In your half equations, you can probably discount sulphur dioxide and elemental sulphur.
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    (Original post by JamesBlood)
    It's a redox reaction.

    Do your half equations, but note that iodide is strong enough to reduce sulphuric acid all the way to hydrogen sulphide.
    This OP...
 
 
 
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