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Quick Question: Half Equations

I'm a little stuck with this question:

"Construct a half equation for the conversion on H2SO4, into H2S and water in the presence of XS acid"

Please could somebody tell me how to do it?!

Thanks!
t-a-r-online
I'm a little stuck with this question:

"Construct a half equation for the conversion on H2SO4, into H2S and water in the presence of XS acid" Please could somebody tell me how to do it?! Thanks!


All of the oxygen from the H2SO4 has to be absorbed by H+ ions to make water. There are 4 of them so this requires 8 H+ ions... but H2SO4 already has 2, leaving only 6 needed... but 2 more are needed to get the 2H+ in the H2S. So a total of 8H+ are needed.
The S from the H2SO4 goes from S(VI) to S(II-) i.e. 8 electrons are transferred.

so this gives:

H2SO4 + 8H+ + 8e --> H2S + 4H2O
Reply 2
Avatar for Borek
Borek
4
When it comes to half equations it is safer to not look at oxidation numbers, but use electrons to balance charge in the equation. Charco get to

H2SO4 + 8H+ --> H2S + 4H2O

just by balancing S, O and H. Now you have 8+ on the left, none on the right - so to balance charge you have to add 8e- on the left.

Check out lecture on half equations method.

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