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Originally Posted by n0nsensical
well for every mole of I- oxidized, 1/2 mole of I2 and 1 mole of electrons are produced. This is represented by the half-reaction
2I- -> I2 + 2e-
The reduction of SO4(2-) is represented by
SO4(2-) + 4H+ + 2e- -> SO2 + 2H2O
You get that by balancing SO4(2-) -> SO2 with H+, e-, and H2O so that charge and mass are conserved. Then add the half-reactions to cancel electrons and get an overall balanced reaction. Thats redox in a nutshell, balancing and adding half-reactions.
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ghoast
ok, take the equations above given by
non: these should be available in the appendix in your chem book. i did a quick google, and found some online tutorials, but not the actual electrochem equations. if you have the table, that's a start.
if you take the reduction equation, 2I- -> I2 + 2e-, first notice that the overall charge balances on both sides. I-*2=2negative charges, and the 2e- on the right hand side.
now take the reduction: SO4(2-) + 4H+ + 2e- -> SO2 + 2 H2O ...once again, the charges balance. zero on the left hand side, and overall zero on the right hand side. Now add them:
2I- -> I2 + 2e-
SO4(2-) + 4H+ + 2e- -> SO2 + 2H2O
the 2e- cancel out, leaving
2I- + SO4(2-) + 4H+ -> SO2 + 2H2O + I2
So I'm getting the mole ratio to be (1:1), using the information
non gave before. These reaction are typically given as standard reduction reactions. The important thing is that the number of e- have to balance. The half reactions are written such that you will always balance them if you multiply both left and right hand sides by a constant factor. So:
4I- -> 2I2 + 4e-
I- -> (1/2)I2 + e-
etc.
you multiply the equations, essentially finding the lowest common denominator for the number of e-, and then add the half reactions to get the overall balanced reaction.
I don't know if that helps, but I've had some experience with this stuff, and will give what help I can. If you can tell me which part is causing the confusion for you, I might be able to help give you some hints. Free of charge
God, that turned into an awful pun..