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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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