I think it is 10e for 2 moles or 5/mole.
When one has I on one side that is both oxidized and reduced, this is the way I handle it.
(from KI) I^- ==> I + e
(from KIO3) IO3^- + 5e ==> I
3CoSO4 + 5KI + KIO3 + 3H2O ==> 3Co(OH)2 + 3K2SO4 + I + 5I (which after balancing can be replaced with 3I2). If you do it another way you almost always end up with double the coefficients and they must be reduced by a factor of two.
CoSO4 + KI + KIO3 + H2O Co(OH)2 + K2SO4 + I2
2KI yield I2 + 2e
2KIO2 10e yield I2
are there 10 electrons being transferred or just 5 only
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