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
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.
To determine the number of electrons being transferred in a chemical reaction, we need to consider the half-reactions involved. In this case, we have two half-reactions:
1. 2KI → I2 + 2e
2. 2KIO3 → I2 + 10e
From the first half-reaction, we can see that 2 moles of KI yield 1 mole of I2 and release 2 moles of electrons (2e).
From the second half-reaction, we can see that 2 moles of KIO3 yield 1 mole of I2 and release 10 moles of electrons (10e).
Therefore, the correct answer is that there are 10 electrons being transferred in the second half-reaction.