Working out number of electrons

How many electrons are transferred in the following reaction?

2ClO3- + 12H+ + 10I- → 5I2 + Cl2 + 6H2O

Thanks for help.

K

you an do this one pretty simply by looking at the I-
There are 10 I-, each of them is going from an oxidation state of -1 (I-) to an oxidation state of 0 (I2)
10 electrons must be being transferred to oxidise 10 I- ions


We can work it out fully using this method
split it into the half equations

oxidation half equation
10I- --------> 5I2

work out the total charge on each side of the arrow
LHS = 10 x I- = -10 charge
RHS = no charged particles
Now, we add as many electrons to the most positive side as are needed to even the charges out.. Add 10e to the RHS

10I- ----> 5I2 + 10e

So 10 electrons are transferred.


Check it by doing the reduction half equation as well

2ClO3^- + 12H+ --------> Cl2 + 6H2O
LHS = (2 x ClO3^-) + 12 H+ = +10
RHS = no charges
add 10 e to the LHS
so we have confirmed that 10 e are transferred

2ClO3^- + 12H+ + 10e --------> Cl2 + 6H2O