Question

Im having trouble balancing this redox rection in acidic solution. I don't know how to balance reactions in which one reactant splits up into two products.

C2H5OH(aq) + I3-(aq) --> I-(aq) + CO2(g) + CHO2-(aq) + H+(aq)

I know I split it into
C2H5OH --> CO2 + CHO2 + H+

and I know the oxidations #'s: for the C's:
C2H5OH C=2-
CO2 C=4+
CHO2- C=2+

But, I don't know where to go from there!

Answer 1

One thing you can do here is to combine the loss of electrons from the C2H5OH. If we rewrite that as
C2H6O ==> CO2 + CH2O^-
Each C on the left is -2. C in CO2 is +4 and C in CH2O^- is -1 so we can combine them this way.
C2H6O ==> CO2 + CH2O^-
1 C on left to CO2 is -2 to +4 which is a loss of 6 electrons.
1 C on the left to CH2O^- is -2 to -1 or a loss of 1 electron. Combining them, we have a loss of 6e + 1e = 7e lost for the duo. So the half equation would look like this.
C2H6O ==> CO2 + CH2O^- + 7e
Probably you can finish but we need to add 8H^+ to the right to balance the charge as in this
C2H6O ==> CO2 + CH2O^- + 7e + 8H^+ and that makes us add H2O to the left (we have 10 H on the right so we need 10 on the left).
2H2O + C2H6O ==> CO2 + CH2O^- + 7e + 8H^+.
Now we check it.
I see 10 H on the left and right.
I see 3 O on the left and right.
I see 2 C on left and right.
I see 0 charge on left and right.
Balanced.
Now you balance the I3^- ==> 3I^-

The final equation is
4H2O + 2C2H6O + 7I3^- ==> 21I^- + 2CO2 + 2CH2O^- + 16H^+
Check it to make sure it balances:
1. By atom
2. By charge
3. By electron change.

Answer 2

Thanks! I actually eventually got the right answer after about 45 gruelling minutes.
But, mine is different than yours

3H2O + C2H5OH + %I3- ==> 15I- + CO2 + CHO2- + 11H+