Redox balancing? 10 pts for best answer??

Question

5) Potassium carbonate + bromine -----> potassium bromide +potassium bromate + carbon dioxide

Answer 1

This one is not too complicated because the oxidation state of K, C, & O don't change; only Br.

As Br2, bromine starts with an oxidation state of zero.

You have, in the product: KBrO3, Br in the +5 oxidation state so you can see that you would need 5 bromines going from +0 to -1 to balance it. When you put in the 5 KBr's that would need to be formed and then adjust the amount of the remainder of the starting materials/products to agree with this, it is now balanced. Here's how I did it, but everybody has their "favorite method"

Starting equation:

K2CO3 + Br2 ---> KBr + KBrO3 + CO2

Note: one Br has gone from Br(0) to Br+5 in the KBrO3, so we'll start by adding 5 "KBr's" which are bromine in the -1 oxidation state (Br-1) to balance the oxidation; so we'd get:

K2CO3 + 3 Br2 ---> 5 KBr + KBrO3 + CO2

The oxidation/reduction part is OK; Now we need to get enough potassiums to make the 5 KBr's and the KBrO3, (6 total). After trippling the amount of K2CO3 we get:

3 K2CO3 + 3Br2 ---> 5 KBr + KBrO3 + CO2

Now we'll check the oxygens and see if everthing is OK:

After we remove the 3 oxygens we need for the KBrO3 all we have left over is 3 CO2's, so the final balanced equation is:

3 K2CO3 + 3 Br2 ---> 5 KBr + KBrO3 + 3 CO2

Answer 2

K2CO3 + Br2 --> KBr + KBrO3 + CO2
Br going from 0 to -1 and from 0 to +5
Right away you know you will get 5 KBr for every 1 KBrO3
then balance the rest by inspection
3K2CO3 + 3Br2 --> 5KBr + KBrO3 + 3CO2

Answer 3

3K2CO3 + 3Br2 >>5KBr + KBrO3 +3 CO2