Balance the following Redox Reaction and show all steps.?

Question

(I can't do subscripts so the big number behind the letters should be read as a subscript)
Equation:

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

Answer 1

ok, so lets´see first which components change oxidation state.
Br2 apparently is one of them, cause is tarts as a pure aubstance and ends up as salts.
so, oxidation state in KBr, since K has +1, is apparently -1
so we have Br+1e->Br-
in KBrO3, since O has -2 and K has +1, Br is left with +5.
so we have Br-5e-> Br+5

so we need at least 5 Bromines to accept the 5 electrons given away by the other bromine, meaning we have a total of 6 bromine atoms, namely 3 Br2 molecules.
as said we need 5Br-, ehere all the bloody electons went to, meaning the compound in which Br- is produced needs a 5 in front of it.
so we have 5KBr.
so on the right we have now 5+1 =6Br, and on the left we have 3x2 =6Br. so far so good.
on the right we have 5+1 K
on the left we have only 2, meaning we need a 3x multiplier.
result so far:
3K2CO3+3Br2->5KBr+KBrO3+?CO2
on the left we have a total of 3 carbons, and on the right only 1.
so we need a 3 there too.
3K2CO3+3Br2->5KBr+KBrO3+3CO2

let´s check all again

left right
3x2K 5+1K
3C 3C
3x3O 3O+3x2O
3x2Br 5+1Br

looks good!

Answer 2

This reaction (which I have never, ever seen before) is an example of a disproportionation.

Start by using Br2 + 2e- -----> 2Br-

and Br2 + 6H2O ------> 2(BrO3)- + 10e- + 12H+

Balance up the electrons as I am sure you have been shown, and then try to get rid of water, H+ ions...

Answer 3

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