ow do I balance this oxidation-reduction equation? KBrO2+KI+HBr-KBr+I2+H2O?

step 1. first write out the oxidation half reaction
step 2. now write out the reduction half reaction sepratley from oxidation
step 3. balance oxygen by adding H2O to opposite side(side that has less oxygen) to both oxidation and reduction half reaction
step 4. now to balance hydrogen, add H+ to opposite side(side that has less hydrogen)
step 5. now add charges on both sides. For both reduction and oxidation
step 6. add electrons to balance charges
step 7. if electron are different numbers for oxidation and reduction half reactions. Then multiply by lowest number they both go into (e.g. if ox has 3e and red has 2e then multiply so they can both become 6e)
step 8. if you multiplyed don't forget to put coefecient in front of compund or element (eg if you started with KBrO2 and then had to multiply by three it now become 3KBrO2)
step 9. now add both half reaction. Cross out the common compound on each side.
step 10. Now you right the everything you have in one big solution.

KBrO2+KI+HBr-KBr+I2+H2O

KBrO2: KI: HBr: > KBr I2 H2O
K= +1 K= +1 H= +1 K= +1 I= 0 H= +1
Br= + 3 I= -1 Br= -1 Br= -1 O = -4
O= -2= -4
+1 + X + -4= 0 +4 - 1

OX: I -1 > I 0 + 1e-
RED: Br + 4e- > Br

2KI > I2 + 1e-
4e- + 2H+ +KBrO2 > KBr + 2H2O

( 2KI > I2 + 1e-) * 4
8KI > 4I2 + 4e-

8KI > 4I2 + 4e-
4e- + 2H+ + KBrO2 > KBr + 2H2O

CROSS OUT 4e-

2H+ + KBrO2 + 8KI > 4I2 + KBr + 2H2O

THIS IS ALL I KNOW UP TO. NOT ALL THAT GOOD AT CHEMISTRY. SO I HOPED A HELPED.

Hum.. maybe:

KBrO2 + 4KI + 4HBr ---> 5KBr + 2I2 + 2H2O

Rate this if its good!

I first balanced the water, which caused a 4HBr, then 5KBr, then to adjust, 4KI and 2I2.