Sb+1 + HClO ------> Sb2O5 + Cl-1

Question

I'm not doing something right. I think I have the charges correct, this is what I have so far . . . .

2Sb^+1 -----> Sb^+5 + 4e-
2(2e- + Cl^+1 ------> Cl^-1
then when I start subbing everything back and start trying to balance everything, it just keeps getting more and more water. Tell me what I am doing wrong.

Answer 1

What you need to realize FIRST is that you have one Sb on the left and two on the right so you need to put a 2 as a coefficient for Sb on the left. That will mean the electron change is for 2 Sb ions. So your half equation would be
2Sb^+1 ==> 2Sb^+5 + 8e
See if that helps. If not let me know.

Answer 2

I am doing that and the equation keeps getting bigger and bigger. Right now my equation looks like this

6H2O + 2Sb^+1 + 4HClO ---> 2SbO5 +4Cl^-1 + 6H^+ + 5H20

right now I have everything equal except I need 5 more Oxygen on the reactant side. See if you can balance this and tell me what I am doing wrong.

Answer 3

2Sb^+1 + 5H2O ==> Sb2O5 + 8e + 10H^+
H^+ + 2e + HClO ==> Cl^- + H2O
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Those are the two half equations.
#1. 2Sb on both sides. 10 H on both sides. 5 O on both sides. Charge of +2 on both sides.

#2. 2H on both sides. 1Cl on both sides. 1 O on both sides. -1 charge on boths sides.

Multiply equation 1 by 1 and equation 2 by 4, then cancel atoms/electrons/etc common to both sides. The final equation is
H2O + 2Sb^+1 + 4HClO ==> 4Cl^- + Sb2O5 + 6H^+

Answer 4

See my post with the two half equations and the full equation(balanced but check it to make sure). I don't know what you're doing wrong since your work isn't shown.