Sb+1 + HClO ------> Sb2O5 + Cl-1
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.
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.
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.
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^+
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.
From what you have written, it seems like you are trying to balance the given chemical equation. Here are some steps you can follow to balance the equation correctly:
Step 1: Write out the unbalanced equation.
Sb+1 + HClO → Sb2O5 + Cl-1
Step 2: Start by balancing the atoms that appear in only one compound on each side of the equation. In this case, we have antimony (Sb), chlorine (Cl), oxygen (O), and hydrogen (H).
Step 3: Balance the antimony (Sb) atoms:
On the left side, we have one antimony atom (Sb+1), and on the right side, we have two antimony atoms (Sb2O5). To balance this, multiply the antimony on the left side by 2 to yield 2Sb^+1.
2Sb+1 + HClO → Sb2O5 + Cl-1
Step 4: Balance the chlorine (Cl) atoms:
On the left side, we have one chlorine atom (Cl-1), and on the right side, we have one chlorine atom (Cl-1). The chlorine atoms are already balanced.
2Sb+1 + HClO → Sb2O5 + Cl-1
Step 5: Balance the oxygen (O) atoms:
On the left side, we have one oxygen atom (O) from HClO, and on the right side, we have five oxygen atoms (O) from Sb2O5. To balance this, multiply HClO on the left side by 5 to yield 5HClO.
2Sb+1 + 5HClO → Sb2O5 + Cl-1
Step 6: Balance the hydrogen (H) atoms:
On the left side, we have five hydrogen atoms (H) from 5HClO, and on the right side, we have none. To balance this, multiply H2O on the right side by 5 to yield 5H2O.
2Sb+1 + 5HClO → Sb2O5 + Cl-1 + 5H2O
Now the equation is balanced. Make sure to double-check that all atoms are balanced on both sides.