Balance equation for reaction occuring in basic solution.
CrO4{2-}+S{2-}gives S+CrO{-}
i don't know how to balance the charges in this way of balancing please teach me how to do it?
Here is the system. I'll do CrO4.
1. Write the half cell.
CrO4^-2 ==>CrO^-
2. Identify the atoms changing oxidation state. Here is a site that will help you do that.
http://www.chemteam.info/Redox/Redox.html
The atom changing is Cr. It changes from +6 on the left to +1 on the right.
3. Balance the atoms that are changing oxidation state preliminarily. This is EXTREMELY important. In this case they are balanced; i.e., there is 1 Cr atom on the left and 1 Cr atom on the right.
4. Add electrons to the appropriate side to balance the change in oxidation state. That means add 5 electrons to the left. The equation now looks like this.
CrO4^-2 + 5e ==> CrO^-
5. Count the charge on the left and right and
a. if acid solution (this one isn't) add H^+ to balance the charge or
b. if basic solution (this one is basic from the problem), add OH^- to balance the charge.
I see a charge of -7 on the left and a charge of -1 on the right.
Since this a basic solution, I add OH^- to balance the charge, in this case, to the right. The equation now looks like this.
6. CrO4^-2 + 5e ==> CrO^- + 6OH^-
7. Add water to balance the H. This means add H2O to the left.
CrO4^-2 + 5e + 3H2O ==> CrO^- + 6OH^-
If we have done things correct, that should balance the O atoms and it doesn.
8. Check everything.
a. atoms balance? yes.
b. charges balance? yes.
c. oxidation change balances? yes.
S is much easier to do.
S^-2 ==> S + 2e.
That site above also has instructions for balancing the entire equation. They may or may not use the same set as above.
Balancing a chemical equation in basic solution requires two main steps:
Step 1: Balance the atoms (excluding H and O)
In the given equation, we have CrO4{2-} and S{2-} as reactants, and S and CrO{-} as products. To start balancing the atoms, we need to make sure there is an equal number of each type of atom on both sides of the equation.
For the reactants:
Cr: 1 Cr on the left side, 1 Cr on the right side (balanced)
O: 4 O on the left side, 1 O on the right side (unbalanced)
S: 1 S on the left side, 1 S on the right side (balanced)
To balance the oxygen atoms, you can add water (H2O) to the right side of the equation since it contains oxygen:
CrO4{2-} + S{2-} ➜ S + CrO{-} + H2O
Now let's count the oxygen atoms again:
O: 4 O on the left side, 5 O on the right side (unbalanced)
Step 2: Balance the hydrogen (H) and oxygen (O) by adding hydroxide ions (OH{-}) on the opposite side.
Since we're in basic solution, we can balance the hydrogen and oxygen atoms by adding OH{-} ions. We add an equal number of OH{-} ions on the side that needs it, and then balance the charge by adding electrons (e{-}) to the other side.
In this case, we need to balance the 4 additional oxygen atoms. Hence, we'll add 4 OH{-} ions to the right side:
CrO4{2-} + S{2-} ➜ S + CrO{-} + H2O + 4 OH{-}
Now let's count the oxygen atoms again:
O: 4 O on the left side, 4 O on the right side (balanced)
Finally, we need to balance the charges. In this case, you're converting CrO4{2-} to CrO{-}. To balance the charge, you'll need to add 2 electrons (e{-}) on the right side:
CrO4{2-} + S{2-} ➜ S + CrO{-} + H2O + 4 OH{-} + 2 e{-}
Now the charges are balanced, and you have successfully balanced the given chemical equation in basic solution.