Balance equation for reaction occuring in basic solution.

Question

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?

Answer 1

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.