Question

I'm having trouble balancing this equation: SO3^2- --> S2O4^2-

It's in a basic solution. Thanks.

Answer 1

To balance this reaction in a basic solution, follow these steps:

1. Balance the elements that are not H or O.
In this case, Sulfur (S) atoms are already balanced: 1 S atom in SO3^2- and 2 S atoms in S2O4^2-.

2. Balance the O atoms by adding H2O to the side with fewer O atoms.
There are 3 O atoms in SO3^2- and 4 O atoms in S2O4^2-. We need to add 1 water molecule (H2O) to the left side to balance the O atoms:

SO3^2- + H2O --> S2O4^2-

3. Balance the H atoms by adding OH- to the side with fewer H atoms.
We added 2 H atoms to the left side in the previous step, so now we need to add 2 hydroxide ions (OH-) to the right side:

SO3^2- + H2O --> S2O4^2- + 2 OH-

4. Balance the charges by adding electrons (e-) to the side with more positive net charge.
The net charge on the left side of the equation is -2, while the net charge on the right side is -4 (from S2O4^2- and two OH- ions). We need to add 2 electrons to the right side to balance the charges:

SO3^2- + H2O --> S2O4^2- + 2 OH- + 2 e-

Now, the equation is balanced. The balanced equation in a basic solution is:

SO3^2- + H2O --> S2O4^2- + 2 OH- + 2 e-