After balancing the following redox equation(using small whole-number coefficients), add the coefficients of all species in the balanced equation.

Bi(OH)3 + SnO2^(2-) → Bi + SnO3^(2-) (basic solution)

--- I separated the equation into two and i know the next step is to balance the e-... i just don't know how to?

6H + Bi(OH)3 → Bi + 3H2O
H20 + SnO2^(2-) → SnO3^(2-) + 2H

Please don't change screen names. Lucy, John Butch, etc won't cut it. We can help better if we can keep one name.

For the Bi half, Bi on the left has an oxidation number of +3 and on the right is 0 so that is a gain of 3e and the 3e goes on the right. That's how you do that part. For the Sn half, on the left Sn is +2 and on the right it is +4 so 2e goes on the left. I can help with these but you need to be explicit about what you don't understand. As for the John Butch post, I'll ignore it.

We all take the same Chemistry class, we are not using different names. We've been using my computer and that's why. You can ignore whatever post you want, it's just sad that I relied on this website for help. To be mad because we supposedly change screen names is something very low of you. What's next, are you going to come at me if my name is not really Lucy or my classmate's name is not John or Addam. If this is a part time job for you then do your job right and stick to answering questions, don't do the questioning yourself.

To balance the redox equation in basic solution, you'll need to follow these steps:

Step 1: Balance the atoms other than hydrogen and oxygen.
Start with balancing the atoms other than hydrogen and oxygen. In this case, you have only Bi on the left side and SnO3^2- on the right side. So, you will balance the Bi atoms.

6H + Bi(OH)3 → Bi + 3H2O
H2O + SnO2^2- → SnO3^2- + 2H

Step 2: Balance oxygen atoms by adding water molecules.
Now, the oxygen atoms are not balanced. On the left side, you have 6 oxygen atoms from the water, and on the right side, you have only 3 oxygen atoms from the Bi(OH)3. To balance oxygen, you will add water molecules.

6H + Bi(OH)3 → Bi + 3H2O
3H2O + SnO2^2- → SnO3^2- + 2H

Step 3: Balance hydrogen atoms by adding H+ ions.
Next, you need to balance the hydrogen atoms. You have 6 hydrogen atoms on the left side and 2 hydrogen atoms on the right side.

To balance the hydrogen atoms, you will add H+ ions to the equation.
Note: The number of H+ ions added should be equal to the number of H atoms that need to be balanced.

6H + Bi(OH)3 → Bi + 3H2O
3H2O + SnO2^2- + 6H+ → SnO3^2- + 2H + 6H+

Step 4: Balance the charge by adding electrons.
At this point, the charges are not balanced. On the left side, you have no charge, while on the right side, you have a -2 charge from the SnO3^2- ion.

To balance the charges, you will add electrons (e-) to the equation.
Note: The number of electrons added should be equal to the difference in charges between the left and right side.

6H + Bi(OH)3 → Bi + 3H2O
3H2O + SnO2^2- + 6H+ + 10e- → SnO3^2- + 2H + 6H+

Step 5: Combine the two half-reactions.
Now that both half-reactions are balanced, you can combine them into a balanced overall redox equation.

Multiply the half-reactions by appropriate factors to ensure the number of electrons canceled out.

6H + Bi(OH)3 + 10e- → Bi + 3H2O
3H2O + SnO2^2- + 6H+ + 10e- → SnO3^2- + 2H + 6H+

Finally, add the two half-reactions together, cancel out common terms, and simplify the equation if necessary.

6H + Bi(OH)3 + 10e- + 3H2O + 3H2O + SnO2^2- + 6H+ + 10e- → Bi + 3H2O + SnO3^2- + 2H + 6H+
Bi(OH)3 + SnO2^2- + 6H+ → Bi + SnO3^2- + 2H

Now you can sum up the coefficients of all species in the balanced equation. In this case, the coefficients are:
6 + 1 + 6 + 1 + 2 = 16