4. Then complete the other half cell.

Question

Oxidation half step:  (Losing electrons = R -3 -> +3) Balance the oxygens with water and the hydrogens with protons. (Determine the numbers that go in the blanks. If zero, record as the number zero "0".) Then balance the oxidation numbers using electrons.

   _____ R-H3 + _____H₂O --> _____HO-R=O + _____H⁺ + _____e^- 

Answer 1

When balancing this reaction, we want to make sure that the left side and the right side have the same amount of R, H, and O, and then we need to look at the overall charge on the left, and add however many electrons onto the right as we need to make them match.

On the left side we currently have 1 R, 5 H, and 1 O.

On the right side we currently have 1 R, 2 H, and 2 O.

The R is already balanced, but the H and O are not. Because H shows up in more than one term on the right side, it's better to start our balancing with O. In order to balance O, we will put the coefficient of 2 on the H₂O term. Now let's see what we have:

_____ R-H3 + 2H₂O --> _____HO-R=O + _____H⁺ + _____e^- 

Left side:

R: 1

H: 7

O: 2

Right side:

R: 1

H: 2

O: 2

So now the only thing not matching is H. Let's add a 6 coefficient to the H⁺.

R-H3 + 2H₂O --> HO-R=O + 6H⁺ + _____e^- 

Now, both sides have 1 R, 7 H, and 2 O. Let's look at the charge.

Left side: R-H₃ is neutral, and H₂O is neutral, so the entire left side is neutral.

Right side: HO-R=O is neutral, the H⁺ term has a positive 1 charge, so 6 of them gives a charge of +6. We need to cancel out the +6 with electrons to match the left side, which is neutral, so we add 6 electrons.

R-H3 + 2H₂O --> HO-R=O + 6H⁺ + 6e^-