Oxidation-Reduction Reactions

Outline

• Tracking Electron Transfers: Oxidation and Reduction
  • Redox Reactions
• Practice with Concepts
• Discussion Questions
• Optional Website Reading

Tracking Electron Transfers: Oxidation and Reduction

Technically, acid-base reactions involve proton (H+) transfers, while oxidation-reduction reactions (redox for short) involve a change in oxidation state. The oxidation state depends on the electron configuration of the atoms involved, whether in a monatomic, molecular compound, or ionic situation. Tracking electron exchanges can sometimes help us understand a reaction that is otherwise puzzling.

Redox Reactions

Redox reactions (as we call them) involve the transfer of electrons between the reactants. Let's look at the reaction that occurs when we drop metallic zinc into hydrochloric acid, a reaction which produces hydrogen gas (H₂):

Zn + HCl + H₂O → ????

First, break down the reaction into the component ions:

>Zn + 2H⁺ + 2Cl^- + H₂O → Zn²⁺ + H₂ + 2Cl^- + H₂O

The water molecules cancel and the chlorine atom is a non-participant. We can look at the pieces which actually react:

>Zn → Zn²⁺ and 2H⁺ → H₂

The zinc atoms lose 2 electrons each, and the hydrogen ions gain an electron each, so this is a redox reaction, because electrons are being exchanged. The element losing electrons is oxidized and causes the other compound to be reduced, so it is the reducing agent. The element gaining electrons is reduced, and causes the other element to lose electrons, so it is the oxidizing agent.

In order to keep track of the electrons involved, we use a special counter, called the oxidation number. We can think of this as related to the charge on an ion or atom in a molecule, and sometimes equal to it. Since it isn't always the same as the charge, the oxidation number is called a "pseudo-charge" number.

The rules for determining oxidation numbers can seem pretty arbitrary at times. But they work when we need to balance the equations, so they are a very useful tool. Here are the rules, one more time:

Ion or molecule	Value of Oxidation Number	Example
Elemental molecule	0	H2 is 0 on each atom
Monatomic ion	Charge on ion	Fe3+ : +3
Group I ions	+1	Na: +1
Group II ions	+2	Ca: +2
Oxygen	-1 in peroxides; -2 otherwise	H2O: O is -2
Hydrogen	-1 in metal hydrides, +1 otherwise	H2O: each H is +1
Fluorine	-1	HFl: H is +1, Fl is -1

For everything else, you have to derive the oxidation number by balancing it to 0 (for a nuetrally charged molecule) or to the charge on the ion. Let's look at H₂SO₄, with a total charge of 0. Each H contributes +1, so the oxidation number of 2 Hs is +2. Each oxygen contributes -2, so the oxidation number of 4 Os is -8. The molecule is neutral and its total oxidation number 0, so the oxidation number of S must be +6.

Now look at chromate: Cr₂O₇^2-. The oxidation of each O is -2, so 7 Os make -14. Since the ion has a charge of -2, the chromium needs to offset only -12 by having a total of +12. There are two chromium atoms, so each is +6.

Why do we play this game? Because once we determine the oxidation numbers for each component in the equation, we can see which elements are oxidized or have increased oxidation numbers, and which are reduced, and have lower oxidation numbers. By balancing the electrons necessary to account for any remaining charge problems, we can balance the entire equation.

One more note, before we do an example. When redox reactions occur in acid solutions, both water molecules and H⁺ are available from the solution to participate in the reaction. When redox reactions occur in base solutions, both water molecules and OH^- are available.

Let's look at the reaction between chromate and iodine in acid solution. We know the primary products and reactants, but we don't have a balanced equation:

Cr₂O₇^2- + I^- → Cr³⁺ + I₂.

1. Determine the oxidization numbers on each component. For Cr₂O₇^2-, we already have shown that the Cr atoms are at +6, the O atoms are at -2. The iodine ion has a charge of -1, so its oxidization number is -1. The chromium ion has a charge of +3, so its oxidization number is +3. Finally, the I₂ product is a neutral elementary molecule, so each atom in it has an oxidization number equal to 0.
2. Now we look at the changes in oxidization numbers. The iodine atoms go from -1 to 0. This is an increase in oxidation number, so the iodine is oxidized, and we can write the oxidation half of the overall equation:

   I^- → I₂.

   Note that we have not yet tried to balance the atoms or charges in the equation.

3. The chromium atoms go from +6 to +3. This is a decrease in oxidation number, so the chromium atoms are reduced, and we can write the reduction half of the overall equation:

   Cr₂O₇^2- → Cr³⁺.

   Note that we have done nothing so far to account for the oxygen atoms, or balance either atoms or charges in the equation.

4. Balance the atoms in the oxidation equation:

   2I^- → I₂.

5. Balance the charges in the oxidation equation. We have to account for 2 negative charges in the reactants that are missing in the product:

   2I^- → I₂ + 2e^-.

6. Balance the chromium atoms in the reduction equation:

   Cr₂O₇^2- → 2Cr³⁺.

7. Account for the oxygen atoms, which go into water (remember that water can be a reactant or a product of a redox reaction occurring in an acid solution):

   Cr₂O₇^2- → 2Cr³⁺ + 7H₂O.

8. Account for the hydrogen we introduced in the last step by recalling that hydrogen ions H⁺ are available in an acid solution:

   Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O.

9. The atoms are now balanced, but the charge is not. Add enough electrons to account for the excess charge on the right side of the equation:

   Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O.

10. Now we can multiply each equation so the electrons will cancel when we add the two half equations together:

    1[2I^- → I₂ + 2e^-] PLUS

    3[Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O] IS

    6I^- + Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O + 3I₂.

    We are done!

Unless, of course, we want to run the reaction in a base solution instead. There are no H⁺ ions available in the base solution, so we have to get rid of them by cancelling them out (turning them into water). 14 H⁺ ions require 14 HO^- ions to make 14 water molecules, so we add 14 HO^- to both sides of the final equation we got from the reaction in acid:

6I^- + Cr₂O₇^2- + 14H⁺ + 14OH^- → 2Cr³⁺ + 7H₂O + 3I₂ + 14OH^-

Combining the hydroxide and hydrogen ions to make water molecules on the left gives

6I^- + Cr₂O₇^2- + 14H₂O → 2Cr³⁺ + 7H₂O + 3I₂ + 14OH^-

Cancelling the water molecules on each side gives the final base solution reaction:

6I^- + Cr₂O₇^2- + 7H₂O → 2Cr³⁺ + 3I₂ + 14OH^-

Practice with the Concepts

Identifying the players

Magnesium reacts with oxygen gas to form an MgO precipitate.

What is the net equation?

Which is the reducing agent? The oxidizing agent?

Discussion Questions

• Why is it important to keep track of where the electrons go in these reactions?
• Which are more likely to be oxidizing agents: metals or halogens? reducing agents? Why?

Optional Readings

Here is a Oxidation state prediction game to help you learn oxidation states.

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