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 (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: Cr2O72-. 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:
Cr2O72- + I- → Cr3+ + I2.
I- → I2.
Note that we have not yet tried to balance the atoms or charges in the equation.
Cr2O72- → Cr3+.
Note that we have done nothing so far to account for the oxygen atoms, or balance either atoms or charges in the equation.
2I- → I2.
2I- → I2 + 2e-.
Cr2O72- → 2Cr3+.
Cr2O72- → 2Cr3+ + 7H2O.
Cr2O72- + 14H+ → 2Cr3+ + 7H2O.
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O.
1[2I- → I2 + 2e-] PLUS
3[Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O] IS
6I- + Cr2O72- + 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- + Cr2O72- + 14H+ + 14OH- → 2Cr3+ + 7H2O + 3I2 + 14OH-
Combining the hydroxide and hydrogen ions to make water molecules on the left gives
6I- + Cr2O72- + 14H2O → 2Cr3+ + 7H2O + 3I2 + 14OH-
Cancelling the water molecules on each side gives the final base solution reaction:
6I- + Cr2O72- + 7H2O → 2Cr3+ + 3I2 + 14OH-
Here is a Oxidation state prediction game to help you learn oxidation states.
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