Redox equations

This is possibly the reason why I hate redox so much. The actual equations themselves aren't too bad, it's just working out whether a reaction occurs or not. My brain always gets mixed up reading that stupid reduction potentials chart... argh...

Anyway. The first two types of redox reactions that I'm going to talk about are metal and halogen displacement reactions.

In metal displacement reactions, a metal reacts with a metal ion in such a way that the originally solid metal becomes an ion and the original metal ion becomes a solid metal. Confused yet?

For example, if you have solid potassium reacting with a solution of sodium chloride (i.e. sodium ions and chloride ions), you'd end up with solid sodium and a solution of potassium chloride (i.e. potassium ions and chloride ions). The net ionic equation would look like this:

K + Na⁺-> Na + K⁺

In this reaction, K has been oxidised and Na has been reduced.

Note that if you had solid sodium and a solution of potassium chloride, they would not react. But how can you tell just by looking at the reactants if you're going to have a reaction or not? This is where the Standard Reduction Potentials table at the back of your data sheet comes in.

The closer an element is to the bottom of the table, the more it wants to be oxidised and become a positive ion. For the most part, groups I and II elements (alkali metals and alkaline earth metals) are at the very bottom and directly above them is mainly transition elements (with exceptions such as aluminium and water which randomly snuck in). I'm very curious as to why this is so, but I'm too lazy to look it up myself.

When you have the reactants of a metal displacement reaction, take a look at the Standard Reduction Potentials chart. Which element has the strongest desire to be a positive ion? If it's already a positive ion, no reaction will occur. If it isn't, then a reaction will occur. That's basically how it works.

Halogen displacement reactions are the opposite of metal displacement reactions. Here's an example:

F₂ + 2Br^- -> 2F^- + Br₂

Now, when working out whether a reaction will occur or not, you have to do the opposite of what you did with the metal displacement reactions as this time we're dealing with negative ions, not positive ones.

If the bottom of the table lists the elements that want to be oxidised and become positive ions, then it also follows that the top of the table lists the elements that want to be reduced to negative ions. When working out halogen displacement reactions, work out which element has the strongest desire to be a negative ion. Then think about whether that element is already a negative ion or not. If it's already a negative ion, then no reaction will occur. If not, then a reaction will occur.

(By the way, there are ways to force the reactants to react even when they don't really want to, but I'm not going to go into that for now.)

Now, next up is half equations. In Chemistry 2AB, you only have to deal with the dead easy half equations, but there's also acidic conditions and stuff that you can learn about too.

You can write half equations for any kind of redox reaction, whether it be metal displacement, halogen displacement, or a myriad of other types of reactions that I don't know the names of. As long as an element is being reduced and another is being oxidised, you can write half equations.

Basically, you write separate equations for the reductant and the oxidant. Then you add in electrons to balance the charges.

F₂ + 2e^- -> 2F^- 

2Br^- -> Br_{2 +} 2e^-

Just make sure that the number of Fs or Brs or whatever are balanced on either side of the equation, and then chuck in some electrons to balance the charges as well.

You can also write balanced net ionic equations as well simply by adding together the two half-equations and then cancelling off the electrons on either side. If the two half-equations involve different numbers of electrons, however, care needs to be taken.

If you have different numbers of electrons in the two equations, what you first need to do is multiply each by a certain number so that the number of electrons are the same. For example, if you have 3 electrons in the first equation and 2 in the second, you can multiply the first equation by 2 and the second by 3 so that you have 6 electrons in both equations. Then you can add the two equations together, cancelling out the electrons on both sides. It's just like simultaneous equations in maths really.

Earlier, I mentioned half equations in acidic conditions. Normally, when you have a question asking for these, it's because you have some crazy polyatomic ion like the chromate ion. The steps for working out these are as follows:

1. Balance the number of atoms on each side (except for hydrogen and oxygen- we'll get to them later).

MnO₄^- -> Mn²⁺

2. Add water in order to balance out the number of oxygen atoms.

MnO₄^- -> Mn²⁺ + 4H₂O

3. Add hydrogen ions (acidic conditions, remember?) to balance out the number of hydrogen atoms.

MnO₄^- + 8H⁺ -> Mn²⁺ + 4H₂O

4. Calculate overall charge on each side.

In above equation, the left hand side has an overall charge of +7 while the right hand side has an overall charge of +2.

5. Add electrons to balance out the charges.

MnO₄^- + 8H⁺ + 5e^- -> Mn²⁺ + 4H₂O

So that's pretty much all you need to know and more on redox equations! Next up- electrolysis!