Redox Reactions

Having covered oxidation numbers, we can now look at redox reactions (remember that reduction and oxidation always occur together).

When we deal with reduction and oxidation we usually use half equations. We usually do this when we set up an electrochemical cell. This is when the sheet that is usually on the back of the Periodic Table comes in very handy. It is a list of reduction potentials. This is the list we use to determine what is reduced and oxidised in reactions.

Handy hint: if you ever get confused as to which way it goes, just remember that fluorine likes gaining an electron sooooooo much more than potassium.

The higher the number (or the biggest positive number) is the species that is going to be reduced.

For the half equations, remember to write the electrons on the right side of the equation.

Fe²⁺ + 2e^– → Fe

This is the reduction half equation for iron (II). Oxidation half equations have the electrons on the other side (all you have to do is reverse the equation given on the sheet).

If you have 2 half equations and need to add them together, you need to make sure that the electrons cancel each other out.

We’ll work through a copper and zinc example.

Cu²⁺ + 2e^– → Cu                            E= 0.16V

Zn²⁺ + 2e^– → Zn                            E= -0.76V

Clearly, the copper half equation is going to be reduced. This means we now flip the zinc half equation and because we reverse it, we also change the reduction potential sign.

Zn → Zn²⁺ + 2e^–                            E = 0.76V

Its nice and easy because the electrons already equal each other but sometime you might need to times one of the equations by something to make them equal (think lowest common multiple stuff from maths).

We add them together now and get our final equation and the voltage generated.

Zn + Cu²⁺ → Zn²⁺+ Cu              E = 0.92V

Easy as, right?

So they are the basics of redox reactions. The redox potential isn’t really that important unless you’re trying to calculate the voltage running through an electrochemical cell.

We’ll try something a little bit harder now. These aren’t so much constructing the redox equations, as balancing them out.

Mn 2+  +  BiO3 –  ===>  MnO4 –  +  Bi 3+

First, we split them up. Our two main species/elements Mn and Bi.

Mn²⁺  →   MnO₄^–

So, now we balance the Mn, which is fine. Next, the oxygen gets balanced. There are 4 oxygen atoms on one side, there needs to be 4 on the other side. This is done by adding water to the other side.

4H₂O + Mn²⁺  →   MnO₄^–

We balance the hydrogen now (quick note – this is only for acidic solutions, basic solutions have a slightly different process). Add hydrogen to the other side, so that it is balanced.

4H₂O + Mn²⁺  →   MnO₄^– + 8H⁺

So, as long as all the species present are balanced, we get to balance the charge now. Each side always need to be equal.

We have a +2 charge on one side and an overall charge of +7 on the other side. So we need 5 electrons to balance it out completely. After you’ve done that, you should have a very pretty equation that looks like this:

4H₂O + Mn²⁺ →    MnO₄^– + 8H⁺ + 5e^–

Have a go at the other half equation with bismuth.

You should end up with something that looks like this:

6H⁺ + BiO₃^– + 2e^–  →  Bi³⁺ + 3H₂O

Now, we add both equations together. No, wait! The electrons need to be balanced first. To do that we have to get the lowest common multiple for the electrons. Two electrons and five electrons; the lowest common multiple is 10.

2x (4H₂O + Mn²⁺ →    MnO₄^– + 8H⁺ + 5e^–)                                  8H₂O + 2Mn²⁺ →    2MnO₄^– + 16H⁺ + 10e^–

5x (6H⁺ + BiO₃^– + 2e^–  →  Bi³⁺ + 3H₂O)                                        30H⁺ + 5BiO₃^– + 10e^–  →  5Bi³⁺ + 15H₂O

 

Cancel out any species that appear on both sides of the equation (like the electrons) and voila!

2Mn²⁺ 14H⁺ + 5BiO₃^–   →  5Bi³⁺ + 7H₂O + 2MnO₄^–

Our equation is now balanced. Well, I’m 95% sure that its balanced.

For basic solutions, its exactly the same with just one extra step. After you’ve added hydrogen but before you add any electrons, add an equal number of hydroxide ions to both sides. The hydrogen and hydroxide (on one side) form water.

That’s all there is to it. Simple, huh? Okay, maybe not at first but after practise (see the website below), it gets pretty easy.

http://wc.pima.edu/~skolchens/C152OL/Ch21/REDOX.htm

http://kurtniedenzu.cmswiki.wikispaces.net/file/view/Redox_Equations_WorkSheet.pdf   – I recommend this one

Click to access redox.pdf

Click to access redox_chem220.pdf

Enjoy!