Oxidation and reduction or, all the things you have forgotten from high school but are expected knowledge at university.

Oxidation/reduction reactions involve the transfer of electrons from one species to another, oxidising one species and reducing the other.

Zn(s) + Cu2+(aq) →  Zn2+(aq) + Cu(s)

In this classic example, the zinc metal gives up two electrons and becomes the zinc (II) ion and the copper (II) ions gain two electrons to become copper metal. The transfer of electrons in redox reactions can be kept track of using oxidation numbers.

Oxidation numbers:

An oxidation number (or state) of a species gives you the hypothetical charge of an atom in a substance, (or actual charge, if it is a monatomic ion) as determined by a few rules. The oxidation numbers will change during an oxidation/reduction reaction, as electrons will be transferred.

The rules for determining the oxidation state are as follows:

  1. The oxidation state of an element (eg. Fe(s), Ca(s), Ar(g)) is always zero.
  2. The oxidation state of an atom in a monatomic ion is the same as the charge on the ion.
  3. The oxidation state of oxygen is always -2, (unless it is a peroxide, in which case it is -1)
  4. The oxidation state of hydrogen is usually +1 (unless it is in a compound such as CaH2, where it is -1)
  5. The oxidation state of fluorine is always -1. Other halogens are usually -1, except when one of the above rules takes precedence.
  6. The sum of the oxidation numbers in a neutral compound is zero. In an ion, the sum of the oxidation numbers is equal to the charge on the ion.

Half reactions:

Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

Shown above is the overall equation for the oxidation of iron with copper (II), but we can also write this reaction in terms of two half reactions. A half reaction is one part of the oxidation/reduction reaction, showing what happens to each species separately. The reaction shown above can be split accordingly:

Fe(s) → Fe2+(aq) + 2e                            electrons lost by Fe

Cu2+(aq) + 2e → Cu(s)                           electrons gained by Cu

The oxidation reaction is the one in which there is a loss of electrons and increase in the oxidation number, (in this case, the iron). The reduction is the gain of electrons and decrease in oxidation number, (the copper). Confusingly, we refer to each species by what it does, not what happens to it. So the oxidising agent, (or oxidant) is the species that causes oxidation in something else but is itself reduced. The reductant is oxidised, but causes reduction in something else. Don’t get this confused.

You may not always be given the full half equation, you may simply be told the redox couple, (eg, Fe/Fe2+, MnO4/Mn2+) and you will have to balance it yourself. Balancing a half equation is fairly simple, you just need to follow the steps:

  1. Assign oxidation numbers so you know what is oxidised/reduced.
  2. Balance the half equation by:
    1. Balance all the atoms except oxygen and hydrogen
    2. Balance the oxygens by adding the appropriate number of water molecules to the other side of the equation.
    3. Balance the hydrogen atoms, (including the ones you have added in the form of water) by adding H+.
    4. Balance the charge by either i) adding the appropriate number of electrons that have been lost/gained as seen by the change in oxidation number or ii) add electrons where necessary to make sure the equation is neutral overall.
  3. Once you’ve done that for both half equations, add them together. You may need to multiply one or both by a factor to make sure that the electrons have cancelled. If you still have electrons in your overall equation, you’ve done something wrong.

Let’s try an example (adapted from Ebbing General Chemistry):

 Zn(s) + NO3(aq) → Zn2+(aq) + NH4+(aq)

 We can split this one into two half equations:

Zn(s) → Zn2+(aq)­

NO3(aq) → NH4+(aq)

The zinc reaction is easy – the oxidation state starts at zero and ends up at +2, so it is an oxidation requiring two electrons:

Zn(s) → Zn2+(aq)­ + 2e

 The nitrogen half reaction is more complex. First, we need to determine the oxidation states of nitrogen:


The oxygens must each have an oxidation number of -2, and there is an overall -1 charge. Therefore the nitrogen must be +5.


The hydrogens are each +1, and there is an overall +1 charge, so nitrogen must be -3 in this case.

NO3(aq) → NH4+(aq)

Now we need to balance the equation – the number of nitrogens is correct, so we move to the oxygens:

NO3(aq) → NH4+(aq) + 3H2O(l)

Now we need to balance the hydrogens, including those on the ammonium:

NO3(aq) + 10H+(aq) → NH4+(aq) + H2O(l)

Finally, the charge:

NO3(aq) + 10H+(aq) + 8e NH4+(aq) + H2O(l)

We now need to multiply the zinc half reaction by 4, so a total of 8 electrons are being transferred in both reactions, and add them together to give:

4Zn(s) + NO3(aq) + 10H+(aq) → 4Zn2+(aq) + NH4+(aq) + 3H2O(l)

These steps work just fine for acidic reactions, but if your reaction takes place under basic conditions, you need to balance all of the protons with hydroxide ions.


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Filed under Chem 2, Inorganic Chemistry

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