Oxidation States, Half Equations and Redox

Oxidation reactions are hugely important in chemistry, in that they relesae large amounts of energy in the form of heat. In biology too, we’ve learned that oxidation reactions form the basis of provide energy for growth (respiration and the production of ATP). But the concepts can be confusin and its easy to get muddled up. This post will hopefully sort it out a little and help you get a clearer picture of how to tackle oxidation states and half-equations.

Many of my students get confused at some point by oxidation, reduction and half equations. The topic gets recapped in CH5 so it is important to get the key points right:

• Remember, oxidation is a process of electron loss and reduction is a process of electron gain, OILRIG (Oxidation Is Loss, Reduction Is Gain).
• The Reducing agent is the electron donor and gets oxidised. the Oxidising agent is the electron acceptor and gets reduced.
• The number of electrons lost (donated) by the reducing agent equals the number of electrons gained (accepted) by the oxidising agent.
• Oxidation number is essentially the amount of charge an element brings to a compound (eg Na brings +1 and Cl brings -1 to NaCl)
• Oxidation numbers of all elements is zero, of all ions alone (Na+, Cl-, Ca2+, O2-) is its charge (+1, -1, 2+, -2 respectively), H is usually +1 and O is ussually -2
• Oxidation numers of elements in ions can be calculated by cancellation of charge, eg Mn in MnO4- must be giving +7 as 7 pluses cancel out seven of the 8 minuses from O (4x-2) leaving – left over.

Then how do we construct half equations?

1. Write the formula of the two species changing from the question across the arrow and identify their oxidation umbers.
2. Check the number of atomic species is the same (both 1 or 2 etc)
3. Balance electrons on the HIGH oxidation number side only, by subtractng High oxidation number – low oxidation number.
4. Balance any oxygen on the opposite side with H2O.
5. Balance any hydrogen on th eopposite side with H+
6. Check the net charge on both sides is equal and feel smug!

Worked example:

Construct a half equation for the oxidation of Cl- to ClO3-

1. Cl- -> ClO3- [Cl in Cl- is -1 and Cl in ClO3- is +5]
2. one chlorine species on either side
3. [High - Low = 5 - -1 = 6 electrons on high side, so Cl- -> ClO3- + 6e-
4. 3H2O + Cl- -> ClO3- + 6e-
5. 3H2O + Cl- -> ClO3- + 6e- + 6H+
6. -1 net charge on left and -1 net charge on right!!

Construct a half equation for the oxidation of Cl- to Cl2

1. Cl- -> Cl2
2. 2Cl- -> Cl2
3. 2Cl- -> Cl2 +2e- [high - low = 0--1 = 1 but it happens twice as there are two chlorine species]
4. END as there are no oxgens or hydrogens

This is a simple and robust method which gets the job done, practice it and it’ll soon become second nature.

Try some common ones like:

1. Cr2O7 2- to Cr3+
2. MnO4- to Mn2+
3. Cl2 to ClO-