IB Chemistry/Oxidation and Reduction
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Oxidation is the loss of electrons, and reduction is the gain of electrons. You can easily remember this with the mnemonics OILRIG (Oxidation Is Loss, Reduction Is Gain) or 'Leo Says Ger' L=loss/O=oxidation and G=gain/R= reduction.
A full reaction has a transfer of electrons - a reaction must have both something that is oxidized and something that is reduced. These reactions are known as a 'Redox' reactions.
Full reactions can be divided into two half reactions; the oxidation and reduction parts. Half-equations contain electrons.
For example: in the reaction 2 Mg + O2 --> 2 MgO we have, (these are known as half-equations)
2 Mg -> 2 Mg+2 + 2 e- Oxidation
O2 + 4 e- -> 2 O-2 Reduction
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[edit] Oxidation Numbers
It is easy to see which element is getting oxidized and which is getting reduced by using Oxidation Numbers, which are assigned by some rules which you must learn.
- The oxidation number of an atom, or a single-atom ion, is the same as the charge:
| Species | Oxidation number |
| Au | 0 |
| Au+ | +1 |
| Au+3 | +3 |
| Au- | -1 |
- Consider every compound to be made of ions. Consider covalent bonds to break with the two electrons assigned to the higher electronegativity atom. If two identical atoms are covalently bonded, consider the bond's electrons to be assigned one to each atom.
| Species | Oxidation number |
| H2 | 0 |
| HCl | +1 (H) and -1 (Cl) |
| H2O | +1 (H) and -2 (O) |
| O2 | 0 |
Consequently:
- The sum of all the oxidation numbers in an ion is the charge of the ion.
- The sum of all the oxidation numbers in a compound is 0.
- Elements not combined with anything have an oxidation number of 0.
- In compounds, the following rules are observed:
Group 1 always form +1 ions, and Group 2 always form +2 ions.
Fluorine is always -1
Oxygen is always -2, (except in peroxides = -1, and oxygen fluoride = +1)
Hydrogen is always +1 (except in hydrides i.e. combined with a low-electronegativity element = -1)
We can quickly assign oxidation states using the observation above:
NaHSO4
Na will be +1 (Group 1 elements only ever form +1 ions).
H will be +1 (It is not a hydride).
O will be -2.
The sum of the oxidation numbers is 0, so S must be +6.
Similarly, Mn is +7 in KMnO4 and Cr is +6 in K2Cr2O7.
Reduction occurs when the oxidation number decreases, and oxidation is when it increases. The change will be the number of electrons involved in the half equation. You should keep in mind that when something is being neutralized or precipitated it is not a redox reaction.
An oxidizing agent is that which oxidizes something, i.e. it takes its electrons, thereby getting reduced. Thus, the oxidizing agent is the thing that gets reduced in the reaction, and vice versa.
there are some special cases like in Ni(CO)4 the oxidation state of Ni is 0. In peroxides 2 oxygens have O.N. -1. CrO5 is an example where 2 peroxo bonds are present, Cr has O.N. +6.
[edit] Reactivity
The more readily a metal loses its outer electrons, the more reactive it is. Therefore, metals higher in the series displace metals lower down. This means that metals high in the reactivity series are better reducing agents as they lose electrons much easier. The electronegativity of the metals therefore increases down the series. The things at the bottom are much better oxidizing agents as they will rip electrons off.
[edit] Reactivity Series
K Na Li Ca Mg Al Zn Fe Sn Pb H Cu Ag
[edit] Voltaic Cells
A half cell is a metal in contact with a solution of its own ions, and a voltaic cell is two half cells connected by a salt bridge and a metal wire. The electrons transfered during the redox reaction are passed through the wire causing a current to flow. The salt bridge allows for the movement of ions so that the half cells do not become charged. the salt bridge is often a gel.
[edit] Electrolysis
Electricity can be used to force a non-spontaneous redox reaction to occur. This is done in the electrolytic cell. Electricity is passed through the electrolyte. If the electrodes are inert, such as graphite, and the electrolye is a molten compound then the products will be the oxidized and reduced forms of each. The anode (positive electrode) is where oxidation occurs as negative ions are attracted and lose electrons. The electron flow is therefore from the positive to the negative electrode (which to me doesn't seem logical -- negative electrodes should be attracted?)
If the electrolyte is aqueous then there is a different, more complicated scenario as there is oxygen and hydrodgen that may also be formed at the electrodes. The lower in the reactivity series a metal is, the greater its ability to gain electrons (high electronegativity). Thus, if hydrogen is lower then it will be reduced at the cathode in preference to another positive ion. This means that copper and silver will be the only metal ions that wil be reduced in preference to hydrogen at the cathode. However, if one ion is more concentrated than another it will be discharged. For example, in dilute solutions of sodium chloride, oxygen gas is given off in preference as there is simply more of it.
If the electrode is made of one of the metal of one of the ions in the electrolyte, ie there are copper electrodes and the electrolyte is CuSO4, then the positive electrode is oxidized, and the negative electrode is reduced. thie smeans that the concentration of the metal ion is constant as it is being removed and deposited at the same rate.
[edit] Electrorefining
Anode made from impure metal, cathode = sheets of pure metal,
the electrolyte is a compound of the metal.
Eg. copper: Oxidation at anode produces copper (II) ions and 2e-
at cathode the impure copper(II) joins with 2e- produced at anode to create copper metal. the impurities drop to the bottom of the cell where they can be removed.
[edit] Factors affecting discharge of ions
- Duration - More time means more electrons can be passed from one electrode to the other.
- Current - More current, more electrons available, more products
- Charge of the ion - Ions with 2+ charge require 2mol of electrons/ mol of products, therefore smaller charge = more products / electron flow.
To find how much is produced = Current*time/96500 = moles of electrons. Then you can work out how many moles of electrons you need per mol of product, and therefore find how much product will be produced, and the weight.
[edit] Standard electrode potentials
The standard electrode potential E0 (a table is found in the IB data booklet), is worked out by using the hydrodgen electrode.
[edit] Hydrogen Electrode
The hydrogen electrode is made up of the H half-cell, which is H2 gas at 1 atm pressure, 298K, 1.0mol/dm3 hydrogen ion content with a Pt electrode. This is assigned 0V.
If the metal in the other half cell is above hydrogen in the reactivity series, then these lose electrons to hydrogen and the electrode potential is negative. If the half-cell has a metal below hydrogen, then electrons will flow from H to the metal and produce a positive value.
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E0cell is the difference between electrode potentials in a cell. If the cell is written with reduction on the right in the standard cell notation (Metal/Ions||Ions/Metal), then the E0 will be the left hand side subtracted from the right hand side. In reality, the easiest way to remember it is that it is always More Positive - More Negative. You should normally get a positive value since a positive E0 means a negative Gibbs free energy, which means the reaction is spontaneous.
ELECTRONS ALWAYS FLOW FROM THE MORE NEGATIVE HALF-CELL TO THE MORE POSITIVE HALF-CELL. Thus you can always tell which is undergoing oxidation and which has reduction.
[edit] HL Material
Topic 19 is the additional HL material for Topic 10.
