IB Chemistry/Oxidation and Reduction
Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. A full reaction will have a net change of 0 electrons, meaning that the amount of electrons lost in one portion of the equation will equal the amount of electrons gained in another. In other words, there will be a reduced portion and an oxidized portion. Collectively, this is referred to as a redox reaction.
Oxidation and Reduction Half Equations
Each full reaction can be broken apart to its component half reactions, the oxidation and reduction parts. In a half reaction, electrons are involved as part of the reaction. We will consider the reaction of magnesium and oxygen to form magnesium oxide. Since this is an ionic compound, it will have a negative ion (O2-) and a positive ion (Mg2+). The first step will be the formation of these ions.
- Mg → Mg2+ + 2e-
- O2 + 4e- → 2O2-
As stated earlier, the loss and gain of electrons must be balanced. In this situation, we will double the entire oxidation reaction (that involving magnesium):
- 2Mg → 2Mg2+ +4e-
If we were to combine these half reactions, we would obtain the full equation:
- 2Mg + O2 → 2MgO
There are several mnemonics for oxidation and reduction, such as OILRIG (Oxidation Is Loss, Reduction Is Gain), and LEO GER (Lose Electrons: Oxidation, Gain Electrons: Reduction).
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:
- 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.
|HCl||+1 (H) and -1 (Cl)|
|H2O||+1 (H) and -2 (O)|
- 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 when in a peroxide (oxidation number is -1), or oxygen fluoride (oxidation number is +1)
- Hydrogen is always +1, except when in a hydride, for example sodium hydride (NaH), (oxidation number is -1)
We can quickly assign oxidation states using the observation above:
- 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 always 0,(1+1+(4*-8)=-6, so S must be +6.
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. Keep this in mind that when something is being neutralized or precipitated it is not a redox reaction.
An oxidizing agent is that which oxidizes something (it takes its electrons, thereby getting reduced). Oppositely, a reducing agent is one that reduces something (it loses its electrons, thereby getting oxidized).
It is possible for an atom to have an oxidation number of 0, for example in Ni(CO)4, due to the ligand nature of the bonds.
The reactivity of a metal is related to how readily it loses its outer electrons. This means that metals higher in the series will displace metals in an ionic reaction lower down. Metals high in the reactivity series are also better reducing agents, due to the fact that they lose their outer electrons more easily. As learned with periodicity, the electronegativity increases down the series. The reactivity series is below, which the better reducing agents are
A half cell is a metal in contact with a solution containing its own ions (for example, Zn in a solution of Zn2+). A voltaic cell is when there are two half cells connected together with a wire and a salt bridge. Electrons are liberated during an oxidation in one half cell, and then used in the reduction in the other half cell. This flow of electrons is the current from the voltaic cell. The salt bridge is used to allow for the ion transfer to eliminate the build up of charge in a half cell. In many simple cases, this can be something such as a porous cup.
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 hydrogen 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. This means that the concentration of the metal ion is constant as it is being removed and deposited at the same rate.
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.
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.
Standard electrode potentials
The standard electrode potential E0 (a table is found in the IB data booklet), is worked out by using the hydrogen electrode. Which is defined as the emf generated when it is connected to the standard hydrogen electrode by an external circuit and a salt bridge, measured under standard conditions. (All solution must have a concentration of 1.0mol/dm3. All gases must be at pressure 100kPa. All substance must be pure. Temperature at 298K.)
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.
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.
Topic 19 is the additional HL material for Topic 10.