Oxidising and reducing agents
Oxidation is the loss of electrons by a substance during a chemical reaction. The substance which accepts these electrons is called the oxidising agent.
Reduction is the gain of electrons by a substance during a chemical reaction. The substance which donates these electrons is called the reducing agent.
The substances being oxidised and reduced, and the oxidising and reducing agents, can be identified in ionic equations.
Example:
Zinc metal can displace copper ions from a solution of copper(II) sulphate. Copper metal and a solution of zinc(II) sulphate are produced.
Ionic equation:
Zn(s) + Cu2+(aq) + SO42-(aq) → Cu(s) + Zn2+(aq) + SO42- (aq)
The equation shows that the zinc atoms (metal) have formed positively charged zinc ions, i.e., they have lost electrons. This means the zinc atoms have beenoxidised.
The positively charged copper ions have formed copper atoms, i.e., they have gained the electrons which the zinc atoms have lost. This means the copper ions have been reduced.
(The sulphate ions have neither lost nor gained electrons; they have not taken part in the reaction. They are spectator ions).
The copper ion is acting as the oxidising agent, because it has accepted electrons from the zinc atom, i.e., caused the zinc atom to be oxidised.
The zinc atom is acting as the reducing agent, because it has donated electrons to the copper ion, i.e., caused the copper ion to be reduced.
Example:
When chlorine reacts with iron(II) ions in solution, chloride ions and iron(III) ions are produced.
Ionic equation:
Cl2(g) + 2Fe2+(aq) → 2Cl-(aq) + 2Fe3+(aq)
The equation shows that the Fe2+ ions have formed Fe3+ ions, i.e., each ion has lost an electron. This means the Fe2+ ions have been oxidised.
The Cl2 molecules have formed Cl- ions, i.e., they have gained the electrons which the Fe2+ ions have lost. This means the Cl2 molecules have been reduced.
The Cl2 molecule is acting as the oxidising agent, because it has accepted electrons from the Fe2+ ions, i.e., caused the Fe2+ ions to be oxidised.
The Fe2+ ions are acting as the reducing agent, because they have donated electrons to the Cl2 molecule, i.e., caused the Cl2 molecule to be reduced.
Although oxidation and reduction happen together, they are often shown separately in ion-electron equations. Many of these equations are shown in the SQA data booklet, page 11. They are written as reductions, but can be reversed to give the oxidation equation.
When copper ions are displaced from solution by zinc atoms, the copper ions are reduced to copper atoms, so the ion-electron equation is written as it is in the data booklet:
Cu2+(aq) + 2e-→ Cu(s)
The zinc atoms are oxidised to zinc ions, so the ion-electron equation in the data booklet has to be reversed:
Zn(s)→Zn2+(aq) + 2e-
The reduction and oxidation equations can be added together to make the overall redox equation. Note: the electrons must be balanced.
Cu2+(aq) + 2e-→ Cu(s) reduction
Zn(s) → Zn2+(aq) + 2e- oxidation
Add: Cu2+(aq) + Zn(s)→Cu(s) + Zn2+(aq) redox
If the electrons in the two ion-electron equations don't balance then the equations have to be multiplied to make them balance, before the two equations are added together.
Example:
When iron(III) nitrate reacts with sodium iodide, the iron(III) ions are reduced to iron(II) ions, and the iodide ions are oxidised to iodine molecules (the nitrate and sodium ions are spectator ions).
From the data booklet, the reduction equation is:
Fe3+(aq) + e-→ Fe2+(aq)
The oxidation equation is:
2I-(aq) → I2(s) + 2e-
The reduction equation has to be multiplied by two to balance the number of electrons in the oxidation equation.
2Fe3+(aq) + 2e-→ 2Fe2+(aq)
The balanced reduction and oxidation equations can now be added to give the redox equation.
2I-(aq)→ I2(s) + 2e- oxidation
2Fe3+(aq) + 2e-→ 2Fe2+(aq) reduction
Add: 2I-(aq)+ 2Fe3+(aq) → I2(s) + 2Fe2+(aq) redox
A number of redox reactions will not take place unless there is acid present, the acid supplies hydrogen ions. During the reaction, these hydrogen ions react to form water.
Example:
Iron(II) ions are oxidised to iron(III) ions, by reaction with acidified potassium permanganate.
The reduction equation is in the data booklet:
MnO4-(aq) + 8H+(aq) + 5e-→ Mn2+(aq) + 4H2O(l)
This can be worked out in a number of steps, if the formulae of the main reactant and product in the reduction are known. In this example:
MnO4-(aq) → Mn2+(aq)
Step 1 Balance all the elements, except hydrogen and oxygen. In this case, there are the same number of 'Mn' on each side of the equation (one):
MnO4-(aq) → Mn2+(aq)
Step 2 If oxygen is unbalanced, add H2O to the side short of oxygen; in this case, the right-hand side (RHS). Multiply until the number of oxygens on each side is the same:
MnO4-(aq) → Mn2+(aq) + 4H2O(l)
Step 3 If hydrogen is unbalanced, add H+(aq) to the side short of hydrogen; in this case, the left-hand side (LHS). Multiply until the number of hydrogens on each side is the same:
MnO4-(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l)
Step 4 The total charge must be the same on both sides of the equation. Only electrons can be added to one side or the other. Balance the charge, by adding the appropriate number of electrons to the side which is short of negative charge; in this case the LHS:
MnO4-(aq) + 8H+(aq) + 5e-→ Mn2+(aq) + 4H2O(l)
The charge on the left-hand side = (one negative) + (eight positive) + (five negative) = two positive.
The charge on the right-hand side = (two positive) + (no charge) = two positive.
Example:
This example is not in the data booklet.
Br2(l) → BrO- (aq)
Step 1 Double the number of bromine oxide molecules on the RHS to balance the bromine on the LHS:
Br2(l) → 2 BrO- (aq)
Step 2 This time add H2O(l) to the LHS to balance the oxygens on the RHS:
Br2(l) + 2H2O(l) → 2 BrO- (aq)
Step 3Now add H+(aq) to the RHS to balance the hydrogens on the LHS:
Br2(l) + 2H2O(l) → 2 BrO- (aq) + 4H+(aq)
Step 4 Two electrons have to be added to the RHS, to make the overall charge zero, i.e., the same as the LHS:
Br2(l) + 2H2O(l) → 2 BrO- (aq) + 4H+(aq) + 2e-
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