Reactions in which both oxidation and reduction occur are called redox reactions.
The acronym OILRIG might be familiar, reminding us that oxidation reactions involve a loss of electrons by a species (oxidation is loss) and reduction reactions involve a species gaining electrons (reduction is gain).
We can split equations such as this into half equations which show us the transfer of electrons more clearly. In the oxidation half equation sodium atoms become positively charged sodium ions by losing an electron. In the reduction half equation, oxygen atoms become negatively charged ions by gaining electrons.
Oxidising agents remove electrons from another substance (and they are reduced themselves). Non-metal elements such as the halogens (F2, Cl2) are very powerful oxidising agents as they are so keen to gain electrons themselves to form anions.
Reducing agents give electrons to another substance (and they are oxidised themselves).
In this example, sodium is the reducing agent, giving electrons to oxygen and sodium is itself oxidised. Oxygen is the oxidising agent, removing electrons from sodium and oxygen is itself reduced.
If I take a piece of zinc and partially submerge it in a blue solution of copper sulfate ,CuSO4, after a short time the zinc becomes coated in an orange-pink powder and eventually the blue solution fades to a greyish colour.
This is an example of a displacement reaction. Zinc is more reactive than copper, so it displaces the copper from its compound of copper sulfate. This happens in two steps theoretically.
- Zinc loses electrons to copper ions from the solution that are in contact with the zinc metal surface. The zinc atoms are oxidised and move into solution.
Zn(s) ⇾ Zn2+(aq) + 2e–
- The copper ions are reduced as they gain electrons and the newly formed copper atoms are deposited on the piece of zinc metal.
Cu2+(aq) + 2e– ⇾ Cu(s)
If we combine these equations such that the electrons cancel out, we have the ionic equation for the reaction …
Zn(s) + Cu2+(aq) ⇾ Zn2+(aq) + Cu(s)
You will note that the sulfate ions don’t feature. Nothing has happened to them so they are the spectator ions. The fully balanced equation looks like:
Zn(s) + CuSO4(aq) ⇾ ZnSO4(aq) + Cu(s)
When we are describing oxidising and reducing agents we name the agents as described in the full equation, rather than in the ionic or half equation. So in this example, the reducing agent is zinc and the oxidising agent is copper sulfate (not just Cu2+).
For non-ionic compounds e.g. NH3 and polyatomic ions e.g. SO42- writing half equations to follow electron movements during a reaction is not possible, so the concept of oxidation states or oxidation numbers was invented. We can deduce whether an element has been oxidised or reduced by following the change in oxidation number during the reaction.
Practice questions
- For each of the following redox reactions, write equations for the two half reactions and identify which is the oxidation half equation and which is the reduction half equation.
- Fe(s) + Pb2+(aq) ⇾ Fe2+(aq) + Pb(s)
- Zn(s) + 2H+(aq) ⇾ Zn2+(aq) + H2(g)
- 2Ag+(aq) + Mg(s) ⇾ Mg2+(aq) + 2Ag(s)
- 2Al(s) + 3Cu2+(aq) ⇾ 2Al3+(aq) + 3Cu(s)
- Cl2 (aq) + 2NaI (aq) ⇾ 2NaCl (aq) + I2 (aq)
- Zn(s) + S(s) ⇾ ZnS(s)
- Write the overall balanced equation for the following pairs of half equations. Identify the oxidising and reducing agents in each example.
- Ni2+(aq) + 2e– ⇾ Ni(s) ; Sr(s) ⇾ Sr2+(aq) + 2e–
- Ag+(aq) + e– ⇾ Ag(s) ; Hg(s) ⇾ Hg2+(aq) + 2e–
- Bi3+(aq) + 3e– ⇾ Bi(s) ; Co(s) ⇾ Co2+(aq) + 2e–
- Br2(aq) + 2e– ⇾ 2Br–(aq) ; Fe2+(aq) ⇾ Fe3+(aq) + e–
Answers
1.
| Oxidation half reaction | Reduction half reaction | |
|---|---|---|
| a. | Fe(s) ⇾ Fe2+(aq) + 2e– | Pb2+(aq) + 2e– ⇾ Pb(s) |
| b. | Zn(s) ⇾ Zn2+(aq) + 2e– | 2H+(aq) + 2e– ⇾ H2(g) |
| c. | Mg(s) ⇾ Mg2+(aq) + 2e– | Ag+(aq) + e– ⇾ Ag(s) |
| d. | Al(s) ⇾ Al3+(aq) + 3e– | Cu2+(aq) + 2e– ⇾ Cu(s) |
| e. | 2I– (aq) ⇾ I2 (aq) + 2e– | Cl2 (aq) + 2e– ⇾ 2Cl– (aq) |
| f. | Zn(s) ⇾ Zn2+(s) + 2e– | S(s) + 2e– ⇾ S2-(s) |
2.
- Ni2+(aq) + Sr(s) ⇾ Ni(s) + Sr2+(aq) ; Ni2+(aq) is the oxidising agent and Sr(s) is the reducing agent
- 2Ag+(aq) + Hg(s) ⇾ 2Ag(s) + Hg2+(aq) ; Ag+(aq) is the oxidising agent and Hg(s) is the reducing agent
- 2Bi3+(aq) + 3Co(s) ⇾ 2Bi(s) + 3Co2+(aq) ; Bi3+(aq) is the oxidising agent and Co(s) is the reducing agent
- Br2(aq) + 2Fe2+(aq) ⇾ 2Br–(aq) + 2Fe3+(aq) ; Br2(aq) is the oxidising agent and Fe2+(aq) is the reducing agent