Redox Equations
e.g.
- Redox reactions involve both oxidation and reduction
- A redox reaction can be constructed from two half-equations; one representing an oxidation process, and the other a reduction process
- To construct a full equation from half equations:
- Balance the electrons
- Combine the equations
- Cancel the electrons
- Check the charge balance and stoichiometry
e.g.
Mg → Mg2+ + 2e-
Cu2++ 2e- → Cu
Cu2++ 2e- → Cu
1). Mg → Mg2+ + 2e-
Cu2+ + 2e → Cu
2). Mg + Cu2++ 2e- →Mg2+ + 2e- + Cu
3). Mg + Cu2+ → Mg2+ + Cu
Cu2+ + 2e → Cu
2). Mg + Cu2++ 2e- →Mg2+ + 2e- + Cu
3). Mg + Cu2+ → Mg2+ + Cu
Redox Reactions
- Reduction: The gain of electrons and decrease in oxidation number of an element
- Oxidation: The loss of electrons and increase in oxidation number of an element
- Redox reactions involve both oxidation and reduction
- Oxidising agents cause oxidation of other species, and so are themselves reduced
- Reducing agents cause reduction of other species, and so are themselves oxidised
- In the reaction below, H is reduced, Na is oxidised.
2 HCl + 2 Na→ 2 NaCl + H2
+1 0 +1 0
+1 0 +1 0
Redox Titrations
- Redox titrations can be carried out to show how much oxidising agent is needed to react exactly with a reducing agent.
- Transition metals have variable oxidation states, so they are often present in either the oxidising or reducing agent
- Manganate(VII) ions are readily reduced to Mn2+ ions under acidic conditions (purple to colourless). This can be used to find the amount of Fe2+ in a solution
Oxidation: Fe2+ → Fe3+ + e-
Reduction: 8H+ + MnO4- + 5e-→ Mn2+ + 4H2O
Overall: 8H+ + MnO4- + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Reduction: 8H+ + MnO4- + 5e-→ Mn2+ + 4H2O
Overall: 8H+ + MnO4- + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
- Fe2+ can also be oxidised by dichromate (VI) ions
Oxidation: Fe2+ → Fe3+ + e-
Reduction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Overall: 6Fe2+ + Cr2O72- + 14H+ → 2Cr3+ + 7H2O + Fe3
Electrode Potentials
- Redox reactions can be used in electrochemical cells to generate electricity (a flow of charge)
- Electrochemical cells consist of two half-cells. At one oxidation occurs, at the other reduction. Electrons flow between the two cells, driving the redox reaction
Zn → Zn2+ + 2e-
Cu2+ + 2e- → Cu
Cu2+ + 2e- → Cu
An equilibrium is reached: Cu2+ (aq) + 2e- ⇌ Cu (s)
- Each one of these beakers is a half-cell. A solution in a standard half-cell will have a concentration of 1.00 mol dm-3
- An electrode is a solid surface which allows the transfer of electrons to and from it.
- The system above describes metal/metal ion half-cells. It is possible to make an ion/ion cell by using the same element with different oxidation states e.g. a mixture of Fe2+ and Fe3+ establishes an equilibrium: Fe3+ (aq) + e- ⇌ Fe2+ (aq)
- In an ion/ion half cell there is no solid metal to transport electrons out of the cell so a platinum electrode is used
- The standard electrode potential, E^θ is the voltage measured under standard conditions when the half-cell is connected to a standard hydrogen electrode.
- Standard conditions includes 298K, 100kPa and 1.00 mol dm-3.
- The voltage measured is also known as the electromotive force of the cell (EMF).
- An electrochemical series is a list of standard electrode potentials of all the possible half-cells.
- The more negative the electrode potential, the more the oxidation (backwards) reaction is favoured.
Electrochemical Cells
- Electrochemical cells can be used as a commercial source of electrical energy
- In a rechargeable battery, when the chemicals have reacted fully, a potential difference can be applied to the cell in the opposite direction, which will regenerate the original chemicals
- Lithium-ion batteries are rechargeable:
At the positive electrode:
Li+ + CoO2 + e- → Li+[CoO2]- Eθ=+0.56 V
Li+ + CoO2 + e- → Li+[CoO2]- Eθ=+0.56 V
At the negative electrode:
Li → Li+ + e- Eθ= -3.04 V
Li → Li+ + e- Eθ= -3.04 V
Eθ cell = +0.56 V (-3.04 V) = + 3.60 V
- Some cells are non-rechargeable and disposed of when the chemicals have fully reacted.
- In fuel cells the chemicals are stored externally and are fed into the cell when electricity is required.
- An alkaline hydrogen fuel cell:
Positive electrode: O2 (g) + 2H2O (l) + 4e- → 4OH- (aq) Eθ= +0.40 V
Negative electrode: 2H2 (g) + 4OH- (aq) → 4H2O (l) + 4e- Eθ= - 0.83 V
Overall reaction: 2H2 (g) + O2 (g) → 2H2O (g) Eθ cell = +1.23 V
Negative electrode: 2H2 (g) + 4OH- (aq) → 4H2O (l) + 4e- Eθ= - 0.83 V
Overall reaction: 2H2 (g) + O2 (g) → 2H2O (g) Eθ cell = +1.23 V
Advantages | Disadvantages |
They are more efficient than burning fossil fuels. | Energy is needed to build the fuel cells and produce hydrogen – this energy comes from fossil fuels. |
They release water, which isn’t harmful. | Hydrogen is highly flammable so needs to be carefully handled. |
They do not need to be recharged – they keep producing electricity for as long as they have fuel. | Highly toxic and can ignite – Li is a very reactive metal. |
Variable Oxidation States
- Vanadium has 4 oxidation states II, III, IV & V
- Vanadium species in oxidation states IV, III and II are formed by the reduction of vanadate(V) ions by zinc in acidic solution
Reduction from V(V) to V(IV)
2VO2+ (aq) + 4H+ (aq) + Zn (s) → 2VO2+ (aq) + Zn2+ (aq) + 2H2O (l)
Reduction from V(IV) to V(III)
2VO2+ (aq) + 4H+ (aq) + Zn (s) → 2V3+ (aq) + Zn2+ (aq) + 2H2O (l)
Reduction from V(III) to V(II)
2V3+ (aq) + Zn (s) → Zn2+ (aq) + 2V2+ (aq)
- Manganate (VII) ions are readily reduced to Mn2+ ions under acidic conditions (purple to colourless). This can be used to find the amount of Fe3+ in a solution
Oxidation: Fe2+ → Fe3+ + e-
Reduction: 8H+ + MnO4- + 5e-→ Mn2+ + 4H2O
Overall: 8H+ + MnO4- + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
- When transition metals are oxidised, the 4s electrons are lost before the 3d electrons