1. Topic 19 Oxidation and reduction
Standard electrode potential
Electrolysis
Electroplating Purification

2. 19.1 Standard electrode potential
You can find tabulated data of potentials of different metals and other redox couples.
They are measured relative the
standard hydrogen electrode:
H+(aq) ½ H2 (g)
Standard conditions: 298 K, concentrations 1 M, gas pressure kPa

3. Cu(s)| Cu2+ || Zn2+ | Zn(s)
Voltaic cell
This cell can also be described in a shorthand way:
Cu(s)| Cu2+ || Zn2+ | Zn(s)
| represent a phase boundary
|| represent a salt bridge

4. Standard electrode potential, Eq Table 14 Data booklet
Eqcell = Eq+ - Eq-

5. You can predict if a reaction will occur or not
In the upper part of the table the redox couple wants to react to the left.
E.g. Zn  Zn2+ + 2e- Eq = V
In the lower part the redox couple wants to react to the right.
E.g Cu2+ + e-  Cu Eq = 0.34 V
=> Eqcell = Eq+ - Eq- = (-0.76) = 1.10 V

6. Oxidised form Reduced form Na+ + e-  Na Good red. agent
Redox couple
Oxidised form Reduced form
Na+ + e  Na Good red. agent
Mg2+ + 2e-  Mg
Fe2+ + 2e-  Fe
2 H+ + 2e-  H2
Cu2+ + 2e-  Cu
I2 + 2e  2 I-
Br2 + 2e-  2 Br-
Good ox. agent F2 + 2e  2 F -

7. In the upper part of the redox reactivity series the redox couple prefer to be on the left side (oxidised form).
Sodium is rather Na+ than Na
In the lower part of the table the redox couple prefer to be on the right side (reduced form)
Fluorine is rather F- than F2

8. You can use the redox reactivity series to predict if a reaction is possible or not by comparing redox couples
If the redox couple standing above in the table is going to the left and the redox couple standing below is going to the right, then a reaction will occur.
Mg2+ + 2e-  Mg
Br2 + 2e-  2 Br-
=> Mg + Br2  MgBr2 is possible
If the above couple going to the right and the pair below to the left no reaction will occur.
E.g. F2 + 2I-  2F- + I2 OK but I2 + F-  no reaction

9. 19.2 Electrolysis of molten NaCl
2 Cl- (l)→ Cl2 (g) + 2 e Na+ (l) + e- → Na (l)

10. Electrolysis in an aqueous solution
The water can be either oxidised or reduced Anode: 2 H2O  O2 + 4 H+ + 4e- Eq = 1.23 V Cathode: 2 H2O + 2e-  H2 + 2 OH- Eq = V

11. Examples of water reacting in electrolysis
Electrolyte (aq)
At the cathode
At the anode
Copper(II) bromide
Copper
Bromine
Sodium iodide
Hydrogen + Hydroxide ion
Iodine
Silver nitrate
Silver
Oxygen + Hydrogen ion
Potassium sulphate

12. Small difference in Eq
If there is a small difference in potential between the water and the ion then the concentration can be the determining factor for the which product you will get
Eg. electrolysis of NaCl
A diluted NaCl solution => Oxygen + Hydrogen ion
Anode: 2 H2O  O2 + 4 H+ + 4e Eq = 1.23 V
A concentrated NaCl solution => Chlorine gas
Anode: 2 Cl- (l)→ Cl2 (g) + 2 e Eq = 1.36 V

13. Other electrode reactions
Sometimes the product formed at the electrode reacts with the electrode, e.g. if you have a copper anode in chloride solution:
2 Cl-  Cl2 + 2 e-
Chlorine is formed
Cu + Cl2  Cu Cl-
Chlorine reacts with copper

14. Electrolysis of Na2SO4(aq)
Water  (Hydrogen + Hydroxide ion) + (Oxygen + Hydrogen ion)
E = 1.23 – (-0.83) = 2.06 V
=> you need at least 2.06 V to split water.

15. The electric current is important when you calculate the quantity of the product
Electrical charge : Coulomb (C) or (As)
Charge = Current (A) * Time (s)
The charge in one electron: 1.6*10-19 C
The charge in one mol electrons: C = Faradays constant
=> moles of electrons =
Current * Time /

16. You electrolyse a zinc ion solution for 3 hour in 2 Amp
You electrolyse a zinc ion solution for 3 hour in 2 Amp. The mass of Zn(s)?
Zn e-  Zn(s)
n of e- = 2*( 3*60*60)/ = 0.22 mol
The amount of substance of Zn = 0.22 mol/2 = 0.11 mol
0.11 mol * 65.4 g/mol = 7.2 g

17. Electroplating- Purification
Copper in nature is often found as oxides or sulphides and the ore also contain other metals as zinc and silver. The copper and the metals are reduced to crude copper which can be purified by electrolysis