Topic 19 Oxidation and reduction

• Standard electrode potential• Electrolysis• Electroplating

Purification

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 101.3 kPa

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

Standard electrode potential, Eq

Eqcell = Eq

+ - Eq

-

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 = -0.76 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

- = 0.34 - (-0.76) = 1.10 V

19.2 Electrolysis of molten NaCl

2 Cl- (l)→ Cl2 (g) + 2 e- Na+ (l) + e- → Na (l)

Electrolysis in an aqueous solution The water can be either oxidised or reducedAnode: 2 H2O O2 + 4 H+ + 4e- Eq = 1.23 V Cathode: 2 H2O + 2e- H2 + 2 OH- Eq = -0.83 V

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 Hydrogen + Hydroxide ion Oxygen + Hydrogen ion

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 ionAnode: 2 H2O O2 + 4 H+ + 4e- Eq = 1.23 V

– A concentrated NaCl solution => Chlorine gasAnode: 2 Cl- (l)→ Cl2 (g) + 2 e- Eq = 1.36 V

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 Cu2+ + 2 Cl-

Chlorine reacts with copper

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.

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: 96 500 C = Faradays

constant

=> moles of electrons = Current * Time / 96 500

You electrolyse a zinc ion solution for 3 hour in 2 Amp. The mass of Zn(s)?Zn2+ + 2 e- Zn(s)

1. n of e- = 2*( 3*60*60)/96 500 = 0.22 mol2. The amount of substance of Zn = 0.22 mol/2

= 0.11 mol3. 0.11 mol * 65.4 g/mol = 7.2 g

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