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

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

Standard electrode potential, Eq Table 14 Data booklet 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

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 -

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

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

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 reduced Anode: 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

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

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 You electrolyse a zinc ion solution for 3 hour in 2 Amp. The mass of Zn(s)? Zn2+ + 2 e-  Zn(s) n of e- = 2*( 3*60*60)/96 500 = 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

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