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Topic 19 Oxidation and reduction

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

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Topic 19 Oxidation and reduction

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  1. Topic 19Oxidation and reduction • Standard electrode potential • Electrolysis • ElectroplatingPurification

  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 101.3 kPa

  3. 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 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= -0.76 V • In the lower part the redox couple wants to react to the right. E.gCu2+ + e- Cu Eq= 0.34 V • => Eqcell = Eq+- Eq- = 0.34 - (-0.76) = 1.10 V

  6. 19.2 Electrolysis of molten NaCl 2 Cl- (l)→ Cl2 (g) +2 e- Na+ (l) + e-→ Na (l)

  7. 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

  8. Examples of water reacting in electrolysis

  9. 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 dilutedNaClsolution => Oxygen + Hydrogen ion Anode: 2 H2O  O2 + 4 H+ + 4e-Eq= 1.23 V • A concentrated NaClsolution => Chlorine gas Anode: 2 Cl- (l)→ Cl2 (g) + 2 e-Eq= 1.36 V

  10. 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

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

  12. 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

  13. 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

  14. 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

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