1. ELECTROCHEMISTRY CHEM171 – Lecture Series Four : 2012/01  Redox reactions  Electrochemical cells  Cell potential  Nernst equation  Relationship between E cell, Δ G and K Brown et al., Chapter 16-1 to 16.5 and 16.8

2. Electrochemistry: study of the relationship between chemical change and electrical energy. Investigated through use of electrochemical cells that incorporate oxidation-reduction (or redox reactions) to produce electrical energy. An electrochemical process releases or absorbs energy, it always involves movement of electrons from one chemical species to another in a redox reaction. CHEM171 – Lecture Series Four : 2012/02

3. OXIDATION NUMBER OR STATES Consider the following compound: KMnO 4 -2 (-8 in total) +1 +1 + Mn + (-8) = 0  Mn +7 Consider the following reaction: Cu 2+ + 2e - → Cu +2 0 Cu has gained 2 electrons – oxidation number has decreased. CHEM171 – Lecture Series Four : 2012/03

4. Now consider Zn → Zn 2+ + 2e - Zn has lost 2 electrons – oxidation number has increased 0 2+ Oxidation refers to any process in which the oxidation number of an atom becomes more positive. Reduction refers to any process that leads to a decrease in the oxidation number of an atom. Oxidation is always accompanied by reduction, these reactions are called redox reactions CHEM171 – Lecture Series Four : 2012/04

5. BALANCING REDOX REACTIONS Redox reaction takes place in acidic media or basic media. Acidic media contains H + /H 2 O Basic media contains OH - /H 2 O CHEM171 – Lecture Series Four : 2012/05

6. OXIDIZING AND REDUCING AGENTS Consider the reaction: A + B m+  A n+ + B A is oxidized by B m+ to A n+ and B m+ is reduced by A to B A - oxidized, therefore reducing agent B m+ - reduced, therefore oxidizing agent An oxidising agent is itself reduced A reducing agent is itself oxidized CHEM171 – Lecture Series Four : 2012/06

7. ELECTROCHEMICAL CELLS Two types of cells based on general thermodynamics: Voltaic cell (or galvanic cell): uses a spontaneous reaction (  G < 0) to generate electrical energy (flashlights, CD player, car) Electrolytic cell: uses electrical energy to drive a nonspontaneous reaction (  G > 0) (electroplating) CHEM171 – Lecture Series Four : 2012/07

8. Two electrodes (conduct electricity between cell and surroundings) are dipped into electrolyte (mixture of ions, usually in aqueous solution) that are involved in the reaction or that carry the charge. Oxidation: anode Reduction: cathode CHEM171 – Lecture Series Four : 2012/08

9. CHEM171 – Lecture Series Four : 2012/09

10. Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) LHS RHS oxidation reduction anode cathode -ve charge +ve charge Mass of Zn rod decreases Mass of Cu rod increases [Zn 2+ ] increases [Cu 2+ ] decreases Zn(s)  Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu(s) CHEM171 – Lecture Series Four : 2012/10

11. Two ½ cells are connected by a salt bridge - tube containing a concentrated solution of electrolyte (KNO 3, KCl) which allows ions of electrolyte to migrate. CHEM171 – Lecture Series Four : 2012/11

12. STANDARD ELECTRODE POTENTIALS Standard electrode potentials help us predict whether a reaction takes place. Consider the following reactions: Cu 2+ + Zn → Zn 2+ + Cu Zn 2+ + Cu → Cu 2+ + Zn One of these reactions will occur spontaneously. CHEM171 – Lecture Series Four : 2012/12

13. Zn 2+ + 2e ‑  ZnE o = ‑ 0.76 V Cu 2+ + 2e ‑  CuE o = +0.34 V Consider the standard electrode potentials for: In the first reaction Cu 2+ + Zn → Zn 2+ + Cu Cu 2+ + 2e ‑  Cu : reduction Zn  Zn 2+ + 2e - : oxidation By convention: E o cell = E o (reduction) – E o (oxidation) CHEM171 – Lecture Series Four : 2012/13

14. Therefore, E o cell = +0.34 V – (-0.76 V) = 1.10 V Similarly, E o cell = -1.10 V for the reaction Zn 2+ + Cu → Cu 2+ + Zn It is a general rule that E o cell will always be positive for a spontaneous reaction. CHEM171 – Lecture Series Four : 2012/14

15. EXAMPLE Which is the direction of spontaneous change for the following reaction – to the left or to the right? 2Cr 3+ + 6Fe 3+ + 7H 2 O  Cr 2 O 7 2- + 14H + + 6Fe 2+ SOLUTION For the forward reaction, the two ½- reactions are: Fe 3+ + e -  Fe 2+ E o = +0.77 V Cr 2 O 7 2- +14H + + 6e -  2Cr 3+ +7H 2 O E o = +1.33 V CHEM171 – Lecture Series Four : 2012/15

16. E o cell = E o (Fe 3+ /Fe 2+ ) - E o (Cr 2 O 7 2- /Cr 3+ ) = +0.77 V – 1.33 V = - 0.56 V The reverse reaction is spontaneous E o cell = E o (Cr 2 O 7 2- /Cr 3+ ) - E o (Fe 3+ /Fe 2+ ) = 1.33 V - 0.77 V = + 0.56 V Cr 2 O 7 2- + 14H + + 6Fe 2+  2Cr 3+ + 6Fe 3+ + 7H 2 O CHEM171 – Lecture Series Four : 2012/16

17. NERNST EQUATION Cell need not operate at standard state conditions. We can calculate potential of a cell in which some or all components are not in standard states. Nernst equation n = number of electrons transferred and Q = reaction quotient CHEM171 – Lecture Series Four : 2012/17

18. EXAMPLE Consider the electrochemical cell represented by: Zn(s) │ Zn 2+ ║ Fe 3+ │ Fe(s) Determine E for the cell when the concentration of Fe 3+ is 10.0 M and that of Zn 2+ is 1.00 ×10 -3 M. SOLUTION Zn  Zn 2+ + 2e - Fe 3+ + 3e -  Fe Zn + Fe 3+  Zn 2+ + Fe CHEM171 – Lecture Series Four : 2012/18 E o cell = -0.04 V - (-0.76 V) = 0.72 V = 0.72 V + 0.109 V = 0.829 V

19. Δ G o = -nFE o cell wheren is the number of electrons transferred in the reaction and F is Faraday’s constant, the charge of 1 mole of electrons which can be found on the data sheet as 96485 C mol ‑ 1 Δ G o = -RTlnK Now from, RELATIONSHIP BETWEEN E o, Δ G o AND K CHEM171 – Lecture Series Four : 2012/19 RTlnK = nFE o R is the gas constant, 8.315 J mol ‑ 1 K ‑ 1 and T is the temperature, which for standard conditions is 298 K (25°C)

20. SOLUTION The reaction equation is MnO 4 - + 5Fe 2+ + 8H + ⇌ Mn 2+ + 5Fe 3+ + 4H 2 O The number of electrons being transferred, n, equals 5. Reduction: E o (MnO 4 - /Mn 2+ ) = 1.51 V Oxidation: E o (Fe 3+ /Fe 2+ ) = 0.77 V EXAMPLE Calculate the equilibrium constant for the reaction between MnO 4 - and Fe 2+ in acidic aqueous solution at 25°C CHEM171 – Lecture Series Four : 2012/20

21. E o cell = 1.51 V – 0.77 V = 0.74 V = 144 K = 3.45 x 10 62 CHEM171 – Lecture Series Four : 2012/21

22. ELECTROLYSIS Non-spontaneous reaction is caused by the passage of an electric current through a solution. Used to isolate metal or nonmetal from molten salt. 2Cl - (l)  Cl 2 (g) + 2e - anode (oxidation) Ca 2+ (l) + 2e -  Ca(s) cathode (reduction) Ca 2+ (l) + 2Cl - (l)  Ca(s) + Cl 2 (g) CHEM171 – Lecture Series Four : 2012/22

23. Quantitative Aspects of Electrolysis: Electroplating: Uses electrolysis to deposit a thin layer of one metal on another metal in order to improve beauty or resistance to corrosion. Faraday’s Law of Electrolysis: the amount of a substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell (and molar mass) CHEM171 – Lecture Series Four : 2012/23

24. EXAMPLE SOLUTION A current of 2.40 A is passed through a solution containing Cu 2+ (aq) for 30.0 minutes, with copper metal being deposited at the cathode. What mass of copper, in grams, is deposited. The reaction at the cathode is: Cu 2+ (aq) + 2e -  Cu(s) Charge (C) = current (A) × time (s) = (2.40 A)(1.8 × 10 3 s) = 4.32 × 10 3 s CHEM171 – Lecture Series Four : 2012/24 1 mol e - will carry a charge of 96500 C  x mol e - will carry a charge of 4320 C x = 4.48 × 10 -2 mol e -

25. Moles of Cu = ½(4.48 × 10 -2 mol) = 2.24 × 10 -2 mol Mass of Cu = (2.24 × 10 -2 mol)(63.55 g mol -1 ) = 1.42 g CHEM171 – Lecture Series Four : 2012/25

26. EXAMPLE How long must a current of 0.800 A flow to form 2.50 g of silver metal in an electroplating experiment? The cathode reaction is Ag + (aq) + e -  Ag(s) SOLUTION Moles of silver = (2.50 g)/(107.9 g mol -1 ) = 2.32 × 10 -2 mol Moles of electrons required = 2.32 × 10 -2 mol CHEM171 – Lecture Series Four : 2012/26 1 mol e - will carry a charge of 96500 C  2.32 × 10 -2 mol mol e - will carry x C x = 2240 C

27. Charge (C) = current (A) × time (s) 2240 C = (0.800 A)(time)  Time = 2.8 × 10 3 s (or 46.5 min) CHEM171 – Lecture Series Four : 2012/27