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For the SHE, we assign 2H + ( aq , 1 M ) + 2e -  H 2 ( g , 1 atm) E  red = 0.

For the SHE, we assign 2H + ( aq , 1 M ) + 2e -  H 2 ( g , 1 atm) E  red = 0. For Zn: E  cell = E  red (cathode) - E  red (anode) 0.76 V = 0 V - E  red (anode). Therefore, E  red (anode) = -0.76 V. Standard reduction potentials must be written as reduction reactions:

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For the SHE, we assign 2H + ( aq , 1 M ) + 2e -  H 2 ( g , 1 atm) E  red = 0.

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  1. For the SHE, we assign • 2H+(aq, 1M) + 2e- H2(g, 1 atm) • Ered = 0. Chapter 20

  2. For Zn: • Ecell = Ered(cathode) - Ered(anode) • 0.76 V = 0 V - Ered(anode). • Therefore, Ered(anode) = -0.76 V. • Standard reduction potentials must be written as reduction reactions: • Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V. Chapter 20

  3. Changing the stoichiometric coefficient does not affect Ered. • Therefore, • 2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V. Chapter 20

  4. Reactions with Ered < 0 are spontaneous oxidations relative to the SHE. • The larger the difference between Ered values, the larger Ecell. Chapter 20

  5. Oxidizing and Reducing Agents • The more positive Ered the stronger the oxidizing agent on the left. • The more negative Ered the stronger the reducing agent on the right. Chapter 20

  6. Chapter 20

  7. More generally, for any electrochemical process Chapter 20

  8. Example: For the following cell, what is the cell reaction and Eocell? Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s) Al3+(aq) + 3e-→ Al(s); EoAl = -1.66 V Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V Chapter 20

  9. Spontaneity of Redox Reactions • EMF and Free-Energy Change • We can show that • G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. • We define • Since n and F are positive, if G > 0 then E < 0. Chapter 20

  10. The Nernst Equation Chapter 20

  11. This gives: • At a temperature of 298 K: Chapter 20

  12. Cell EMF and Chemical Equilibrium • A system is at equilibrium when G = 0. • E = 0 V and Q = Keq: Chapter 20

  13. In the problem presented before, calculate ΔGo and Keq Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s) Al3+(aq) + 3e-→ Al(s); EoAl = -1.66 V Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V Chapter 20

  14. Example: For the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Calculate the Ecell given that Eocell = +1.10 V, [Cu2+] = 5.0 M and [Zn2+] = 0.050 M Chapter 20

  15. Exercise: Calculate the emf generated by the cell that employs the following cell reaction: 2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq) Take Eocell to be +2.20 V; [Al3+] = 4.0 x 10-3M and [I-] = 0.010M Ans: +2.36 V Chapter 20

  16. Batteries • Battery- self-contained electrochemical power source with one or more voltaic cell. • When the cells are connected in series, greater emfs can be achieved. • Primary cell – nonrechargeable • Secondary cell-rechargeable Chapter 20

  17. 2 of 2 Prev Next How Hydrogen Fuels Cells Work • Lead-acid battery: Pb and PbO2 are electrodes and are immersed in sulphuric acid. Electrodes are separated by glass fibres or wood. • Alkaline battery: anode is Zn in contact with a concentrated solution of KOH. The cathode is a mixture of MnO2(s) and graphite separated from the anode by a porous fabric Chapter 20

  18. 2 of 2 Prev Next How Hydrogen Fuels Cells Work • Nickel-cadmium, Ni-metal hydride and Li-ion batteries Chapter 20

  19. 2 of 2 Prev Next How Hydrogen Fuels Cells Work Fuel cells Fuel cells differ from batteries in that they are not self-contained systems. They use conventional fuels e.g. H2 and CH4 to produce electricity. Chapter 20

  20. 2 of 2 Prev Next How Hydrogen Fuels Cells Work The most common fuel cell is the one involving the reaction of H2(g) and O2(g) to produce H2O(l). This is based on that if H2O can be split by electricity, then combining H2 and O2 should produce water and electricity. Chapter 20

  21. 2 of 2 Prev Next How Hydrogen Fuels Cells Work Chapter 20

  22. Corrosion • Corrosion of Iron • Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen. • Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l). • Anode: Fe(s)  Fe2+(aq) + 2e-. • Dissolved oxygen in water usually causes the oxidation of iron. • Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s). Chapter 20

  23. 2 of 2 Prev Next How Hydrogen Fuels Cells Work Galvanised iron, i.e. Fe coated with Zn uses the principle of electrochemistry to protect the iron from corrosion even after the surface coat is broken. The standard red. Pot. for Fe and Zn are: Fe2+(aq) + 2e- → Fe(s) Eored = -0.44 V Zn2+(aq) + 2e- → Zn(s) Eored = -0.76 V Chapter 20

  24. 2 of 2 Prev Next How Hydrogen Fuels Cells Work Protecting a metal from corrosion by making it the cathode in an electrochemical cell is called cathodic protection. The metal that is oxidised while protecting the cathode is called the sacrificial anode. Chapter 20

  25. Electrolysis of Aqueous Solutions • Nonspontaneous reactions require an external current in order to force the reaction to proceed. • Electrolysis reactions are nonspontaneous. • In voltaic and electrolytic cells: • reduction occurs at the cathode, and • oxidation occurs at the anode. • However, in electrolytic cells, electrons are forced to flow from the anode to cathode. Chapter 20

  26. In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.) Chapter 20

  27. Example, decomposition of molten NaCl. • Cathode: 2Na+(l) + 2e- 2Na(l) • Anode: 2Cl-(l)  Cl2(g) + 2e-. • Industrially, electrolysis is used to produce metals like Al. Chapter 20

  28. Electroplating • Active electrodes: electrodes that take part in electrolysis. • Example: electrolytic plating. Chapter 20

  29. Chapter 20

  30. Electroplating • Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO4: • Anode: Ni(s)  Ni2+(aq) + 2e- • Cathode: Ni2+(aq) + 2e- Ni(s). • Ni plates on the inert electrode. • Electroplating is important in protecting objects from corrosion. Chapter 20

  31. Quantitative Aspects of Electrolysis • We want to know how much material we obtain with electrolysis. • Consider the reduction of Cu2+ to Cu. • Cu2+(aq) + 2e- Cu(s). • 2 mol of electrons will plate 1 mol of Cu. • The charge of 1 mol of electrons is 96,500 C (1 F). • Since Q = It, the amount of Cu can be calculated from the current (I) and time (t) taken to plate. Chapter 20

  32. Example: When an aqueous solution of CuSO4 is electrolysed, Cu metal is deposited: Cu2+(aq) + 2e-→ Cu(s) If a constant current was passed for 5.00 h and 404 mg of Cu metal was deposited, what was the current? Ans: 6.81 x 10-2 A Chapter 20

  33. Example: The half reaction for formation of Mg metal upon electrolysis of molten MgCl2 is: Mg2+(aq) + 2e-→ Mg(s) a) Calculate the mass of Mg formed upon passage of a current of 60.0A for a period of 4.00 x 103s b) How many seconds would be required to produce 50.0g of Mg from MgCl2 if the current is 100.0A? Ans: a) 30.2g, b) 3.97 x 103s Chapter 20

  34. Electrical Work ∆G is a measure of maximum useful work, wmax, that can be extracted from the process: ∆ G = wmax = -nFE For an electrolytic cell, an external source of energy is used to bring about a non-spontaneous electrochemical process. In this case, w = nFEext Chapter 20

  35. Electrical work can be expressed in energy units of watts x time: 1 W = 1 J/s Usually, the kWh is used Example: Calculate the no. of kWh of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl2 if the applied emf is 5.00 V Ans: 11.0 kWh Chapter 20

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