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Electrolysis: An Introduction VCE Chemistry Unit 4: Chemistry at Work Area of Study 2 – Using Energy.

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Presentation on theme: "Electrolysis: An Introduction VCE Chemistry Unit 4: Chemistry at Work Area of Study 2 – Using Energy."— Presentation transcript:

1 Electrolysis: An Introduction VCE Chemistry Unit 4: Chemistry at Work Area of Study 2 – Using Energy

2 During electrolysis, electrical energy is converted into chemical energy. This is the reverse of the process that occurs in a galvanic cell where spontaneous reactions are a source of electrical energy. Electrolysis has a variety of industrial applications. These include electroplating and electrorefining as well as the recharging secondary cells. Non-spontaneous reactions are forced to occur in electrolytic cells by passing an electric current from an external power source through an electrolyte. The potential energy of the system is increased as electrical energy is forced into the cell. Electrolysis

3 In galvanic cells the reactions are separated so electrons can be used as they are transferred. Electrolytic cells have two electrodes and an electrolyte. Electrolytic cells differ from galvanic cells in that the anode is the positive electrode while the cathode is the negative electrode. The polarities of the anode and cathode change because in galvanic cells the polarity is imposed by the reactions themselves. At the anode, electrons are produced and at the cathode electrons are consumed. In electrolytic cells the polarity is imposed by the power source. This pulls electrons from the anode and forces electrons onto the cathode. Galvanic & Electrolytic Cells - A Comparison

4 Regardless of which cell, electron flow is always the same – from anode to cathode! Oxidation always occurs at the anode and reduction always occurs at the cathode! Electrolytic Cells

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6 When electrons are forced into the cell, the oxidants present at the surface of the electrode compete with one another to accept them. The strongest of the oxidants present preferentially accepts the electrons and is discharged, i.e. the oxidant highest on the electrochemical series accepts the electrons. Similarly, at the anode where the electrons are being withdrawn, the reductants present at the anode surface compete to donate electrons. The strongest of the reductants is preferentially discharged. Predicting ElectrolyticReactions Predicting Electrolytic Reactions

7 Note that if two competitors have very similar E° values, they will probably both be discharged simultaneously. In aqueous solutions, the water may be involved in either the oxidation or the reduction reaction or both. Predicting Electrolytic Reactions

8 Predict the products at each electrode during electrolysis for the following 1 M solutions (unreactive electrodes used). The nitrate ion is not involved in any of the reactions. a. Copper(II) bromide b. Sodium iodide c. Lead nitrate d. Zinc chloride e. Aluminium nitrate Predicting Electrolytic Reactions - Questions

9 a. 1 M copper(II) bromide Species present at the electrodes are Cu 2+ (aq), Br - (aq) and H 2 O (l) Possible reductions: Cu 2+ (aq) + 2e -  Cu (s)+0.34 V 2H 2 O (l) + 2e -  H 2 (g) + 2OH - (aq)-0.83 V Here the strongest oxidant (most positive E o value) is Cu 2+. Thus, at the cathode Cu forms. Possible oxidations: Br 2 (g) + 2e -  2Br - (aq) V O 2 (g) + 4H + (aq) + 4e -  2H 2 O (aq) V Here the strongest reductant (least positive E o value) is 2Br -. Thus at the anode Br 2 forms. Overall equation: Cu 2+ (a) + 2Br - (aq)  Cu (s) + Br 2 (g) Predicting Electrolytic Reactions - Solutions

10 b.1 M sodium iodide Species present at the electrodes are Na + (aq), I - (aq) and H 2 O (l) Possible reductions: Na + (aq) + e -  Na (s)-2.71 V 2H 2 O (l) + 2e-  H 2 (g) + 2OH - (aq)-0.83 V The strongest oxidant (most positive E o value) is H 2 O. Thus at the cathode H 2 and 2OH - form. Possible oxidations: I 2 (g) + 2e -  2I - (aq) V O 2 (g) + 4H + (aq) + 4e -  2H 2 O (aq) V The strongest reductant (least positive E o value) is 2I -. Thus at the anode I 2 forms. Overall equation: 2H 2 O (l) + 2I - (aq)  H 2 (g) + 2OH - (aq) + I 2 (g) Predicting Electrolytic Reactions - Solutions

11 c.1 M lead(II) nitrate Species present at the electrodes are Pb 2+ (aq) and H 2 O (l) Possible reductions: Pb 2+ (aq) + 2e -  Pb (s)+0.13 V 2H 2 O (l) + 2e -  H 2 (g) + 2OH - (aq)-0.83 V The strongest oxidant (most positive E o value) is Pb 2+. Thus, at the cathode Pb forms. Possible oxidations: O 2 (g) + 4H + (aq) + 4e -  2H 2 O (aq) V The strongest reductant (least positive E o value) is 2H 2 O. Thus at the anode O 2 and 4H + form. Overall equation: 2Pb 2+ (aq) + 2H 2 O (l)  2Pb (s) + O 2 (g) + 4H + (aq ) Predicting Electrolytic Reactions - Solutions

12 d.1 M zinc chloride Species present at the electrodes are Zn 2+ (aq), Cl - (aq) and H 2 O (l) Possible reductions: Zn 2+ (aq) + 2e -  Zn (s)-0.76 V 2H 2 O (l) + 2e -  H 2 (g) + 2OH - (aq)-0.83 V The strongest oxidant (most positive E o value) is Zn 2+. Thus, at the cathode Zn forms. Possible oxidations: Cl 2 (g) + 2e -  2Cl - (aq) V O 2 (g) + 4H + (aq) + 4e -  H 2 O (l) V The strongest reductant (least positive E o value) is 2H 2 O. Thus at the anode O 2 and 4H + form. Overall equation: 2Zn 2+ (aq) + 2H 2 O (aq)  2Zn (s) + O 2 (g) + 4H + (aq) Predicting Electrolytic Reactions - Solutions

13 e.1 M aluminium nitrate Species present at the electrodes are Al 3+ (aq) and H 2 O (l) Possible reductions: Al 3+ (aq) + 3e -  Al (s)-1.71 V 2H 2 O (l) + 2e -  H 2 (g) + 2OH - (aq)-0.83 V The strongest oxidant (most positive E o value) is 2H 2 O. Thus at the cathode H 2 and 2OH - form. Possible oxidations: O 2 (g) + 4H + (aq) + 4e -  2H 2 O (aq) V The strongest reductant (least positive E o value) is 2H 2 O. Thus at the anode O 2 and 4H + form. Overall equation: 6H 2 O (l)  2H 2 (g) + 4OH - (aq) + O 2 (g) + 4H + (aq) Predicting Electrolytic Reactions - Solutions

14 In this application, the cathode is coated with a thin layer of metal from a solution containing ions of the metal. This is done to improve the appearance or to prevent corrosion by the application of a protective layer. The object to be plated is attached to the negative terminal of a power supply and becomes the cathode. It is then placed in an electrolyte solution containing ions of the metal that forms the plating. The anode is made from the metal that is to form the plating. To ensure the coating on the surface is strongly bonded and smooth, the electrolyte is a complex mixture of ions and the conditions (voltage, current, time) carefully controlled. Applied Electrolysis - Electroplating

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16 Metals can be purified by electrolysis. The impure metal is used as the anode and oxidised. Metals that are stronger reductants will also be discharged at the electrode surface. Any constituents of the anode that are weaker reductants are not oxidised but fall as uncharged metal atoms to the bottom of the cell. This leaves an electrolyte solution containing ions of the metal to be refined and a small amount of other ions. The mixture of metal ions in solution compete for electrons at the cathode surface. As the dissolved impure ions are weaker oxidants, the metal to be refined is preferentially reduced at the cathode, and is in a much purer form. Applied Electrolysis - Electrorefining

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18 Applied Electrolysis Sodium from Molten Sodium Chloride – The Downs Cell

19 ProductsSodium metal and chlorine gas must be kept separate otherwise they will spontaneously react Anode reaction Cathode reaction Overall reaction 2Cl - (l)  Cl 2 (g) + 2e - Na + (l) + e -  Na (l) 2Na (l) + 2Cl - (l)  2Na (l) + Cl 2 (g) Anode (+) Cathode (-) Carbon in centre of cell Cylindrical iron near edge of the cell ConditionsMolten NaCl and CaCl 2 electrolyte added to the melt to lower the melting point from 801 o C to below 600 o C 7 Volts, Amps, electric current melts electrolyte Applied Electrolysis

20 Aluminium from Molten Alumina The Hall-Héroult Process Applied Electrolysis

21 Aluminium from Molten Alumina The Hall-Héroult Process ProductsAluminium metal and carbon dioxide gas must be kept separated otherwise they will spontaneously react Anode reaction Cathode reaction Overall reaction C (s) + 2O 2- (l)  CO 2 (g) + 4e - Al 3+ (l) + 3e -  Al (l) 2Al 2 O 3 (l) + 3C (s)  4Al (l) + 3CO 2 (g) Anode (+) Cathode (-) Graphite, lowered into top of cell and consumed during reaction Graphite lining and aluminium metal at the bottom of the cell ConditionsAl 2 O 3 dissolved in cryolite, Na 3 AlF 6, which lowers the melt point of alumina from 2045 o C to ~1000 o C 5 Volts, Amps, electric current melts electrolyte Applied Electrolysis

22 Sodium Hydroxide and Chlorine from Brine The Membrane Cell Applied Electrolysis

23 Sodium Hydroxide and Chlorine from Brine The Membrane Cell ProductsSodium hydroxide, chlorine gas and hydrogen gas must be kept separated otherwise they will spontaneously react Anode reaction Cathode reaction Overall reaction 2Cl - (l)  Cl 2 (g) + 2e - 2H 2 O (l) + 2e -  H 2 (g) + 2OH - (aq) 2H 2 O (l) + 2Cl - (l)  H 2 (g) + 2OH - (aq) + Cl 2 (g) Anode (+) Cathode (-) Graphite Steel mesh ConditionsCells separated by plastic semipermeable membrane that prevents mixing of products and allows Na + ions to migrate but neither Cl - or OH - ions. Ensures a purer NaOH solution. 3.5 – 5 Volts, up to Amps, electric current heats electrolyte Applied Electrolysis

24 If you wished to electroplate an object, you might ask yourself the following questions: How can I determine how much metal is being plated? How long should I leave the object being plated in the electroplating cell? What size electric current should be used? Faraday’s Laws of Electrolysis

25 Michael Faraday was a 19th century English chemist. His studies yielded two laws relating amount of electric current and the chemicals produced by the current or used to produce it. These laws enable us to: Work out the amount of energy required to discharge a metal ion and place it out on an object or To calculate the amount of metal produced in an electrolytic cell. Faraday’s Laws of Electrolysis

26 The mass of metal produced at the cathode is directly proportional to the quantity of electricity passed through the cell m  Q Electric charge, Q, is measured using the unit coulomb. The electric charge passing through a cell may be calculated from measurements of the current, I, through the cell and the time, t, for which the current flows. Charge (coulombs) = current (amps) x time (seconds) Q = I t Faraday’s Laws of Electrolysis – First Law

27 In order to produce one mole of metal, one two or three moles of electrons must be consumed. Faraday found that there was a certain charge associated with one mole of electrons. This amount of charge is now called the Faraday and is equivalent to (96 500) Coulomb. 1 Faraday = Coulomb Faraday’s Laws of Electrolysis – Second Law

28 In order to produce one mole of metal, one two or three moles of electrons must be consumed. Ag + (aq), Cu 2+ (aq) and Cr 3+ (aq) require 1, 2 and 3 moles of electrons for discharge. The quantity of electricity required will be: Amount of charge = no. of moles of metal ions x charge on an ion x 1 Faraday Q = n z F and Q = I t I t = n z F Faraday’s Laws of Electrolysis – Second Law

29 1.Calculate the number of mole of copper produced in an electrolytic cell if a current of 5.0 A at a voltage of 6.0 V flows through a solution of copper ions for 10 minutes. 2.Calculate the time taken to deposit 1.00 g of copper onto an object that is placed in a solution of copper nitrate, Cu(NO 3 ) 2, and has a current of 2.50 A flowing through it. 3.In an operating Hall-Héroult cell, a current of A is used at 5.0 V. Calculate the mass of aluminium that would be produced if this cell operates continuously for 1 day. 4.The electrolysis of a solution of chromium ions using a current of 2.2 A for 25 minutes produced 0.60 g of chromium. Calculate the charge on the chromium ion. 5.Calculate the masses of metal produced when 600 Faraday of charge is used to reduce the ions of aluminium, silver and zinc. Faraday’s Laws of Electrolysis – Questions

30 1.n (Cu) = I t / z F n (Cu) = 5.0 x 10 x 60 / 2 x n (Cu) = 1.6 x mol That is, 1.6 x mole of copper would be produced 2.t = n (Cu) z F / I t = (1.00 / 63.5) x 2 x / 2.5 t = 1216 seconds That is, it takes 20 minutes 15 seconds to deposit 1.00 g of copper 3.n (Al) = I t / z F n (Al) = x 24 x 60 x 60 / 3 x n (Al) = mol m (Al) = x 27 = g That is, 1.2 tonne of aluminium is produced per day Faraday’s Laws of Electrolysis – Solutions

31 4.z (Cr) = I t / n F z (Cr) = 2.2 x 25 x 60 / (0.60 / 52) x z (Cr) = 2.96 As the charge on an ion is a small integer, it must be 3+. The ion is Cr Q = m z F / M As Q = 600 F, then m = 600 M / z m (Al) = 600 x 27 / 3 = 5400 g m (Ag) = 600 x / 1 = g m (Zn) = 600 x 65.4 / 2 = g The same 600 F of charge would produce different masses of these metals; 5.4 kg of Al, nearly 20 kg of Zn and almost 65 kg of Ag. Faraday’s Laws of Electrolysis – Solutions

32 Q.Australia leads the world in the per capita recycling of aluminium. Explain why the production of aluminium uses so much energy and therefore why it is so much better to recycle aluminium. A.Aluminium is a reactive metal - it’s low on the electrochemical series. The only method of reducing Al ions is electrolytically, hence the energy costs will be quite high. Because Al is so low on the electrochemical series, most other reactions will occur more easily - any water present would be reduced in preference to Al. From Faraday’s Second Law, no. of moles of Al is inversely related to the charge of Al. Thus more charge is required to produce Al than almost any other metal. Mass of Al produced per mole of charge depends directly on the molar mass of Al. Al has the lowest molar mass of the commonly used industrial metals. Faraday’s Laws of Electrolysis – Application


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