Houghton Mifflin Company and G. Hall. All rights reserved. 1 Lecture 25 Electrolysis  Define electrolysis?  Some examples.  What are the values of  G and E cell ?  Electrolysis of water.  Some industrial applications.  Corrosion

Houghton Mifflin Company and G. Hall. All rights reserved. 2 Two Types of Cells  Cell 1: does work by releasing free energy from a spontaneous reaction to produce electricity such as a battery.  Cell 2: does work by absorbing free energy from a source of electricity to drive a non-spontaneous reaction such as electroplating.

Houghton Mifflin Company and G. Hall. All rights reserved. 3 A voltaic (Galvanic) cell can power an electrolytic cell

Houghton Mifflin Company and G. Hall. All rights reserved. 4 Electrolysis  The splitting (lysing) of a substance or decomposing byforcing a current through a cell to produce a chemical change for which the cell potential is negative.

Houghton Mifflin Company and G. Hall. All rights reserved. 5 Electrolysis  Previously our lectures on electrochemistry were involved with voltaic cells i.e. cells with E cell > 0 and  G < 0 that were spontaneous reactions.  Today we discuss electrochemical cells where E cell 0 that are non- spontaneous reactions and require electricity for the reactions to take place. We can take a voltaic cell and reverse the electrodes to make an electrochemical cell.

Houghton Mifflin Company and G. Hall. All rights reserved. 6 Voltaic Electrolytic

Houghton Mifflin Company and G. Hall. All rights reserved. 7

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9 Fig

Houghton Mifflin Company and G. Hall. All rights reserved. 10 Fig : Car battery, both voltaic and electrochemical cell.

Houghton Mifflin Company and G. Hall. All rights reserved. 11 Increase oxidizing power Increase reducing power

Houghton Mifflin Company and G. Hall. All rights reserved. 12 A standard electrolytic cell. A power source forces the opposite reaction

Houghton Mifflin Company and G. Hall. All rights reserved. 13 Electrolysis

Houghton Mifflin Company and G. Hall. All rights reserved. 14 (a) A silver-plated teapot. (b) Schematic of the electroplating of a spoon.

Houghton Mifflin Company and G. Hall. All rights reserved. 15 Schematic of the electroplating of a spoon. AgNO 3 (aq)

Houghton Mifflin Company and G. Hall. All rights reserved. 16 The electrolysis of water produces hydrogen gas at the cathode (on the right) and oxygen gas at the anode (on the left).

Houghton Mifflin Company and G. Hall. All rights reserved. 17 Fig Electrolysis of water

Houghton Mifflin Company and G. Hall. All rights reserved. 18 Electrolysis of water  At the anode (oxidation):  2H 2 O(l) + 2e - = H 2 (g) + 2OH - (aq) E=-0.42V  At the cathode (reduction):  2H 2 O(l) = O 2 (g) + 4H + (aq) + 4e - E= 0.82V  Overall reaction after multiplying anode reaction by 2,  2H 2 O(l) = 2H 2 (g) + O 2 (g)  E o cell = = V

Houghton Mifflin Company and G. Hall. All rights reserved. 19 Electrolysis: Consider the electrolysis of a solution that is 1.00 M in each of CuSO 4 (aq) and NaCl(aq)  Oxidation possibilities follow.  2Cl – (aq) = Cl 2 (g) + 2e – E° = –1.358 V  2SO 4 2– (aq) = S 2 O 8 2– (aq) + 2e – E° = –2.01 V  2H 2 O = 4H + (aq) + O 2 (g) + 4e – E° = –1.229 V  Reduction possibilities follow:  Na + (aq) + e – = Na(s)E° = –2.713 V  Cu 2+ (aq) + 2e – = Cu(s)E° = V  2H 2 O + 2e – = H 2 (g) + 2OH – (aq)E° = V

Houghton Mifflin Company and G. Hall. All rights reserved. 20 Electrolysis  We would choose the production of O 2 (g) and Cu(s).  But the voltage for producing O 2 (g) from solution is considerably higher than the standard potential, because of the high activation energy needed to form O 2 (g).  The voltage for this half cell seems to be closer to –1.5 V in reality.  The result then is the production of Cl 2 (g) and Cu(s). anode, oxidation: 2Cl – (aq) = Cl 2 (g) + 2e – E° = –1.358 V  cathode, reduction: Cu 2+ (aq) + 2e – : Cu(s) E° = V  overall: CuCl 2 (aq) : Cu(s) + Cl 2 (g) E = –1.021 V  We must apply a voltage of more than V to cause this reaction to occur.

Houghton Mifflin Company and G. Hall. All rights reserved. 21 E = V

Houghton Mifflin Company and G. Hall. All rights reserved. 22 E = 1.07 V E cell = = -3.44V

Houghton Mifflin Company and G. Hall. All rights reserved. 23

Houghton Mifflin Company and G. Hall. All rights reserved. 24 Prob. 21.9

Houghton Mifflin Company and G. Hall. All rights reserved. 25 Stoichiometry of electrolysis: Relation between amounts of charge and product  Faraday’s law of electrolysis relates to the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell (half reaction).  Each balanced half-cell shows the relationship between moles of electrons and the product.

Houghton Mifflin Company and G. Hall. All rights reserved. 26 Application of Faraday’s law  1. First balance the half-reactions to find number of moles of electrons needed per mole of product.  2. Use Faraday constant (F = 9.65E4 C/mol e - ) to find corresponding charge.  3. Use the molar mass of substance to find the charge needed for a given mass of product.  1 ampere = 1 coulomb/second or 1 A = 1 C/s  A x s = C

Houghton Mifflin Company and G. Hall. All rights reserved. 27 Stoichiometry of Electrolysis 4 How much chemical change occurs with the flow of a given current for a specified time?  current and time  quantity of charge   moles of electrons  moles of analyte   grams of analyte

Houghton Mifflin Company and G. Hall. All rights reserved. 28 Fig

Houghton Mifflin Company and G. Hall. All rights reserved. 29 Doing work with electricity.

Houghton Mifflin Company and G. Hall. All rights reserved. 30

Houghton Mifflin Company and G. Hall. All rights reserved. 31 Industrial Applications of Electrolysis

E_cells.ppt.. All rights reserved. 32 What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)? Na + Cl - Let’s examine the electrolytic cell for molten NaCl.

All rights reserved. E_cells.ppt battery Na (l) electrode half-cell Molten NaCl Na + Cl - Na + Na + + e -  Na2Cl -  Cl 2 + 2e - Cl 2 (g) escapes Observe the reactions at the electrodes NaCl (l) (-) Cl - (+)

E_cells.ppt. All rights reserved battery e-e- e-e- NaCl (l) (-)(+) cathode anode Molten NaCl Na + Cl - Na + Na + + e -  Na 2Cl -  Cl 2 + 2e - cations migrate toward (-) electrode anions migrate toward (+) electrode At the microscopic level

E_cells.ppt. All rights reserved. 35 Molten NaCl Electrolytic Cell cathode half-cell (-) REDUCTION Na + + e -  Na anode half-cell (+) OXIDATION2Cl -  Cl 2 + 2e - overall cell reaction 2Na + + 2Cl -  2Na + Cl 2 X 2 Non-spontaneous reaction!

Houghton Mifflin Company and G. Hall. All rights reserved. 36 The Downs Cell for the Electrolysis of Molten Sodium Chloride

Houghton Mifflin Company and G. Hall. All rights reserved. 37 If the products are mixed, the result is household bleach. 2 NaOH(aq) + Cl 2 (g) = NaCl(aq) + NaOCl(aq) + H 2 O

Houghton Mifflin Company and G. Hall. All rights reserved. 38 The Mercury Cell for Production of Chlorine and Sodium Hydroxide

Houghton Mifflin Company and G. Hall. All rights reserved. 39 A schematic diagram of an electrolytic cell for producing aluminum by the Hall- Heroult process.

Houghton Mifflin Company and G. Hall. All rights reserved. 40 Fig A schematic diagram of an electrolytic cell for producing aluminum by the Hall-Heroult process.

E_cells.ppt. 41 The Hall Process for Aluminum  Electrolysis of molten Al 2 O 3 mixed with cryolite – lowers melting point  Cell operates at high temperature – 1000 o C  Aluminum was a precious metal in  A block of aluminum is at the tip of the Washington Monument!

42 carbon-lined steel vessel acts as cathode CO 2 bubbles Al (l) Al 2 O 3 (l) Draw off Al (l) - + Cathode: Al e -  Al (l) Anode: 2 O -2 + C (s)  CO 2 (g) + 4e - from power source Al +3 O -2 Al +3 O -2 graphite anodes  e- e- e- e-  From:

E_cells.ppt. 43 The Hall Process Cathode: Al e -  Al (l) Anode: 2 O -2 + C (s)  CO 2 (g) + 4e - 4 Al O C (s)  4 Al (l) + 3 CO 2 (g) x 4 x 3 The graphite anode is consumed in the process.

Houghton Mifflin Company and G. Hall. All rights reserved. 44 Fig : Production of solid Mg

Houghton Mifflin Company and G. Hall. All rights reserved. 45 Corrosion  Electrochemistry plays a major role in corrosion and protection against it.

46 Calculating the cell potential, E o cell, at standard conditions Fe e -  Fe E o = v O 2 (g) + 2H 2 O + 4e -  4 OH - E o = v This is corrosion or the oxidation of a metal. Consider a drop of oxygenated water on an iron object Fe H 2 O with O 2 Fe  Fe e - -E o = v2x 2Fe + O 2 (g) + 2H 2 O  2Fe(OH) 2 (s) E o cell = v reverse From:

Houghton Mifflin Company and G. Hall. All rights reserved. 47 Cathodic Protection Against Corrosion Underground steel pipes offer the strength to transport fluids at high pressures, but they are vulnerable to corrosion driven by electrochemical processes. A measure of protection can be offered by driving a magnesium rod into the ground near the pipe and providing an electrical connection to the pipe. Since the magnesium has a standard potential of volts compared to -.41 volts for iron, it can act as a anode of a voltaic cell with the steel pipe acting as the cathode. With damp soil serving as the electrolyte, a small current can flow in the wire connected to the pipe. The magnesium rod will be eventually consumed by the reactioncorrosionvoltaic cell Mg(s) -> + Mg 2+ (aq) + 2e - while the steel pipe as the cathode will be protected by the reaction O 2 (g) + 2H 2 O(l) + 4e - -> 4OH - (aq). From:

Houghton Mifflin Company and G. Hall. All rights reserved. 48 Lecture summary  Electrolysis is often the reverse of voltaic cell in that E cell 0 and reaction is non-spontaneous.  Electrolysis of water and to produce O 2 and H 2.  Faraday’s law allows us to determine how much current is needed to produce a certain amount of an element.  Industrial applications are numerous for producing a variety of solid elements (Al, Mg, Na, etc).