17.7-8 Electrolysis & Applications

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17.7-8 Electrolysis & Applications17.7-8 Electrolysis & Applications

• Since chemical oxidation-reduction involves Since chemical oxidation-reduction involves the transfer of electrons from one substance the transfer of electrons from one substance to another, it should be possible to harness to another, it should be possible to harness the flow of electrons to produce electricity. the flow of electrons to produce electricity. We do this with voltaic cells.We do this with voltaic cells.

• Electricity can also be used to cause non-Electricity can also be used to cause non-spontaneous chemical reactions (i.e. spontaneous chemical reactions (i.e. recharging batteries). This process is called recharging batteries). This process is called electrolysiselectrolysis (carried out in electrolytic cells) (carried out in electrolytic cells)

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17.7 Electrolysis17.7 Electrolysis

• Electrolysis is used for isolating Electrolysis is used for isolating active elements, purifying metals, active elements, purifying metals, and electroplating.and electroplating.

• Pure compounds: HPure compounds: H22O, molten O, molten

saltssalts

• Use inert electrodes in the liquid Use inert electrodes in the liquid and pass electricity through the and pass electricity through the systemsystem

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17.7 Electrolysis17.7 Electrolysis

• The negative The negative electrode electrode (cathode) (cathode) attracts cations; attracts cations; reduction occurs.reduction occurs.

• The positive The positive electrode (anode) electrode (anode) attracts anions; attracts anions; oxidation occurs.oxidation occurs.

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Electrolysis of Molten NaClElectrolysis of Molten NaCl

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Electrolysis of NaClElectrolysis of NaCl

• Cathode:Cathode:

NaNa++(l) + e(l) + e-- Na(l)Na(l)

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• Anode:Anode:

2Cl2Cl--(l) (l) ClCl22(g)+ 2e(g)+ 2e--

anode.exe.lnk

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• 2Na2Na++(l) + 2Cl(l) + 2Cl--(l) (l) 2Na(l) + Cl2Na(l) + Cl22(g) E(g) Eoo = -4.07 V = -4.07 V

• Must supply at least 4.07 V to electrolyze Must supply at least 4.07 V to electrolyze molten sodium chloride. molten sodium chloride.

• NaCl melts at 804NaCl melts at 804ooC, where Na vaporizes and C, where Na vaporizes and burns. burns.

• Lower the temperature by adding CaClLower the temperature by adding CaCl22. .

(Why does this work?)(Why does this work?)

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• Other problem, Na reacts Other problem, Na reacts with Clwith Cl22, even at room , even at room

temperature.temperature.

• Commercial operations use Commercial operations use a Downs Cell. (described a Downs Cell. (described in 17.8, pg 859)in 17.8, pg 859)

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Downs CellDowns Cell

• How does the Downs Cell solve the problem How does the Downs Cell solve the problem of reaction between Na and Clof reaction between Na and Cl22? ?

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Applications of ElectrolysisApplications of Electrolysis

• Electrolysis can be used in a variety of Electrolysis can be used in a variety of applications:applications:• Chemical recovery of elements in mixturesChemical recovery of elements in mixtures• Industrial recovery of elements, miningIndustrial recovery of elements, mining• Plating out of metals, electroplating.Plating out of metals, electroplating.• And many more!And many more!

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Industrial ProcessesIndustrial Processes

• Purification of Copper: Recovered from its Purification of Copper: Recovered from its ores by chemical reduction.ores by chemical reduction.

• Purified by electrolysis.Purified by electrolysis.• Recover impurities:Recover impurities:• Mo (25%)Mo (25%)• Se (93%)Se (93%)• Te (96%)Te (96%)• Au (32%)Au (32%)• Ag (28%)Ag (28%)

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Electroplating of NickelElectroplating of Nickel

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Electrolytic Processes with Metals

Electrolytic Processes with Metals

• A variety of metals can be prepared by A variety of metals can be prepared by electrolysis, if a cheap source of electricity electrolysis, if a cheap source of electricity is available. In addition, some metals* are is available. In addition, some metals* are purified by electrolysis.purified by electrolysis.• aluminumaluminum cadmiumcadmium• calciumcalcium copper*copper*• gold*gold* lead*lead*• magnesiummagnesium sodiumsodium• zinczinc

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Faraday’s LawFaraday’s Law

• Recall…Recall…• FF = charge on 1 mol e = charge on 1 mol e-- = 96500 coul/mol = 96500 coul/mol

• and Electrical Current = charge / timeand Electrical Current = charge / time• 1 ampere = 1 coulomb of charge / second1 ampere = 1 coulomb of charge / second• 1 A = 1 coul / s1 A = 1 coul / s

• Use these relationships to analyze Use these relationships to analyze electrolytic processeselectrolytic processes

• 77 a, 79 a, 8177 a, 79 a, 81

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17.7 Faraday’s Law17.7 Faraday’s Law

• Faraday’s Law: the mass of product Faraday’s Law: the mass of product produced by a given amount of current is produced by a given amount of current is proportional to the number of electrons proportional to the number of electrons transferred.transferred.

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• If we electrolyze molten NaCl with a current If we electrolyze molten NaCl with a current of 50.0 A for 30. min (or 1800 s), what mass of of 50.0 A for 30. min (or 1800 s), what mass of Na is produced?Na is produced?

• NaNa++ + e + e-- Na Na

• 50.0 50.0 CC x 1800 s x x 1800 s x 1 mol 1 mol ee-- ss 96500 C96500 C

= 0.9326 mol e= 0.9326 mol e--

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• moles Na = 0.9326 mol emoles Na = 0.9326 mol e-- x 1 mol Na/1 mol e x 1 mol Na/1 mol e--

= 0.9326 mol= 0.9326 mol

• mass Na = 0.9326 mol x 22.99 g/mol = 21.44 gmass Na = 0.9326 mol x 22.99 g/mol = 21.44 g

• We can also calculate how much electrical We can also calculate how much electrical energy it will take for an electrolysis. We energy it will take for an electrolysis. We will not pursue these calculations.will not pursue these calculations.

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“Stoich Map”…“Stoich Map”…

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Revisit Electrolysis of KI(aq) Introduction

Revisit Electrolysis of KI(aq) Introduction

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Electrolysis of H2OElectrolysis of H2O

• Anode (oxidation):Anode (oxidation):2H2H22O O OO22(g) + 4H(g) + 4H++ + 4e + 4e-- EEoxox

oo = -1.23 V = -1.23 V• Cathode (reduction):Cathode (reduction):

2H2H22O + 2eO + 2e-- HH22(g) + 2OH(g) + 2OH-- EEredredoo = -0.83 V = -0.83 V

2H2H22O O 2H 2H22(g) + O(g) + O22(g) (g) EEcellcelloo = -2.06 V = -2.06 V

• Must supply at least 2.06 V to electrolyze water (if Must supply at least 2.06 V to electrolyze water (if anode [Hanode [H++] = 1.0 M and cathode [OH] = 1.0 M and cathode [OH--] = 1.0 M) ] = 1.0 M)

• In pure water, [HIn pure water, [H++] = [OH] = [OH--] = 10] = 10-7-7 M and the overall M and the overall potential is –1.23 Vpotential is –1.23 V

• An electrolyte is usually added to increase electrical An electrolyte is usually added to increase electrical conductivityconductivity

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Electrolysis of Aqueous Solutions


• Products depend on whether it is easier to Products depend on whether it is easier to oxidize or reduce the dissolved ions or oxidize or reduce the dissolved ions or water.water.

• Sample Problem: Consider a solution of Sample Problem: Consider a solution of NiClNiCl22 under standard conditions. under standard conditions.

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• Possible Anode Oxidations:Possible Anode Oxidations:2Cl2Cl-- Cl Cl22 + 2e + 2e-- EEoo = -1.36 V = -1.36 V

2H2H22O O O O22 + 4H + 4H++ + 4e + 4e-- EEoo = -1.23 V = -1.23 V

Because of more positive voltage, we would predict HBecause of more positive voltage, we would predict H22O will O will oxidize before Cloxidize before Cl--.*.*

• Possible Cathode Reductions:Possible Cathode Reductions:NiNi22++ + 2e + 2e-- Ni Ni EEoo = -0.25 V = -0.25 V

2H2H22O + 2eO + 2e-- H H22 + 2OH + 2OH-- EEoo = -0.83 V = -0.83 V

Because of more positive voltage, NiBecause of more positive voltage, Ni2+ 2+ will reduce before Hwill reduce before H22O.O.

• Products are OProducts are O22* and Ni.* and Ni.

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Group WorkGroup Work

1.1. What are the products of electrolysis What are the products of electrolysis of an aqueous NiBrof an aqueous NiBr22 solution? solution?

2.2. What are the products of electrolysis What are the products of electrolysis of an aqueous CuFof an aqueous CuF22 solution? solution?

3.3. What are the products of electrolysis What are the products of electrolysis of a mixture of aqueous CuBrof a mixture of aqueous CuBr22 and and NiFNiF22??

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Group Work 1Group Work 1

• What are the products of electrolysis of a What are the products of electrolysis of a NiBrNiBr22 solution? solution?• Anode reactions:Anode reactions:

2Br2Br-- Br Br22 + 2e + 2e-- E Eoo = -1.07 V = -1.07 V


• Cathode reactions:Cathode reactions:

NiNi22++ + 2e + 2e-- Ni Ni E Eoo = -0.25 V = -0.25 V

2H2H22O + 2eO + 2e-- H H22 + 2OH + 2OH-- E Eoo = -0.83 V = -0.83 V

• Ni and BrNi and Br22

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• What are the products of electrolysis of a What are the products of electrolysis of a CuFCuF22 aqueous solution? aqueous solution?

• Anode:Anode:2F2F-- F F22 + 2e + 2e-- E Eoo = -2.87 V = -2.87 V

2H2H22O O O O22 + 4H + 4H++ + 4e + 4e-- E Eoo = -1.23 V = -1.23 V

• Cathode:Cathode:CuCu22++ + 2e + 2e-- Cu Cu E Eoo = 0.34 V = 0.34 V

2H2H22O + 2eO + 2e-- H H22 + 2OH + 2OH-- E Eoo = -0.83 V = -0.83 V

• Cu and OCu and O22

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• What are the products of electrolysis of a mixture of What are the products of electrolysis of a mixture of aqueous CuClaqueous CuCl22 and NiCl and NiCl22??

• Anode:Anode:2Br2Br-- Br Br22 + 2e + 2e-- EEoo = -1.07 V = -1.07 V

2F2F-- F F22 + 2e + 2e-- EEoo = -2.87 V = -2.87 V


• Cathode:Cathode:NiNi22++ + 2e + 2e-- Ni Ni EEoo = -0.25 V = -0.25 VCuCu22++ + 2e + 2e-- Cu Cu EEoo = 0.34 V = 0.34 V

2H2H22O + 2eO + 2e-- H H22 + 2OH + 2OH-- EEoo = -0.83 V = -0.83 V

• Cu and BrCu and Br22

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17.7-8 Electrolysis & Applications