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Chemistry 1011

Chemistry 1011. TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18. 18.5 Electrolytic Cells. YOU ARE EXPECTED TO BE ABLE TO: Construct a labelled diagram to show the structure of an electrolytic cell and describe its operation

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18 Chemistry 1011 Slot 5

  2. 18.5 Electrolytic Cells YOU ARE EXPECTED TO BE ABLE TO: • Construct a labelled diagram to show the structure of an electrolytic cell and describe its operation • Identify half reactions that take place at the anode and cathode of an electrolytic cell • Predict the most likely products of electrolysis of aqueous solutions and melts of ionic compounds • Calculate the minimum voltage necessary for the electrolysis of a solution under standard conditions • Carry out calculations related to the current passing through an electrolytic cell and the quantities of product formed Chemistry 1011 Slot 5

  3. Electrolytic Cells • In an electrolytic cell, a non-spontaneous redox reaction is made to occur by pumping electrical energy into the system • DC Source + • Could be Voltaic cell CATHODE - + ANODE Reduction occurs at the cathode M+ + e- M Oxidation occurs at the anode X- X + e- Chemistry 1011 Slot 5

  4. Electrolytic Cells • The simple electrolytic cell consists of • two electrodes, dipping into • a solution containing positive and negative ions • The external battery or DC source acts as a pump • It pushes electrons to the cathode and removes electrons from the anode • Metal ions in solution travel to the cathode and gain electrons (reduction) • Non-metal ions travel to the anode and give up electrons (oxidation) • The process is called electrolysis Chemistry 1011 Slot 5

  5. Electrolytic Cell Reactions • In aqueous solution, a number of anode and cathode half reactions are possible • At the cathode: • Reduction of a metal cation to the metal: Cu2+(aq) + 2e- Cu(s) Eored =+0.339V Common when transition metal cations present; does not occur if cation is difficult to reduce (Eored <-0.828V) • Reduction of a water molecule to hydrogen gas: 2H2O + 2e- H2(g) + 2OH-(aq) Eored =-0.828V Occurs if cation is difficult to reduce Chemistry 1011 Slot 5

  6. Electrolytic Cell Reactions • At the anode: • Oxidation of an anion to the non-metal: 2I-(aq) I2(s) + 2e-Eoox =-0.534V Does not occur if anion is difficult to oxidize (Eored <-1.229) • Oxidation of a water molecule to oxygen gas: 2H2O  O2(g) + 4H+(aq) + 4e- Eored =-1.229V Occurs if anion is difficult to reduce • Oxidation of the anode material Cu(s) Cu2+(aq) + 2e-Eored =-0.339V Chemistry 1011 Slot 5

  7. Electrolysis of Sodium Chloride • Possible anode half reactions: 2Cl-(aq) Cl2(g) + 2e-Eoox =-1.360V 2H2O  O2(g) + 4H+(aq) + 4e- Eored =-1.229V • At high concentrations, the chlorine oxidation half reaction is favoured • Possible cathode half reactions: 2H2O + 2e- H2(g) + 2OH-(aq) Eored =-0.828V Na+(aq) + e- Na(s) Eored =-2.714V • The water reduction half reaction has a much lower Eored value and is favoured • Overall cell reaction: 2Cl-(aq) + 2H2O Cl2(g) + H2(g) + 2OH-(aq) Eocell =-2.188V Chemistry 1011 Slot 5

  8. Electrolysis of Sodium Chloride • Chlorine gas is produced at the anode • Hydrogen gas is produced at the cathode • Sodium ions remain in solution • Hydroxide ions are formed in the solution • Commercially, this process is used to manufacture • Chlorine • Hydrogen • Sodium hydroxide Chemistry 1011 Slot 5

  9. Electrolysis of Sodium Chloride Chemistry 1011 Slot 5

  10. Electroplating • In many electrolytic cells, metal is deposited at the cathode • An object can be electroplated by making it the cathode • Copper, chromium and silver are often plated onto brass, nickel, or other metals by this method • The process is also used to purify copper, as the last step in the manufacture of copper from copper ore Chemistry 1011 Slot 5

  11. Electrolysis of melts • Salts will conduct electricity when melted, as well as when in solution, e.g.: 2NaCl(l) 2Na (l) + Cl2(g) • Aluminum is produced by electrolysis of alumina, Al2O3 • The melting point of Al2O3 is too high for direct electrolysis, so it is dissolved in molten Na3AlF6 (cryolite) at 980oC • The cathode is iron • The anode is carbon • Cathode reaction: Al3+ + 3e-Al(l) • Anode reaction: 2O2- O2(g) + 4e- • (The oxygen reacts with the carbon anode to form carbon dioxide gas) Chemistry 1011 Slot 5

  12. Quantitative Relationships • The amount of a substance produced or used in an electrochemical cell will be determined by the amount of electricity flowing in the cell • Cathode reaction: Ag+(aq) + e- Ag(s) • One mole of electrons will produce one mole of silver Cu2+(aq) + 2e- Cu(s) • Two moles of electrons will produce one mole of copper Chemistry 1011 Slot 5

  13. Electrical Units Quantity Unit Potential Volt (V) Current Ampere (A) Rate of flow of electrons Charge Coulomb (C) 1 Amp for 1 second Energy Joules (J) #Volts  #Coulombs 1 mole of electrons = 96500 Coulombs = 1 Faraday Chemistry 1011 Slot 5

  14. Faraday’s Laws • In an electrolysis, the amount of product produced or reactant consumed is proportional to the length of time that a constant current is passed in the circuit • To produce one mole of product, or to consume one mole of reactant, requires n moles of electrons, (where n is the number of electrons gained or lost by one atom or ion). This is 96,500n Coulombs Chemistry 1011 Slot 5

  15. Faraday’s Laws • Faraday’s Laws provide a relationship between • The current flowing in an electrolytic cell (Amperes) • The time that the current flows (seconds) • The amount of material deposited or consumed (moles) Chemistry 1011 Slot 5

  16. Applying Faraday’s Laws 1 mole e- - + AgNO3 solution cathode CuSO4 solution cathode 1 mole Ag (108g) deposited on cathode ½ mole Cu (32g) deposited on cathode Chemistry 1011 Slot 5

  17. Applying Faraday’s Laws • What current will deposit 0.155g of silver from a solution of silver ions in 11.0 minutes ( 660 seconds)? Ag+(aq) + e- Ag(s) # moles silver produced = 0.155g  108 g/mol = 1.44  10-3 mol # moles of electrons = 1.44  10-3 mol Total charge = 1.44  10-3 mol  96500 Coulombs/mol = 139C • Current = Charge (Coulombs)  time (seconds) • Current = 139C  660 seconds = 0.211 Amps Chemistry 1011 Slot 5

  18. Applying Faraday’s Laws How much copper will be deposited if a current of 0.150A is passed through a solution of copper sulfate for 20.0 minutes? Cu2+(aq) + 2e- Cu(s) Two moles of electrons will be required to deposit each mole of copper • Total charge (C) = current (A)  time (s) = 0.150A  1200s = 180C • # moles (Faradays) = 180C  96500C/mol = 1.87 x 10-3 mol • #moles of copper deposited = ½ 1.87  10-3mol = 9.33 x 10-4mol • Mass of copper = 9.33 x 10-4mol  63.5 g/mol = 5.92  10-2g Chemistry 1011 Slot 5

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