Discovering Electrochemical Cells

Transcription

1 Discovering Electrochemical Cells Part I Electrolytic Cells Many important industrial processes PGCC CHM 102 Cell Construction e e power conductive medium What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)? vessel () () Sign or polarity of s inert s Let s examine the electrolytic cell for molten NaCl. Molten NaCl Observe the reactions at the s Molten NaCl At the microscopic level Cl 2 (g) escapes e NaCl (l) Na (l) halfcell () NaCl (l) halfcell e Na 2 Cl 2 2e () cations migrate toward () cathode e () () anions migrate toward () anode e Na 2 Cl 2 2e 1

2 Molten NaCl Electrolytic Cell cathode halfcell () REDUCTION e Na anode halfcell () OXIDATION 2 Cl 2 2e overall cell reaction Cl 2 X 2 Definitions: CATHODE REDUCTION occurs at this ANODE OXIDATION occurs at this Nonspontaneous reaction! What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)? H 2 O Will the halfcell reactions be the same or different? Aqueous NaCl What could be reduced at the cathode? cathode different halfcell power e e () () H 2 O NaCl (aq) anode 2 Cl 2 2e Aqueous NaCl Electrolytic Cell possible cathode halfcells () REDUCTION e Na 2H 2 0 2e H 2 2OH possible anode halfcells () OXIDATION 2 Cl 2 2e 2H 2 O O 2 4H 4e overall cell reaction 2 2H 2 0 H 2 Cl 2 2OH Ag e Ag For every electron, an atom of silver is plated on the. Ag e Ag Electrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO 3 (aq) under standard conditions, will deposit silver at the rate of g Ag/sec 1 amp = g Ag/sec 2

3 Faraday s Law The mass deposited or eroded from an depends on the quantity of electricity. Quantity of electricity coulomb (Q) Q is the product of current in amps times time in seconds coulomb Q = It current in amperes (amp) time in seconds 1 coulomb = 1 ampsec = g Ag Ag e Ag 1.00 mole e = 1.00 mole Ag = g Ag g Ag/mole e = 96,485 coul/mole e g Ag/coul 1 Faraday (F ) mole e = Q/F mass = mole metal x MM mole metal depends on the halfcell reaction Examples using Faraday s Law How many grams of will be deposited in 3.00 hours by a current of 4.00 amps? 2 2e The charge on a single electron is x coulomb. Calculate Avogadro s number from the fact that 1 F = 96,487 coulombs/mole e. A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au 3, 2, and Ag, and Au,, and Ag were plated out respectively, calculate the amount of metal deposited at each anode. e e e e 1.0 M Au M M Ag Au 3 3e Au 2 2e Ag e Ag 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! graphite anodes CO 2 bubbles Al 2 O 3 (l) Al 3 O 2 Al 3 O 2 Al (l) O 2 carbonlined steel vessel acts as cathode Cathode: Al 3 3e Al (l) e Anode: 2 O 2 C (s) CO 2 (g) 4e from power e Draw off Al (l) 3

4 The Hall Process Cathode: Al 3 3e Al (l) x 4 Anode: 2 O 2 C (s) CO 2 (g) 4e x 3 Part II Galvanic Cells 4 Al 3 6 O 2 3 C (s) 4 Al (l) 3 CO 2 (g) Batteries and corrosion The graphite anode is consumed in the process. Cell Construction Observe the s to see what is occurring. Salt bridge KCl in agar Provides conduction between halfcells What about halfcell reactions? cathode halfcell 2 2e What about the sign of the s? Why? anode halfcell 2 2e 1.0 M SO M SO 4 plates out or deposits on 1.0 M SO 4 What happened at each? 1.0 M SO 4 erodes or dissolves Galvanic cell cathode halfcell () REDUCTION 2 2e anode halfcell () OXIDATION 2 2e overall cell reaction 2 2 Now for a standard cell composed of / 2 and / 2, what is the voltage produced by the reaction at 25 o C? Standard Conditions Temperature 25 o C All solutions 1.00 M All gases 1.00 atm Spontaneous reaction that produces electrical current! 4

5 Now replace the light bulb with a volt meter. cathode halfcell 2 2e 1.0 M SO volts 1.0 M SO 4 anode halfcell 2 2e We need a standard to make measurements against! The Standard Hydrogen Electrode (SHE) 25 o C 1.00 M H 1.00 atm H 2 H 2 input 1.00 atm Halfcell 2H 2e H 2 E o SHE = 0.0 volts Pt 1.00 M H inert metal Now let s combine the copper halfcell with the SHE E o = 0.34 v cathode halfcell 2 2e 0.34 v H atm anode halfcell H 2 2H 2e Now let s combine the zinc halfcell with the SHE E o = 0.76 v anode halfcell 2 2e 0.76 v H atm cathode halfcell 2H 2e H 2 KCl in agar Pt KCl in agar Pt 1.0 M SO M H 1.0 M SO M H Al 3 3e Al E o = 1.66 v 2 2e Assigning the E o Write a reduction halfcell, assign the voltage measured, and the sign of the to the voltage. E o = 0.76 v 2H 2e H 2 E o = 0.00 v 2 2e E o = 0.34 Ag e Ag E o = 0.80 v Increasing activity The Nonactive Metals Metal H no reaction since E o cell < Db Bh 5

6 Calculating the cell potential, E o cell, at standard conditions H 2 O with O 2 Consider a drop of oxygenated Fe water on an iron object Fe 2 2e Fe E o = 0.44 v reverse Is iron an active metal? What would happen if iron is exposed to hydrogen ion? 2x Fe 2H Fe 2 H 2 (g) E o cell = 0.44 V Fe Fe 2 2e E o = 0.44 v 2x Fe Fe 2 2e E o = 0.44 v O 2 (g) 4H 4e 2H 2 0 E o = 1.23 v O 2 (g) 2H 2 O 4e 4 OH E o = 0.40 v 2Fe O 2 (g) 2H 2 O 2Fe(OH) 2 (s) E o cell= 0.84 v This is corrosion or the oxidation of a metal. 2Fe O 2 (g) 4H 2Fe 2 2H 2 O E o cell= 1.67 v How does acid rain influence the corrosion of iron? Enhances the corrosion process What happens to the potential if conditions are not at standard conditions? The Nernst equation adjusts for nonstandard conditions For a reduction potential: ox ne red at 25 o C: E = E o log (red) n (ox) Calculate the E for the hydrogen where 0.50 M H and 0.95 atm H 2. Free Energy and the Cell Potential 2x 2 2e E o = 0.34 Ag e Ag E o = 0.80 v 2Ag 2 2Ag E o cell= 0.46 v G o = nfe o cell where n is the number of electrons for the balanced reaction What is the free energy for the cell? 1F = 96,500 J/v from thermodynamics: G o = 2.303RT log K and the previous relationship: G o = nfe o cell nfe o cell at 25 o C: E o cell = 2.303RT log K = log K n Comparison of Electrochemical Cells galvanic produces electrical two current s anode () cathode () salt bridge G < 0 conductive medium vessel electrolytic need power anode () cathode () G > 0 where n is the number of electrons for the balanced reaction 6