unit 15 electrochemistry

Unit 15Electrochemistry

Chapter 20 Problem Set: p. 890-8983, 11, 13, 17, 23, 27, 31, 35, 43, 45, 47, 53, 61, 63, 69, 72, 75, 79, 85, 87, 95

Oxidation NumbersPlease assign oxidation numbers to the elements in each substance.1) HNO3

2) HNO2

3) PbSO4

4) (NH4)2Ce(SO4)3

5) Al(C2H3O2)3

Electrochemistry: Relationship between electricity and redox reactionsElectricity = movement of electrons

The energy released in a spontaneous redox reaction that is used to perform electrical work is harnessed in Voltaic CellsAlso called Galvanic CellsElectrons transfer through an external pathway rather than directly between reactantsElectrodes: metals used in the circuit

Voltaic Cells

Electrochemical Cell

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

Using a Porous Disk as a Salt Bridge

Components of Electrochemical CellsAnode: electrode at which oxidation occursCathode: electrode at which reduction occursSalt Bridge: a connection between two half-cells that allows for the flow of ions

Composed of a strong electrolyteMust be a solution of a compound that will not produce a precipitate with any other ion in solutionSalt anions migrate toward the anode; cations migrate toward the cathode

**Electrons flow from the anode to the cathode**

Pneumonic DevicesKeep your vowels together and your consonants togetherAnode = oxidationCathode = reduction

An ox and Red catsAnode is oxidation, Cathode is reduction

Electrons flow from A to C (in alphabetical order)

Standard Cell Potentials

Used to determine which metal will be the cathode/anode for the reaction to be spontaneousCell EMF: Electromotive ForceCaused by a difference in potential energy between the different electrodes.

Allows electrons to be pushedDenoted Ecell , measured in volts, VAlso called cell potential

Cell PotentialCell potential is

positive for spontaneous reactionsnegative for nonspontaneous reactions

Depends on1. Reactions Occurring (Metals Used)2. Concentrations of Solutions (Molarity)3. Temperature (normally 25oC)

Eocell cell potential at standard conditions

EMF can then be used to determine what will happen in a chemical equation

The more positive the Eored value for a half

reaction, the greater the tendency for the reactant of the half reaction to be reduced and, therefore, to oxidize another speciesThe half reaction with the smallest reduction potential is most easily reversed as an oxidationThe Eo

red table acts as an activity series for which substances act as oxidizers and reducers

Calculating Cell PotentialEo

cell based on standard reduction potentialsPotential associated with each electrode is the ability for reduction to occur at the electrode.Eo

red = standard reduction potential for metal (based on Standard Hydrogen Electrode, SHE – use Appendix E p. 1128!)

Eocell = Eo

red (cathode) – Eored (anode)

If the stoichiometry of the reaction changes DO NOT multiply the value by the the cell potential for that metalThe more positive the Eo

red for a metal, the greater the ability for reduction

Thus the metal with a larger Eored will be the better

cathode in a voltaic cell

Example: A voltaic cell is based on the following half cells:Cd2+ + 2e- Cd Sn2+ + 2e- Sn

Determine voltage produced in spontaneous reaction.

Example: A voltaic cell is based on the following half cells:Cd2+ + 2e- Cd Sn2+ + 2e- SnDetermine voltage produced in spontaneous reaction.Eo

red (Cd2+ /Cd ) = -0.403 VEo

red (Sn2+ /Sn ) = -0.136 VCathode: Sn2+ + 2e- SnAnode: Cd Cd2+ + 2e-

Eocell = Eo

cathode – Eoanode

= -0.136 V – (-0.403 V) = 0.267 V

Example: What would be the electrical potential for the reaction:PbO2 + Na Pb2+ + Na+

Example: What would be the electrical potential for the reaction:PbO2 + Na Pb2+ + Na+

Na Na+ + 1e- + 2.71 VPbO2 Pb2+ + 1.50 V

(Na Na+ + 1e-)x2 *DO NOT MULTIPLY Eored by 2*

4H+ + 2e- + PbO2 Pb2+ + 2H2O4H+ + PbO2 + 2Na 2Na+ + Pb2+ + 2H2O

Overall potential for the battery = 2.71 + 1.50 = 4.3V

Spontaneous Redox ReactionsReduction Potentials can be used to calculate energies of reactions when using redox processesRemember:E = + SpontaneousE = - NonspontaneousE can be converted into G for Gibbs Free Energy or other thermochemical quantities

∆Go = -nFEo

n= number of electrons transferredF = Faraday’s constant = 96500 C/mol or J/KmolEo = Calculated from Eo

red values

Calculate ∆G for the following reaction: Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq)

Calculate ∆G for the following reaction: Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq)

Start with Half-ReactionsFe Fe2+ + 2e-

Cu Cu2+ + 2e-

Flip the copper reaction so it matches the overall equationFe Fe2+ + 2e- 0.44 VCu2+ + 2e- Cu 0.34 VCu2+(aq) + Fe(s) Cu(s) + Fe2+(aq) 0.78V

Solve for ∆Go = -nFEo

∆Go = -(2)(96500)(0.78) = -150517 J/mol = -151 kJ/molWhat is the Keq for this reaction? ∆Go = -RTlnKeq

Warm-Up : Do another page on the net ionic equations! The whole packet is due TODAY!The 2008 FRQ was your Homework. Please be ready to grade it and the crossword page.

Effect of Concentration on Cell EMFAs voltage is produced, concentrations vary. Reactants are consumed and products are produced.When E = 0, the cell is dead, the concentrations cease to change due to equilibrium being reached.For different concentrations, the Nernst equation is used:

To calculate voltage, use the Nernst Equation

Cell EMF and Equilibrium∆G = 0, ∆E = 0 at EquilibriumUsing Nernst

Will be zero

Solving for the equilibrium constant

Example: What is the electrical potential for the cell with the following concentrations?VO2

1+ + Zn Zn2+ + VO2+

[VO21+] = 2.0 M [H+] = 0.50 M

[VO2+] = 1.0 x 10 M [Zn2+] = 0.10 M

Example: What is the electrical potential for the cell with the following concentrations?VO2

1+ + Zn Zn2+ + VO2+

[VO21+] = 2.0 M [H+] = 0.50 M

[VO2+] = 1.0 x 10-2 M [Zn2+] = 0.10 MStart with the half reactions2(2H+ + 1e- + VO2

1+ VO2+ + H2O ) 1.00 VZn Zn2+ +2e- 0.76 V 4H+ + 2VO2

1+ + Zn Zn2+ + 2VO2+ + 2H2O 1.76 VWrite the equilibrium expression

4 x 10-5

Solve for E

=1.76 – (-0.13)

E = 1.89 V

The voltage went up!

Use LeChatelier’s Principle to explain why

What will the voltmeter read for the following electrochemical cell?Anode Reaction – Oxidation - dilute solution becomes more concentrated.Ag(s) → Ag+ (0.1 M) + e- +0.8 VCathode Reaction – Reduction-concentrated solution becomes more dilute.Ag+ (1.0 M) + e- → Ag (s) -0.8 VNet Cell ReactionAg+ (1.0 M) → Ag+ (0.1 M) 0.0V

Concentration Cells: an electrochemical cell where the same electrodes (same half-reactions) are used, but the solutions have different concentrations

V

Use the Nernst Equation to solve for the potential

You DON’T need your calculator for this one!!!E = 0.0592 V

BatteriesSelf-contained electrochemical power source consisting of 1 or more galvanic cells.Voltages are additive when batteries are connected due to a continuation of the flow of electrons

Fuel CellsA voltaic cell that converts chemical energy into electrical energyNot self-containedProduces current by combustion and electron-excitation

CorrosionA spontaneous redox reaction in which a metal is attacked by some substance in the environment and converted to an unwanted compoundIs often prevented, or reversed, by use of a sacrificial anode (a more active metal)

ElectrolysisThe use of electricity to cause nonspontaneous reactions to occur by driving the reaction in the opposite direction (basis of rechargeable batteries Take place in electrolytic cellsCathode is connected to – terminal to accept e-, anode is attached to + terminal to donate e- (opposite of voltage)

Electrolytic Cell

Calculations with Electrolytic CellsAmpere = units to measure current in Coulombs/second

Measures the rate of electron flowFaraday’s Constant, F = 96485 Coulomb/mol

Example: Calculate the amount of time required to produce 1000 g of magnesium metal by electrolysis of molten MgCl2 using a current of 50 A.

Calculations with Electrolytic CellsAmpere = units to measure current in Coulombs/second

Measures the rate of electron flowFaraday’s Constant, F = 96485 Coulomb/mol

Example: Calculate the amount of time required to produce 1000 g of magnesium metal by electrolysis of molten MgCl2 using a current of 50 A.Use a t-table!

Time = 158823 sec = 2647 min = 44 hours = 1.84 days

Example: a Cr3+ (aq) solution is electrolyzed using a current of 7.60 A. What mass of Cr (s) is plated out after 2.00 days?

Example: a Cr3+ (aq) solution is electrolyzed using a current of 7.60 A. What mass of Cr (s) is plated out after 2.00 days?

3 e + Cr3+ CrUse a t-table!

Mass of Cr = 2.36 x 102 g

Electrolysis can be used to electroplate metals

Thin coating is plated onto an existing metal to help protect itMetal to be plated is attached to the cathode.Metal coating is produced by the metal attached to anode and a “like” solution

NO MORE AP CHEM NOTES!GOOD LUCK ON YOUR EXAM!