The study of the relationships between electricity and chem rxns Electrochemistry The study of the relationships between electricity and chem rxns

20.1 Identify the substances that are oxidized, reduced, the oxidizing agent, and the reducing agent. Cd(s) + NiO2(s) + 2H2O(l)  Cd(OH)2(s) + Ni(OH)2(s)

Don’t forget LEO says GER

Balancing Oxidation-Reduction Equations To balance redox rxns consider the law of cons of mass and the gain and loss of e’s

Balancing Oxidation-Reduction Equations Half-Reactions show oxidation and reduction as two separate equations

Balancing Oxidation-Reduction Equations In the overall redox rxn the number of e’s lost in the ox half-rxn must equal the number of electrons gained in the red half-rxn

Balancing Oxidation-Reduction Equations Ox Sn2+  Sn4+ + 2e- Red 2Fe3+ + 2e-  2Fe2+ Overall Sn2+ + 2Fe3+  Sn4+ + 2Fe2+

Balancing Equations by the method of Half-Reactions (for acidic solutions) 1. Divide the equation into two incomplete half-reactions, ox and red

Balancing Equations by the method of Half-Reactions (for acidic solutions) 2. Balance each half rxn a. Balance elements other than H or O b. Balance the O atoms by adding H2O

Balancing Equations by the method of Half-Reactions (for acidic solutions) c. Balance the H atoms by adding H+ d. Balance the charge by adding e-s to the side with the greater overall positive charge

Balancing Equations by the method of Half-Reactions (for acidic solutions) 3. Multiply each half rxn by an integer so that the number of electrons lost in one half rxn equals the electrons gained in the other

Balancing Equations by the method of Half-Reactions (for acidic solutions) 4. Add the two half rxns together and simplify where possible by canceling species appearing on both sides of the equation

Balancing Equations by the method of Half-Reactions (for acidic solutions) 5. Check atoms and charge

20.2 Complete and balance the following equation by the method of half-rxns Cr2O72- + Cl-  Cr3+ + Cl2 (acidic solution)

For Basic Solutions Balance initially as if they were balanced in acidic solution Neutralize H+ by adding OH- to both sides of the equation and creating and canceling water molecules.

20.3 Complete and balance the following equation CN- + MnO4-  CNO- + MnO2 (basic solution)

Voltaic Cells A voltaic or galvanic cell uses spontaneous oxidation-reduction reactions to generate electricity

Voltaic Cells In a voltaic cell the oxidation and reduction half-reactions occur in separate compartments

Voltaic Cells Each compartment has a solid surface called an electrode, where the half-reaction occurs

Voltaic Cells Oxidation occurs at the anode(-) Reduction occurs at the cathode (+)

Voltaic Cells The electrons released at the anode flow through an external circuit where they do electrical work and move to the cathode

Voltaic Cells For a voltaic cell to work it must be electrically neutral

Voltaic Cells This is maintained by the migration of ions between the two compartments through a device such as a salt bridge or a porous glass barrier

Voltaic Cells A salt bridge consists of a u-shaped tube that contains an electrolyte solution such as NaNO3 whose ions will not react with the ions involved in the cell

Voltaic Cells Anions always migrate toward the anode and cations to the cathode

20.4 Voltaic Cells Cr2O72- + 14H+ + 6I-  2Cr3+ + 3I2 + 7H2O A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate (1) the reaction occurring at the anode (2) the reaction at the cathode (3) the direction of electron migration (4) the direction of ion migration (5) the signs of the electrodes

Voltaic Cells simulation http://content.blackgold.ca/ict/Division4/Science/Div.%204/Voltaic%20Cells/Voltaic.htm

Cell EMF Electrons flow from the anode to the cathode in a voltaic cell because the potential energy of the electrons in the anode is higher than the cathode so they spontaneously flow this way

Cell EMF The difference in potential energy per electrical charge between two electrodes is measured in volts (V) 1 V = 1 J/C

Cell EMF Electromotive force (emf) is the potential difference between electrodes. (synonyms: Ecell or cell potential) Eocell = Eored(cathode) – Eored(anode) Ecell will be positive for a spontaneous rxn

Cell EMF The emf depends on the reactions that occur at the cathode and anode, concentration of reactants and products, and temperature (25oC) Standard cell potential Eocell is at standard conditions of 1M and 1atm

Standard Reduction (Half-Cell) Potentials Appendix E lists standard reduction potentials Stoichiometric coefficients added to rxns do not change Eored values

Standard Reduction (Half-Cell) Potentials

Standard Reduction (Half-Cell) Potentials By convention, the potential associated with each electrode is the potential for reduction to occur at that electrode, Eored Eored is not measured directly rather it is determined based on a reference 2H+(aq, 1M) + 2e-  H2 (g,1 atm) Eored = 0 V

20.6 Calculate the standard emf for Cr2O72- + 14H+ + 6I-  2Cr3+ + 3I2 + 7H2O

Standard Reduction (Half-Cell) Potentials The more positive value for Eored, the greater the driving force for the reduction In any voltaic cell the reaction at the cathode has a more positive value of Eored than does the reaction at the anode The half-rxn with the smallest Eored is most easily reversed as an oxidation

20.7 Determine the half-rxn at cathode and anode and Eocell Cd2+ +2e-  Cd Sn2+ +2e-  Sn

Spontaneity of Redox Rxns For any ox-red process (not just voltaic cells) Eo = Eored(red rxn) – Eored(ox rxn) E > 0 spontaneous E < 0 nonspontaneous

Sample Exercise 20.9 Spontaneous or Not? Using standard reduction potentials in Table 20.1, determine whether the following reactions are spontaneous under standard conditions.

Spontaneity of Redox Rxns G = -nFE n is the number of moles e- transferred in the rxn F is Faraday’s constant – the quantity of charge on 1 mole of electrons 1F=96,500C/mol e- E is the emf

Spontaneity of Redox Rxns Standard states Go = -nFEo

Sample Exercise 20.10 Determining ΔG° and K (a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, , and the equilibrium constant, K, at 298 K for the reaction (b) Suppose the reaction in part (a) was written What are the values of E°, ΔG°, and K when the reaction is written in this way?

The Nernst Equation (nonstandard conditions) E = Eo –(0.0592/n) logQ (T=298K) As the conc. of products of a redox rxn increases, the voltage decreases. As the conc. of reactants of a redox rxn increases, voltage increases.

Sample Exercise 20.11 Voltaic Cell EMF under Nonstandard Conditions Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 When

Electrolysis Electrical energy can be used to drive a nonspontaneous redox rxn to occur in an electrolysis reaction

Electrolysis This occurs in an electrolytic cell which consists of two electrodes in a molten salt or solution

Electrolysis A battery or some other source of direct electrical current acts as an electron pump, pushing electrons into one electrode and pulling them from the other

Electrolytic Cell

Electrolytic Cell http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/electroChem/electrolysis10.html

Electrolysis The electrode of the electrolytic cell that is connected to the negative terminal of the voltage source is the cathode of the cell, it receives electrons and reduces the substance

Electrolysis The electrons that are removed during the oxidation process at the anode travel to the positive terminal of the voltage source, completing the circuit of the cell

Electrolysis Consider the electrolysis of molten sodium chloride: Cl-(l)  1/2Cl2(g) + e- Eo = -1.359V Na+(l) + e-  Na(l) Eo = -2.71V

Electrolysis Consider the electrolysis of water using an inert electrolyte: 2H2O(l) + 2e-  H2(g) + 2OH-(aq) Eo = -0.83 V H2O(l)  1/2O2(g) +2H+(aq) +2e- Eo= -1.23

Electrolysis Consider the electrolysis of aqueous sodium chloride: 2H2O(l) + 2e-  H2(g) + 2OH-(aq) Eo = -0.83 V Na+(aq) + e-  Na(l) Eo=-2.71 V 2Cl-(aq)  Cl2(g) + 2e- Eo=-1.359V

Electrolysis Active electrodes participate in the electrolysis as in electroplating where a thin layer of metal is deposited to another to resist corrosion or ornamental purposes. .

Quantitative aspects of Electrolysis A coulomb is the quantity of charge passing a point in a circuit in 1 s when the current is 1 ampere (A) 1 C = 1 A s Coulombs = ampere x seconds

Quantitative aspects of Electrolysis

20.17 Calculate the number of grams Al produced in 1.00hr by the electrolysis of molten AlCl3 if the electrical current is 10.0A.

20.17 practice How many seconds would be required to produce 50.0g of Mg from MgCl2 if the current is 100.0A?