CELL POTENTIAL, E. Terms Used for Galvanic Cells. Uses of E o Values CELL POTENTIAL, E. Galvanic Cell. Organize halfreactions

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1 Electrons move from anode to cathode in the wire. Anions & cations move thru the salt bridge. Terms Used for Galvanic Cells Galvanic Cell We can calculate the potential of a Galvanic cell using one of the following E cell = E cathode E anode (use Red/Ox potentials) E cell = - [E anode E cathode ] (use Red potentials) E cell = E cathode E anode (use Red potentials and 2, anode CELL POTENTIAL, E??? V wire salt bridge Cu 2 ions Cu 2 ions 1.0 M 1.0 M Cu and Cu 2, cathode Electrons are driven from anode to cathode by an electromotive force or emf. For /Cu cell, this is indicated by a voltage of??? V at 25 C and when [ 2 ] and [Cu 2 ] = 1.0 M. 2 ions wire salt bridge Cu Cu 2 ions CELL POTENTIAL, E For /Cu cell, potential is 1.10 V at 25 C and when [ 2 ] and [Cu 2 ] = 1.0 M. This is the STANDARD CELL POTENTIAL, E o a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 C. Uses of E o Values Organize halfreactions by relative ability to act as oxidizing agents Table 17.1 / A5.5 Use this to predict cell potentials and direction of redox reactions. 2 ions wire salt bridge Cu Cu 2 ions

2 Standard Redox Potentials, E o Think of it this way oxidizing ability of ion E o (V) Cu 2 2e- Cu H 2e- H e reducing ability of element Any substance on the right will reduce any substance higher than it on the left. can reduce H and Cu 2. H 2 can reduce Cu 2 but not 2 Cu cannot reduce H or 2. What happens when you place Mg(s) in a solution of Cu 2 (aq)? (with a little acid) Mg goes into solution Cu plates out. In terms of the Reduction Potentials The more negative the value the more active the substance. So, they tend to want to go into solution. Therefore those substances are the Anode Using Standard Potentials, E o In which direction do the following reactions go? Cu(s) 2 Ag (aq) ---> Cu 2 (aq) 2 Ag(s) 2 Fe 2 (aq) Sn 2 (aq) ---> 2 Fe 3 (aq) Sn(s) What is E o net for the overall reaction? E o for a Voltaic Cell Cd Cd 2 Cd --> Cd 2 2eor Cd 2 2e- --> Cd Volts Salt Bridge Fe 2 Fe Fe --> Fe 2 2eor Fe 2 2e- --> Fe All ingredients are present. Which way does reaction proceed? Cd Cd 2 E o for a Voltaic Cell Volts Salt Bridge Fe 2 Fe Overall reaction Fe Cd 2 ---> Cd Fe 2 E o = E cathode -E anode = (-0.40 V) - (-0.44 V) = 0.04 V From the table, you see Fe is a better reducing agent than Cd Cd 2 is a better oxidizing agent than Fe 2 More About Calculating Cell Voltage Assume I - ion can reduce water. 2 H 2 O 2e- ---> H 2 2 OH - 2 I - ---> I 2 2e- Anode I - 2 H 2 O --> I 2 2 OH - H 2 Assuming reaction occurs as written, E net = E cathode -E anode = ( V) - (0.535 V) = V Minus E means rxn. occurs in opposite direction

3 CELL POTENTIALS, E o Can t measure 1/2 reaction E o directly. Therefore, measure it relative to a STANDARD HYDROGEN CELL, SHE. 2 H (aq, 1 M) 2e- <----> H 2 (g, 1 atm) E o = 0.0 V Negative electrode Supplier of / 2 half-cell hooked to a SHE. E o for the cell = 0.76 V - Volts Salt Bridge 2 H --> 2 2 2e- 2e- OXIDATION Oxidation ANODE Anode Positive electrode 2 H 2e- --> H 2e- 2 Reduction REDUCTION CATHODE H 2 Acceptor of Which is the Anode/? Standard notation has the Anode on the left and the on the right. In a Galvanic cell, you always want to have a positive E cell. Out of the two components (in terms of the standard reduction potential) The more negative one is the ANODE The more positive one is the CATHODE Is E related to G? YES! Michael Faraday Originated the terms anode, cathode, anion, cation, electrode. Discoverer of electrolysis magnetic props. of matter electromagnetic induction benzene and other organic chemicals Was a popular lecturer. E o and G o E o is related to G o, the free energy change for the reaction. G o = - n F E o where F = Faraday constant = x 10 4 J/V mol and n is the number of moles of transferred Michael Faraday

4 E o and G o G o = - n F E o For a product-favored reaction Reactants ----> Products G o < 0 and so E o > 0 E o is positive For a reactant-favored reaction Reactants <---- Products G o > 0 and so E o < 0 E o is negative E at Nonstandard Conditions RT E = E o ln(q) nf TheNERNST EQUATION E = potential under nonstandard conditions n = no. of exchanged Electrolysis Using electrical energy to produce chemical change. Sn 2 (aq) 2 Cl - (aq) ---> Sn(s) Cl 2 (g) SnCl 2 (aq) Electrolysis of Aqueous NaOH Electric Energy ----> Chemical Change Anode () 4 OH - ---> O 2 (g) 2 H 2 O 4e- 4 H 2 O 4e- --->2 H 2 4 OH - E o for cell = V Cl 2 Sn Electrolysis Electric Energy ---> Chemical Change Electrolysis of Molten NaCl Electrolysis of molten NaCl. Here a battery pumps from Cl - to Na. Polarity of electrodes is reversed from batteries. Anode Cl - Na

5 Anode Electrolysis of Molten NaCl Cl - Na Anode () 2 Cl - ---> Cl 2 (g) 2e- Na e- ---> Na E o for cell (in water) = E c -E a = V (1.36 V) = V (in water) External energy needed because E o is (-). Electrolysis of Aqueous NaCl Anode () 2 Cl - ---> Cl 2 (g) 2e- 2 H 2 O 2e- ---> H 2 2 OH - E o for cell = V Note that H 2 O is more easily reduced than Na. Anode Cl - Na H 2 O Also, Cl - is oxidized in preference to H 2 O because of kinetics. Electrolysis of Aqueous NaCl Cells like these are the source of NaOH and Cl 2. In 1995: 25.1 x 10 9 lb Cl 2 and 26.1 x 10 9 lb NaOH Electrolysis of Aqueous NaI Also the source of NaOCl for use in bleach. Anode (): 2 I - ---> I 2 (g) 2e- : 2 H 2 O 2e- ---> H 2 2 OH - E o for cell = V Electrolysis of Aqueous CuCl 2 Anode () 2 Cl - ---> Cl 2 (g) 2e- Cu 2 2e- ---> Cu E o for cell = V Note that Cu is more easily reduced than either H 2 O or Na. Anode Cl - Cu 2 H 2 O Electrolytic Refining of Copper Impure copper is oxidized to Cu 2 at the anode. The aqueous Cu 2 ions are reduced to Cu metal at the cathode.

6 Producing Aluminum 2 Al 2 O 3 3 C ---> 4 Al 3 CO 2 Charles Hall ( ) developed electrolysis process. Founded Alcoa. Alcoa World's leading producer of primary aluminum, fabricated aluminum, and alumina. Active in all major aspects of the industry technology, mining, refining, smelting, fabricating and recycling. Alcoa's aluminum products and components are used worldwide in aircraft, automobiles, beverage cans, buildings, chemicals, sports and recreation, and a wide variety of industrial and consumer applications, including such Alcoa consumer brands as Alcoa wheels, Reynolds Wrap aluminum foil, and Baco household wraps. Vital statistics: 120,000 Alcoans in 41 countries, $21.5 billion in revenues Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag (aq) e- ---> Ag(s) 1 mol e- ---> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-? charge passing Current = time I (amps) = coulombs I (amps) Quantitative Aspects of Electrochemistry = coulombs Current = charge passing time But how is charge related to moles of? Charge on 1 mol e - = 1.60 x C 6.02 x e- e- mol = 96,500 C/mol e- = 1 Faraday Quantitative Aspects of Electrochemistry I (amps) = coulombs 1.50 amps flow thru a Ag (aq) solution for 15.0 min. What mass of Ag metal is deposited? (a) Calc. charge Charge (C) = current (A) x time (t) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C Quantitative Aspects of Electrochemistry I (amps) = coulombs 1.50 amps flow thru a Ag (aq) solution for 15.0 min. What mass of Ag metal is deposited? (a) Charge = 1350 C (b) (c) Calculate moles of e- used 1350 C 1 mol e - = mol e - 96, 500 C Calc. quantity of Ag mol e - 1 mol Ag = mol Ag or 1.51 g Ag 1 mol e -

7 Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) HSO 4- (aq) ---> PbSO 4 (s) H (aq) 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e mol Pb 2 mol e - 1 mol Pb = 4.38 mol e - c) Calculate charge 4.38 mol e- 96,500 C/mol e- = 423,000 C Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) HSO 4- (aq) ---> PbSO 4 (s) H (aq) 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol c) Charge = 423,000 C d) Calculate time Time (s) = 423, 000 C 1.50 amp = 282, 000 s Time (s) = Charge (C) I (amps) About 78 hours