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1 CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf.Electrons are “driven” from anode to cathode by an electromotive.

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Presentation on theme: "1 CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf.Electrons are “driven” from anode to cathode by an electromotive."— Presentation transcript:

1 1 CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf.Electrons are “driven” from anode to cathode by an electromotive force or emf. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M.For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. Zn and Zn 2+, anode Cu and Cu 2+, cathode 1.10 V 1.0 M

2 2 CELL POTENTIAL, E For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M.For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. This is the STANDARD CELL POTENTIAL, E oThis 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.—a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

3 3 Calculating Cell Voltage Balanced half-reactions can be added together to get overall, balanced equation.Balanced half-reactions can be added together to get overall, balanced equation. Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e- ---> Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e- ---> Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) If we know E o for each half-reaction, we could get E o for net reaction.

4 4 CELL POTENTIALS, E o STANDARD HYDROGEN CELL, SHE. 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

5 5 Zn/Zn 2+ half-cell hooked to a SHE. E o for the cell = +0.76 V Zn/Zn 2+ half-cell hooked to a SHE. E o for the cell = +0.76 V Negative electrode Supplier of electrons Acceptor of electrons Positive electrode 2 H + + 2e- --> H 2 ReductionCathode Zn --> Zn 2+ + 2e- OxidationAnode

6 6 Reduction of H + by Zn Figure 20.10

7 7 Overall reaction is reduction of H + by Zn metal. Zn(s) + 2 H + (aq) --> Zn 2+ + H 2 (g) E o = +0.76 V Therefore, E o for Zn ---> Zn 2+ (aq) + 2e- is +0.76 V Zn is a (better) (poorer) reducing agent than H 2.

8 8 Cu/Cu 2+ and H 2 /H + Cell E o = +0.34 V Acceptor of electrons Supplier of electrons Cu 2+ + 2e- --> Cu ReductionCathode H 2 --> 2 H + + 2e- OxidationAnode Positive Negative

9 9 Cu/Cu 2+ and H 2 /H + Cell Overall reaction is reduction of Cu 2+ by H 2 gas. Cu 2+ (aq) + H 2 (g) ---> Cu(s) + 2 H + (aq) Measured E o = +0.34 V Therefore, E o for Cu 2+ + 2e- ---> Cu is +0.34 V

10 10 Zn/Cu Electrochemical Cell Zn(s) ---> Zn 2+ (aq) + 2e-E o = +0.76 V Cu 2+ (aq) + 2e- ---> Cu(s)E o = +0.34 V --------------------------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) E o (calc’d) = +1.10 V Cathode, positive, sink for electrons Anode, negative, source of electrons +

11 11 Uses of E o Values Organize half- reactions by relative ability to act as oxidizing agentsOrganize half- reactions by relative ability to act as oxidizing agents Table 20.1Table 20.1 Use this to predict cell potentials and direction of redox reactions.Use this to predict cell potentials and direction of redox reactions.

12 12 TABLE OF STANDARD REDUCTION POTENTIALS 2 E o (V) Cu 2+ + 2e- Cu+0.34 2 H + + 2e- H0.00 Zn 2+ + 2e- Zn-0.76 oxidizing ability of ion reducing ability of element

13 13 Potential Ladder for Reduction Half-Reactions Figure 20.11

14 14

15 15 Using Standard Potentials, E o Table 20.1 Which is the best oxidizing agent: O 2, H 2 O 2, or Cl 2 ? _________________ Which is the best reducing agent: Hg, Al, or Sn? ____________________

16 16 Standard Redox Potentials, E o Cu 2+ + 2e- Cu +0.34 + 2 H + 2e- H 2 0.00 Zn 2+ + 2e- Zn -0.76 Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner (anode) and oxidizing agent at northwest corner (cathode). Any substance on the right will reduce any substance higher than it on the left.

17 17 Standard Redox Potentials, E o Any substance on the right will reduce any substance higher than it on the left. Zn can reduce H + and Cu 2+.Zn can reduce H + and Cu 2+. H 2 can reduce Cu 2+ but not Zn 2+H 2 can reduce Cu 2+ but not Zn 2+ Cu cannot reduce H + or Zn 2+.Cu cannot reduce H + or Zn 2+.

18 18 Using Standard Potentials, E o Table 20.1 In which direction do the following reactions go?In which direction do the following reactions go? Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s)Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) 2 Fe 2+ (aq) +2 Fe 2+ (aq) + Sn 2+ (aq) ---> 2 Fe 3+ (aq) + Sn(s) What is E o net for the overall reaction?

19 19 Standard Redox Potentials, E o E˚ net = “distance” from “top” half-reaction (cathode) to “bottom” half-reaction (anode) E˚ net = E˚ cathode - E˚ anode E o net for Cu/Ag+ reaction = +0.46 V

20 20 Cd --> Cd 2+ + 2e- or Cd 2+ + 2e- --> Cd Fe --> Fe 2+ + 2e- or Fe 2+ + 2e- --> Fe E o for a Voltaic Cell All ingredients are present. Which way does reaction proceed?

21 21 From the table, you see Fe is a better reducing agent than CdFe is a better reducing agent than Cd Cd 2+ is a better oxidizing agent than Fe 2+Cd 2+ is a better oxidizing agent than Fe 2+ E o for a Voltaic Cell Overall reaction Fe + Cd 2+ ---> Cd + Fe 2+ E o = E˚ cathode - E˚ anode = (-0.40 V) - (-0.44 V) = +0.04 V

22 22 More About Calculating Cell Voltage Assume I - ion can reduce water. 2 H 2 O + 2e- ---> H 2 + 2 OH - Cathode 2 I - ---> I 2 + 2e- Anode ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 2 H 2 O + 2e- ---> H 2 + 2 OH - Cathode 2 I - ---> I 2 + 2e- Anode ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 Assuming reaction occurs as written, E˚ net = E˚ cathode - E˚ anode = (-0.828 V) - (+0.535 V) = -1.363 V Minus E˚ means rxn. occurs in opposite direction

23 23 Is E˚ related to ∆G? YES!

24 24 Michael Faraday 1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of electrolysiselectrolysis magnetic props. of mattermagnetic props. of matter electromagnetic inductionelectromagnetic induction benzene and other organic chemicalsbenzene and other organic chemicals Was a popular lecturer.

25 25 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 = 9.6485 x 10 4 J/Vmol and n is the number of moles of electrons transferred Michael Faraday 1791-1867

26 26 E o and ∆G o ∆G o = - n F E o For a product-favored reaction Reactants ----> Products Reactants ----> Products ∆G o 0 E o is positive For a reactant-favored reaction Reactants <---- Products Reactants <---- Products ∆G o > 0 and so E o 0 and so E o < 0 E o is negative

27 27 E at Nonstandard Conditions The NERNST EQUATIONThe NERNST EQUATION E = potential under nonstandard conditionsE = potential under nonstandard conditions n = no. of electrons exchangedn = no. of electrons exchanged ln = “natural log”ln = “natural log” If [P] and [R] = 1 mol/L, then E = E˚If [P] and [R] = 1 mol/L, then E = E˚ If [R] > [P], then E is ______________ than E˚If [R] > [P], then E is ______________ than E˚ If [R] < [P], then E is ______________ than E˚If [R] < [P], then E is ______________ than E˚


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