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CELL POTENTIAL, E

1.10 V. 1.0 M. 1.0 M. CELL POTENTIAL, E. 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. Zn and Zn 2+ , anode. Cu and Cu 2+ , cathode.

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CELL POTENTIAL, E

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  1. 1.10 V 1.0 M 1.0 M CELL POTENTIAL, E • 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 [Zn2+] and [Cu2+] = 1.0 M. Zn and Zn2+, anode Cu and Cu2+, cathode

  2. CELL POTENTIAL, E • For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M. • This is the STANDARD CELL POTENTIAL, Eo • —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

  3. Calculating Cell Voltage • Balanced half-reactions can be added together to get overall, balanced equation. Zn(s) ---> Zn2+(aq) + 2e- Cu2+(aq) + 2e- ---> Cu(s) -------------------------------------------- Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) If we know Eo for each half-reaction, we could get Eo for net reaction.

  4. CELL POTENTIALS, Eo Can’t measure 1/2 reaction Eo directly. Therefore, measure it relative to a STANDARD HYDROGEN CELL, SHE. 2 H+(aq, 1 M) + 2e- <----> H2(g, 1 atm) Eo = 0.0 V

  5. Negative electrode Positive electrode Zn/Zn2+ half-cell hooked to a SHE. Eo for the cell = +0.76 V Supplier of electrons Acceptor of electrons Zn --> Zn2+ + 2e- Oxidation Anode 2 H+ + 2e- --> H2 Reduction Cathode

  6. Reduction of H+ by Zn Figure 20.10

  7. Overall reaction is reduction of H+ by Zn metal. Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Eo = +0.76 V Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is +0.76 V Zn is a (better) (poorer) reducing agent than H2.

  8. Cu/Cu2+ and H2/H+ Cell Eo = +0.34 V Positive Negative Acceptor of electrons Supplier of electrons Cu2+ + 2e- --> Cu Reduction Cathode H2 --> 2 H+ + 2e- Oxidation Anode

  9. Cu/Cu2+ and H2/H+ Cell Overall reaction is reduction of Cu2+ by H2 gas. Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq) Measured Eo = +0.34 V Therefore, Eo for Cu2+ + 2e- ---> Cu is +0.34 V

  10. + Zn/Cu Electrochemical Cell Zn(s) ---> Zn2+(aq) + 2e- Eo = +0.76 V Cu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 V --------------------------------------------------------------- Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) Eo (calc’d) = +1.10 V Anode, negative, source of electrons Cathode, positive, sink for electrons

  11. Uses of Eo Values • Organize half-reactions by relative ability to act as oxidizing agents • Table from book • Use this to predict cell potentials and direction of redox reactions.

  12. oxidizing o ability of ion E (V) 2+ Cu + 2e- Cu +0.34 + 2 H + 2e- H 0.00 2+ Zn + 2e- Zn -0.76 reducing ability of element TABLE OF STANDARD REDUCTION POTENTIALS 2

  13. Potential Ladder for Reduction Half-Reactions Figure 20.11

  14. Using Standard Potentials, Eo • Which is the best oxidizing agent: O2, H2O2, or Cl2? _________________ • Which is the best reducing agent: Hg, Al, or Sn? ____________________ H2O2, Cl2, O2 Al, Sn, Hg

  15. 2+ Cu + 2e- Cu +0.34 + 2 H + 2e- H 0.00 2 2+ Zn + 2e- Zn -0.76 Standard Redox Potentials, Eo Any substance on the right will reduce any substance higher than it on the left. Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner (anode) and oxidizing agent at northwest corner (cathode).

  16. Standard Redox Potentials, Eo Any substance on the right will reduce any substance higher than it on the left. • Zn can reduce H+ and Cu2+. • H2 can reduce Cu2+ but not Zn2+ • Cu cannot reduce H+ or Zn2+.

  17. Using Standard Potentials, Eo • In which direction do the following reactions go? • Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) • 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s) • What is Eonet for the overall reaction? • See board for example

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

  19. Eo for a Voltaic Cell Cd --> Cd2+ + 2e- or Cd2+ + 2e- --> Cd Fe --> Fe2+ + 2e- or Fe2+ + 2e- --> Fe All ingredients are present. Which way does reaction proceed?

  20. Eo for a Voltaic Cell From the table, you see • Fe is a better reducing agent than Cd • Cd2+ is a better oxidizing agent than Fe2+ Overall reaction Fe + Cd2+ ---> Cd + Fe2+ Eo = E˚cathode - E˚anode = (-0.40 V) - (-0.44 V) = +0.04 V

  21. More About Calculating Cell Voltage 2 H2O + 2e- ---> H2 + 2 OH- Cathode 2 I----> I2 + 2e- Anode ------------------------------------------------- 2 I- + 2 H2O --> I2 + 2 OH- + H2 Assume I- ion can reduce water. 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

  22. YES! Is E˚ related to ∆G?

  23. Michael Faraday1791-1867 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 and one wild dude

  24. Eo and ∆Go Eo is related to ∆Go, the free energy change for the reaction. ∆Go = - n F Eo where F = Faraday constant = 9.6485 x 104 J/V•mol and n is the number of moles of electrons transferred do you understand moles e- transferred? (board) Michael Faraday 1791-1867

  25. Eo and ∆Go ∆Go = - n F Eo For a product-favored reaction Reactants ----> Products ∆Go < 0 and so Eo > 0 Eo is positive For a reactant-favored reaction Reactants <---- Products ∆Go > 0 and so Eo < 0 Eo is negative

  26. E at Nonstandard Conditions • The NERNST EQUATION (please see p.827 equation 17.1) • E = potential under nonstandard conditions • n = no. of electrons exchanged • ln = “natural log” • 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˚ More Less

  27. E at Nonstandard Conditions • The NERNST EQUATION is the cell potential before any reaction occurs. What will happen to E as reaction continues? • E will decrease • In fact, cell will continue to produce lower amounts of V until the r’xn reaches equilibrium, at that point the reaction has lost its driving force and Q = K and E = 0 • At this point your battery is D… E… A… D!

  28. Example problem • Determine the cell potential at 25 oC for Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) when • [Fe2+] = 0.010M and [Cd2+] = 1.0M and 2. [Fe2+] = 1.0M and [Cd2+] = 0.010M. Also comment on the favorability of the reaction

  29. Example problem • Determine the cell potential at 25 oC for Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) 1. [Fe2+] = 0.010M and [Cd2+] = 1.0M 2. [Fe2+] = 1.0M and [Cd2+] = 0.010M. Solution: First solve for the standard potential Oxidation: Fe(s) Fe2+(aq) + 2 e-Eo = + 0.44 V Reduction: Cd2+(aq) + 2 e-  Cd(s) Eo = - 0.40 V Overall: Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s)Eo = + 0.04 V

  30. Example problem • Determine the cell potential at 25 oC for Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) 1. [Fe2+] = 0.010M and [Cd2+] = 1.0M 2. [Fe2+] = 1.0M and [Cd2+] = 0.010M. Solution: Next solve problem 1 by substituting in the concentrations (n = 2 because there are 2 moles of electrons exchanged as seen in the ½ reaction) E = + 0.10 V

  31. Example problem • Determine the cell potential at 25 oC for Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) 1. [Fe2+] = 0.010M and [Cd2+] = 1.0M 2. [Fe2+] = 1.0M and [Cd2+] = 0.010M. Solution: Now solve problem 2 by substituting in the concentrations E = - 0.02 V

  32. Example problem • Determine the cell potential at 25 oC for Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) 1. [Fe2+] = 0.010M and [Cd2+] = 1.0M 2. [Fe2+] = 1.0M and [Cd2+] = 0.010M. Solution: Favorability: For reaction 1. Does the forward r’xn occur? The reaction under these conditions is (more / less) favorable.

  33. Example problem • Determine the cell potential at 25 oC for Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) 1. [Fe2+] = 0.010M and [Cd2+] = 1.0M 2. [Fe2+] = 1.0M and [Cd2+] = 0.010M. Solution: Favorability: For reaction 2. Does the forward r’xn occur?

  34. E and Equilibrium Constants • The NERNST EQUATION (please see p.827 equation 17.1) • As reactants are converted to products E declines and eventually will reach 0 • E = 0 is an indication that no net reaction is occurring • Equilibrium is reached • Sub 0 for E and K for [P]/[R] …

  35. E and Equilibrium Constants • This equation allows us to find K from Eo Problem: Calculate the equilibrium constant for the following reaction Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) Solution:

  36. E and Equilibrium Constants Problem: Calculate the equilibrium constant for the following reaction. Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s) Solution:

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