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Anode

Basic Concepts of Electrochemical Cells. Electrifying!. Anode. Cathode. CHEMICAL CHANGE ---> ELECTRIC CURRENT. With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”. Zn is oxidized and is the reducing agent Zn(s) ---> Zn 2+ (aq) + 2e-

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Anode

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  1. Basic Concepts of Electrochemical Cells Electrifying! Anode Cathode

  2. CHEMICAL CHANGE --->ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” • Zn is oxidized and is the reducing agent Zn(s) ---> Zn2+(aq) + 2e- • Cu2+ is reduced and is the oxidizing agentCu2+(aq) + 2e- ---> Cu(s)

  3. Electrons are transferred from Zn to Cu2+, but there is no useful electric current. CHEMICAL CHANGE --->ELECTRIC CURRENT Oxidation: Zn(s) ---> Zn2+(aq) + 2e- Reduction: Cu2+(aq) + 2e- ---> Cu(s) -------------------------------------------------------- Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

  4. CHEMICAL CHANGE --->ELECTRIC CURRENT • To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. This is accomplished in a GALVANIC or VOLTAIC cell. A group of such cells is called a battery.

  5. Zn --> Zn2+ + 2e- Cu2+ + 2e- --> Cu Oxidation Anode Negative Reduction Cathode Positive •Electrons travel thru external wire. • Salt bridge allows anions and cations to move between electrode compartments. <--Anions Cations-->

  6. The Cu|Cu2+ and Ag|Ag+ Cell Electrons move from anode to cathode in the wire. Anions & cations move thru the salt bridge.

  7. Anode, site of oxidation, negative Cathode, site of reduction, positive

  8. 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

  9. 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. • This means pure solids or in solution at a concentration of 1M!!!!

  10. 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. • Lets revisit my haiku!

  11. Oxidation Haiku! • Lost an electron • But now feeling positive • Oxidized is cool! • What is that? You want a reduction Haiku?

  12. Reduction Haiku!!! • Gained some electrons • Gave me a negative mood! • Now I can say Ger! • Thank you… Enjoy the buffet… Don’t eat the chemicals or furniture kids!

  13. 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

  14. 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

  15. Reduction of H+ by Zn Figure 20.10

  16. 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.

  17. 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

  18. 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

  19. + 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

  20. Yes It is finally time for a DEMO!! • Do you feel like bridging that salt?

  21. 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

  22. Potential Ladder for Reduction Half-Reactions Figure 20.11

  23. Table 21.1 Page 970

  24. 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).

  25. 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+.

  26. Using Standard Potentials, EoTable 20.1 • 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?

  27. 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

  28. 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?

  29. 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

  30. 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

  31. If you have reached this far, you need a break!

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