Lecture 21- Electrochemical cells

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Chapter 21 Lecture for Honors & Prep Chemistry

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Lecture 21- Electrochemical cells

  1. 1. BELLWORK Balance the following redox reaction using the “half-reaction” method HNO 2 + I -  NO + I 2 2e - + 2H + + 2 HNO 2  2 NO + 2H 2 O 2 I -  I 2 + 2e - 2H + + 2HNO 2 + 2I -  2NO + 2H 2 O + I 2
  2. 2. Electrochemical cells produce an electric current from redox reactions CAN ALSO BE CALLED A Voltaic cell Galvanic cell Battery
  3. 3. A simple battery separates the reduction reaction from the oxidation reaction so that the electrons must travel through a wire <ul><li>Oxidation e - e - e - e - e - e - e - Reduction </li></ul><ul><li>Lose e - e - e - e - e - e - e - e - e - e - e - e - Gain e - </li></ul>
  4. 4. Redox reaction Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) <ul><li>Separate into 2 half-reactions and balance </li></ul><ul><li>Zn(s)  Zn 2+ + 2e - oxidation </li></ul><ul><li>2e - + Cu 2+ (aq)  Cu(s) reduction </li></ul>
  5. 5. How to make a battery or an electrochemical cell <ul><li>Create each half reaction in a separate chamber called a half cell and connect them by a wire </li></ul>e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - Oxidation Reduction half cell half cell
  6. 6. To make the Zinc half cell Add a strip of Zinc metal to a solution containing Zn 2+ 1M ZnSO 4 Zn(s) Zinc metal
  7. 7. To make the copper half cell Add a strip of copper metal to a solution containing Cu 2+ 1M CuSO 4 Cu(s) Copper metal
  8. 8. The metal strips are electrodes The solutions are electrolytes <ul><li>oxidation occurs at the anode </li></ul><ul><li>Zn(s)  Zn 2+ + 2e - metal is lost </li></ul><ul><li>reduction occurs at the cathode </li></ul><ul><li>2e - + Cu 2+ (aq)  Cu(s) metal is plated </li></ul><ul><li>An Ox Red Cat </li></ul>
  9. 10. When a wire connects the half cells an electrical current is created. This can’t occur for long because positive charge will build up at the anode and negative charge will build up at the cathode A salt bridge is used to keep the two cell compartments neutral
  10. 11. The salt bridge contains a salt (positive and negative ions) that flows into each beaker as needed to keep the charge neutral <ul><li>The salt bridge completes the circuit and the current will continue to flow </li></ul>SALT BRIDGE Gotta have it!!!
  11. 12. How do you determine the voltage of a battery? <ul><li>Use the table of reduction potentials . </li></ul><ul><ul><ul><li>All of the half-reactions are for reduction </li></ul></ul></ul><ul><ul><ul><li>E º is the electrical potential of the ½ reaction in Volts </li></ul></ul></ul><ul><ul><ul><li>If you need the oxidation potential, flip the equation and change the sign of  E º. All reduction potentials are reversible. </li></ul></ul></ul><ul><ul><ul><li>The top of the table = elements that are easily reduced, and not easily oxidized. They take electrons easily. </li></ul></ul></ul><ul><ul><ul><li>The bottom of the table= elements that are easily oxidized and not easily reduced. They give electrons easily. </li></ul></ul></ul><ul><li>To determine the voltage of a battery you add the voltages of the two half-reactions </li></ul>
  12. 13. <ul><li>Flip the reaction for the oxidized element and change the sign of its voltage. </li></ul><ul><li>Ex. What is the voltage of the Zn-Cu battery? </li></ul><ul><li>Cu 2+ + 2e -  Cu(s) 0.34V </li></ul><ul><li>Zn 2+ + 2e -  Zn(s) -0.76V oxidized because it is the lowest on the table of reduction potentials </li></ul><ul><li>Flip it </li></ul><ul><li>Zn(s)  Zn 2+ + 2e - 0.76V </li></ul>The element at the top of the table will be reduced. The lower one will be oxidized. ADD THESE FOR VOLTAGE
  13. 14. Describe the voltaic cell represented as: Al (s) | Al 2 (SO 4 ) 3 (aq) || NiSO 4 (aq) | Ni (s)

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