1. Activity Series lithiumpotassiummagnesiumaluminumzincironnickelleadHYDROGENcoppersilverplatinumgold Oxidizes easily Reduces easily Less active More active

2. Activity series More active metals will displace less active metal ions. More active metals will displace less active metal ions. Hydrogen is the reference point. Hydrogen is the reference point. Electrons will flow spontaneously from the stronger reducing agent to the stronger oxidizing agent. Electrons will flow spontaneously from the stronger reducing agent to the stronger oxidizing agent.

3. lithiumpotassiummagnesiumaluminumzincironnickelleadHYDROGENcoppersilverplatinumgold Will the following reactions occur ? Ag + + Cu Cu 2+ + Ag Mg 2+ + Zn Zn 2+ + Mg Zn 2+ + Cu Mg 2+ + Li No Yes The metal must be above (more active than) the ion for it to be a spontaneous reaction. No Yes

4. Electrochemistry Chemical energy becoming electrical energy

5. Current- the flow of electricity (amps) Current- the flow of electricity (amps) Potential difference- the force pushing the current (volts) Potential difference- the force pushing the current (volts) Electrolyte- an ionic solution that can carry electric current. Electrolyte- an ionic solution that can carry electric current. Spontaneous redox reactions produce this electric energy. Spontaneous redox reactions produce this electric energy.

6. Voltaic Cell Anode attracts anions attracts anions where oxidation occurs where oxidation occursCathode attracts cations attracts cations where reduction occurs where reduction occurs Salt bridge connects the two half cells connects the two half cells contains a strong electrolyte contains a strong electrolyte

7. Zn Zn + Cu 2+ Cu in shorthand Two half cells connected by a salt bridge

8. Reduction half reactions F 2 is the strongest oxidizing agent Li is the strongest reducing agent

9. Reduction Potentials (E) E o is the standard electrode potential E o is the standard electrode potential all ions are 1M and gases are 1 atm all ions are 1M and gases are 1 atm The net E o is the sum of the E o of the half reactions The net E o is the sum of the E o of the half reactions The stronger oxidizing agent reduces. The stronger oxidizing agent reduces. Reverse the sign of the substance oxidized. Reverse the sign of the substance oxidized. If E is positive, the reaction is spontaneous. If E is positive, the reaction is spontaneous. If E is negative, the reverse reaction is spontaneous. If E is negative, the reverse reaction is spontaneous.

10. What is the voltage produced from the reaction of Zn metal with Cu 2+ ions? Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn 2+ + 2e -  Zn -0.7628 Cu 2+ + 2e -  Cu 0.3402 -(-0.7628) + 0.3402 = 1.103 volts Will happen spontaneously Zn 2+ + Ni(s)  Zn(s) + Ni 2+ -(-0.23) + (-0.7628) = -.53 Will not occur spontaneously Cu 2+ Cu 2+ will reduce!

12. Faraday’s Law Coulombs = amperes x seconds Coulombs = amperes x seconds 1 C = 1amp·1sec 1 C = 1amp·1sec 96,485 coulombs = 1 mole e - 96,485 coulombs = 1 mole e -

13. What mass of copper will be deposited by a current of 7.89 amps flowing for 1200 seconds? Cu 2+ + 2e -  Cu at the cathode Cu 2+ + 2e -  Cu at the cathode 7.89A x 1200s x 1C x 1 mol e - =.0981 mol e - A·s 96,485C A·s 96,485C.0981 mol e - x 1 mol Cu x 63.5g Cu = 3.1g Cu 2 mol e - 1 mole Cu 2 mol e - 1 mole Cu

14. What mass of Cr 3+ is produced by a current of 0.713A flowing for 12,800 seconds? 14H 3 O + + 6Fe 2+ + Cr 2 O 7 2-  6Fe 3+ + 2Cr 3- +21H 2 O.713A x 12,800s x 1C x 1 mol e - =.0946 mol e - A·s 96,485C A·s 96,485C.0946 mol e - x 2mol Cr 3+ x 52.0g Cr 3+ = 1.64gCr 3+ 6mol e - 1mol Cr 3+ 6mol e - 1mol Cr 3+

15. Nernst Equation E o Voltage under standard conditions E o Voltage under standard conditions (1M solutions at 25°C and 101.3kPa) (1M solutions at 25°C and 101.3kPa) At non-standard conditions, use Nernst equation E = E° - 0.05916 log [products] n [reactants] n [reactants] n = no. of electrons transferred n = no. of electrons transferred Coefficients in front of reactants or products are used as powers of their concentrations.

16. Determine the voltage of a Zn Zn 2+ (.75M) Ni 2+ (2.0M) Ni cell at standard temperature and pressure. E o = +0.7626V + -.257V =.506V E =.506V -.0592 log.75M 2 2.0M 2 2.0M E =.506V – 0.0296 (log 0.375) E =.506V – (-0.0126) E =.519V