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Valdosta State University Experiment 11 Oxidation-Reduction Chemistry Valdosta State University.

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Presentation on theme: "Valdosta State University Experiment 11 Oxidation-Reduction Chemistry Valdosta State University."— Presentation transcript:

1 Valdosta State University Experiment 11 Oxidation-Reduction Chemistry Valdosta State University

2 Purpose To observe and predict the EMF of a voltaic cell and a concentration cell. Also, to balance redox equations in acidic and alkaline solutions. Valdosta State University

3 Background Valdosta State University Voltaic Cells An electrochemical cell that uses spontaneous chemical reactions to produce a voltage. The voltage is produced by potential difference between two substances. The transfer of electrons from one ion or molecule to another occurs: Oxidation – when one substance loses electrons. Reduction – when one substance gains electrons. Remember – LEO the lion goes GER (Loss of Electrons is Oxidation, Gain of Electrons is Reduction) Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) The interesting part is converting this equation to a workable cell.

4 Valdosta State University Background Valdosta State University Voltaic Cells

5 Valdosta State University Background Valdosta State University Voltaic Cells Current is carried by: Electrons (through wire) Ions (solution, salt bridge) Both pathways are required to form a completed circuit.

6 Valdosta State University Background Valdosta State University Voltaic Cells Notice that each beaker (half-cell) contains the complete half-reaction. Anode (Oxidation) Zn(s)  Zn 2+ (aq) + 2 e - Cathode (Reduction) Cu 2+ (aq) + 2 e -  Cu(s) Remember the phrase Red Cat (Reduction occurs at the Cathode) to help you remember this. V

7 Valdosta State University Background Valdosta State University Voltaic Cells – Cell EMF Cell electromotive force (E cell ) – the voltage a voltaic cell generates. The voltage generated by a voltaic cell depends on a number of factors, a few of which are: the half-cells used the concentrations of the reagents the temperature of the cell.

8 Valdosta State University Background Valdosta State University Voltaic Cells – Standard Cell Potential Cell electromotive force (E cell ) – the voltage a voltaic cell generates. If the cell is operated at standard state (298K, 1M solution concentrations and 1 atmosphere pressure) the voltage generated is referred to as a standard cell potential ( )

9 Valdosta State University Background Valdosta State University Voltaic Cells – Standard Cell Potential Cell electromotive force (E cell ) – the voltage a voltaic cell generates. If the cell is operated at standard state (298K, 1M solution concentrations and 1 atmosphere pressure) the voltage generated is referred to as a standard cell potential ( ) Zn(s) + Cu 2+ (aq)  Zu 2+ (aq) + Cu(s)

10 Valdosta State University Background Valdosta State University Voltaic Cells – Standard Reduction Potential Reduction potentials are summarized in a table. These potentials can be used to determine the cell potential. If you do the calculation for E o cell and get a negative number, then you chose the wrong anode and wrong cathode and need to switch both of them.

11 Valdosta State University Background Valdosta State University Voltaic Cells – Concentration Cell Solution concentration has an impact on cell emf. The Nernst equation can be used to predict the emf. Simplifying the equation yields (at T = 298.15K): *Ions move towards diluted solution Products/Reactants Diluted/Concentrated

12 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions 1.Break up the overall reaction into two half reactions omitting H +, OH -, and H 2 O. 2.For each half reaction, balance all of the elements except O and H. 3.Balance for oxygen. For each half reaction, balance the O's by adding H 2 O to the side of the reaction that is deficient in oxygen. 4.Balance for hydrogen. For each half reaction, add H + to the side of the reaction that is deficient in hydrogen. If you are balancing a reaction in acid solution, skip to step 5 BASE ONLY a)For each half reaction, add OH  to both sides of the equation. Add enough OH- to neutralize the H + ions added in step 4. b)For each half reaction, take the H + and OH  and combine them to form water. Cancel out water molecules that appear on both sides of the reaction. 5.For each half reaction, balance the charge by adding electrons. The electrons are added to the more positive side, and the difference in charge gives the number of electrons to add. 6.Multiply one (or both) half reactions by a whole number so that both half reactions have the same number of electrons. 7.Add the half reactions. The electrons will cancel, as will some of the H +, OH , and H 2 O.

13 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 1.Break up the overall reaction into two half reactions omitting H +, OH -, and H 2 O. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g)H 2 S(aq)  S(s) Oxidation half reaction Reduction half reaction

14 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 2.For each half reaction, balance all of the elements except O and H. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g) H 2 S(aq)  S(s)

15 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 3.Balance for oxygen. For each half reaction, balance the O's by adding H 2 O to the side of the reaction that is deficient in oxygen. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g) H 2 S(aq)  S(s)

16 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 4.Balance for hydrogen. For each half reaction, add H + to the side of the reaction that is deficient in hydrogen. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g) H 2 S(aq)  S(s)

17 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 5.For each half reaction, balance the charge by adding electrons. The electrons are added to the more positive side, and the difference in charge gives the number of electrons to add. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g)H 2 S(aq)  S(s)

18 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 6.Multiply one (or both) half reactions by a whole number so that both half reactions have the same number of electrons. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g)H 2 S(aq)  S(s)

19 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 6.Multiply one (or both) half reactions by a whole number so that both half reactions have the same number of electrons. H 2 S(aq) + NO 3 - (aq)  NO(g) + S(s) NO 3 - (aq)  NO(g)H 2 S(aq)  S(s)

20 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 7.Add the half reactions. The electrons will cancel, as will some of the H +, OH , and H 2 O. 3H 2 S(aq)  3S(s) + 6H + (aq) + 6e - 6e - + 8H + (aq) + 2NO 3 - (aq)  2NO(g) + 4H 2 O(l)

21 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Acid) 7.Add the half reactions. The electrons will cancel, as will some of the H +, OH , and H 2 O. 3H 2 S(aq)  3S(s) + 6H + (aq) + 6e - 6e - + 8H + (aq) + 2NO 3 - (aq)  2NO(g) + 4H 2 O(l) 2H + (aq) + 2NO 3 - (aq) + 3H 2 S(aq)  2NO(g) + 3S(s) + 4H 2 O(l)

22 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 1.Break up the overall reaction into two half reactions omitting H +, OH -, and H 2 O. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq)MnO 2 (s)  Mn 2+ (aq) Oxidation half-reaction Reduction half-reaction

23 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 2.For each half reaction, balance all of the elements except O and H. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq)MnO 2 (s)  Mn 2+ (aq)

24 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 3.Balance for oxygen. For each half reaction, balance the O's by adding H 2 O to the side of the reaction that is deficient in oxygen. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq) MnO 2 (s)  Mn 2+ (aq)

25 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 4.Balance for hydrogen. For each half reaction, add H + to the side of the reaction that is deficient in hydrogen. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq)MnO 2 (s)  Mn 2+ (aq)

26 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 4.Balance for hydrogen. For each half reaction, add H + to the side of the reaction that is deficient in hydrogen. a)For each half reaction, add OH  to both sides of the equation. Add enough OH- to neutralize the H + ions added in step 4. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq)MnO 2 (s)  Mn 2+ (aq)

27 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 4.Balance for hydrogen. For each half reaction, add H + to the side of the reaction that is deficient in hydrogen. b)For each half reaction, take the H + and OH  and combine them to form water. Cancel out water molecules that appear on both sides of the reaction. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq)MnO 2 (s)  Mn 2+ (aq)

28 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 4.Balance for hydrogen. For each half reaction, add H + to the side of the reaction that is deficient in hydrogen. b)For each half reaction, take the H + and OH  and combine them to form water. Cancel out water molecules that appear on both sides of the reaction. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq)  CrO 4 2- (aq)MnO 2 (s)  Mn 2+ (aq)

29 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 5.For each half reaction, balance the charge by adding electrons. The electrons are added to the more positive side, and the difference in charge gives the number of electrons to add. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq) + 8OH - (aq)  CrO 4 2- (aq) + 4H 2 O(l)MnO 2 (s) + 2H 2 O(l)  Mn 2+ (aq) + 4OH - (aq)

30 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 6.Multiply one (or both) half reactions by a whole number so that both half reactions have the same number of electrons. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq) + 8OH - (aq)  CrO 4 2- (aq) + 4H 2 O(l)MnO 2 (s) + 2H 2 O(l)  Mn 2+ (aq) + 4OH - (aq)

31 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 6.Multiply one (or both) half reactions by a whole number so that both half reactions have the same number of electrons. Cr 3+ (aq) + MnO 2 (s)  Mn 2+ (aq) + CrO 4 2- (aq) Cr 3+ (aq) + 8OH - (aq)  CrO 4 2- (aq) + 4H 2 O(l)MnO 2 (s) + 2H 2 O(l)  Mn 2+ (aq) + 4OH - (aq)

32 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 7.Add the half reactions. The electrons will cancel, as will some of the H +, OH , and H 2 O. 2Cr 3+ (aq) + 16OH - (aq)  2CrO 4 2- (aq) + 8H 2 O(l) + 6e - 6e - + 3MnO 2 (s) + 6H 2 O(l)  3Mn 2+ (aq) + 12OH - (aq)

33 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 7.Add the half reactions. The electrons will cancel, as will some of the H +, OH , and H 2 O. 2Cr 3+ (aq) + 16OH - (aq)  2CrO 4 2- (aq) + 8H 2 O(l) + 6e - 6e - + 3MnO 2 (s) + 6H 2 O(l)  3Mn 2+ (aq) + 12OH - (aq) 6e - + 2Cr 3+ (aq) + 3MnO 2 (s) + 16OH - (aq) + 6H 2 O(l)  2CrO 4 2- (aq) + 3Mn 2+ (aq) + 12OH - (aq) + 8H 2 O(l) + 6e -

34 Valdosta State University Background Valdosta State University Balancing Oxidation-Reduction Reactions (Alkaline) 7.Add the half reactions. The electrons will cancel, as will some of the H +, OH , and H 2 O. 2Cr 3+ (aq) + 16OH - (aq)  2CrO 4 2- (aq) + 8H 2 O(l) + 6e - 6e - + 3MnO 2 (s) + 6H 2 O(l)  3Mn 2+ (aq) + 12OH - (aq) 2Cr 3+ (aq) + 3MnO 2 (s) + 4OH - (aq)  2CrO 4 2- (aq) + 3Mn 2+ (aq) + 2 H 2 O(l)

35 Valdosta State University Procedure – Experiment 11 - For this experiment, work in pairs.

36 Valdosta State University Procedure – Experiment 11 Voltaic Cells Computer Set-Up 1. Hook the rectangular end of the gray cord into the back of the laptop computer and the round end into Science Workshop black box interface. 2. Plug in and turn on the Pasco Science Workshop black box interface. You should see a green LED on the front face of the computer. 3. Plug the DIN plug of the voltage sensor into Channel A on the Scientific Workshop black box interface. 4.Turn on the computer. When the computer has booted-up, double click on the “Data Studio” icon. The program should start. 5. On the opening screen, select “Create Experiment”. 6. Click and drag the “Voltage” sensor to Channel A of the interface box as shown in the experimental setup window. 7. Click and drag the "digits" icon labeled “Voltage Ch A (V)”. 8. To have the computer measure the voltage, press “Start”.

37 Valdosta State University Procedure – Experiment 11 Voltaic Cells Cell Set-Up (Cu / Zn) 1.Get Zn, Cu, and Mg electrodes. Polish each electrode with sandpaper. 2.Pour approximately 50 mL of 0.1 M CuSO 4 into a 150 mL beaker. Pour 0.1 M Zn(NO 3 ) 2 into the porous cup until the cup is about 2/3 full. Put the porous cup into the CuSO 4 solution. 3.Place a Zn electrode into the Zn(NO 3 ) 2 solution and a Cu electrode into the CuSO 4 solution. 4.Hook one of the alligator clips to the Zn electrode, and the other alligator clip to the Cu electrode. 5.Record the voltage on your report sheet. If the voltage is negative, reverse the alligator clips.

38 Valdosta State University Procedure – Experiment 11 Voltaic Cells

39 Valdosta State University Procedure – Experiment 11 Voltaic Cells Cell Set-Up (Cu / Mg) 6.Remove the alligator clips and empty the contents of the porous cup into the waste container. Rinse the inside of the cup with distilled water 7.Place 0.1 M Mg(NO 3 ) 2 into the porous cup until the cup is about 2/3 full. Put the Mg electrode into the Mg(NO 3 ) 2 solution. Place the porous cup into the CuSO 4 solution. 8.Hook one of the alligator clips to the Cu electrode, and the other alligator clip to the Mg electrode. 9.Record the voltage on your report sheet. If the voltage is negative, reverse the alligator clips.

40 Valdosta State University Procedure – Experiment 11 Voltaic Cells Cell Set-Up (Zn / Mg) 10.Remove the alligator clips. DO NOT EMPTY ANY CONTAINERS!! 11.Get another 150 mL beaker and put 50 mL of 0.1 M Zn(NO 3 ) 2 in the beaker. Put the porous cup containing the Mg(NO 3 ) 2 solution into the beaker containing the Zn(NO 3 ) 2. 12.Place the Zn electrode into the Zn(NO 3 ) 2 solution and make sure the Mg electrode is in the Mg(NO 3 ) 2 solution. 13.Hook one of the alligator clips to the Zn electrode, and the other alligator clip to the Mg electrode. 14.Record the voltage on your report sheet. If the voltage is negative, reverse the alligator clips.

41 Valdosta State University Procedure – Experiment 11 Concentration Cells Cell Set-Up (Concentration Cell) 15.Remove the alligator clips and put both the Mg(NO 3 ) 2 solution and the Zn(NO 3 ) 2 solution into the waste container. SAVE THE 0.1 M CuSO 4 SOLUTION. YOU WILL USE IT NEXT. 16.Wash the porous cup with distilled water and fill it approximately 2/3 full with 0.001 M CuSO 4 solution. 17.Hook each of the alligator clips to the Cu electrodes in each solution. 18.Record the voltage on your report sheet. If the voltage is negative, reverse the electrodes. 19.To the porous cup containing the 0.001 M CuSO 4 solution, add 10 drops 6 M NH 3 and mix well. Record the voltage on your report sheet.

42 Valdosta State University Safety The 6 M NH 3 gives off irritating fumes. Valdosta State University

43 Waste Disposal All solutions used in this experiment must be placed in the container marked "Recovered Metals and Metal Ions". Electrode materials should be returned to the “Used Electrodes” container. Valdosta State University


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