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Voltaic Cells Harnessing the power of Redox Redox Basics Redox Reactions happen when –one substance loses electrons (oxidized) and –one gains electrons.

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Presentation on theme: "Voltaic Cells Harnessing the power of Redox Redox Basics Redox Reactions happen when –one substance loses electrons (oxidized) and –one gains electrons."— Presentation transcript:

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3 Voltaic Cells Harnessing the power of Redox

4 Redox Basics Redox Reactions happen when –one substance loses electrons (oxidized) and –one gains electrons (reduced) Electrons are flowing Cl 2 2Na 2Cl - 2Na + 2 Electrons Half equation #1: 2Na  2Na+ + 2e - Half equation #2: Cl 2 + 2e -  2Cl - nett: 2Na + Cl 2  2Na + 2Cl -

5 Example Reaction Zn + Cu 2+  Cu + Zn 2+ Gains 2 electrons- reduced Loses 2 electrons- oxidized Cu 2+ Zn Zn 2+ Cu Zn  Zn 2+ = +0.76 Cu 2+  Cu = +0.34 Overall =+1.10

6 Harness Electrons If the reactants are separated The electrons cannot travel directly between reactants Those electrons can be used to do work Note: the electrons will always flow in the direction which gives a +E o cell E o cell = E o (red)-E o (ox)

7 SO 4 2- Zn 2+ CuZn Reaction: Zn + CuSO 4  Cu + ZnSO 4 Build a Cell 1 (Ch 20, sect 3) SO 4 2- Cu 2+ Electrode- carries electrons, may or may not be part of the reaction Reaction: Zn + Cu 2+  Cu + Zn 2+

8 Zn 2+ CuZn Reaction: Zn + Cu 2+  Cu + Zn 2+ Electrons! Build a Cell 2 Cu 2+

9 Zn 2+ CuZn Reaction: Zn + Cu 2+  Cu + Zn 2+ Electrons! Build a Cell 3 Cu 2+ Salt BridgeKCl K+K+ Cl - Cu Zn Build up of + ions Build up of - ions

10 Terminology Electrode: conductor which allows electrons to enter or leave a cell (any conductive but unreactive material, sometimes part of the redox reaction) –Anode: oxidation electrode, negative (Zn) –Cathode: reduction electrode, positive (Cu) Half-cell: half of the electrochemical cell where ox. or red. takes place. Electrolyte: salt solution in a half-cell, always contains oxidized and reduced form

11 Cl - Zn 2+ H2H2 Zn Reaction: Zn + 2H +  H 2 + Zn 2+ Electrons! Other Cells 1 SO 4 2- H+H+ Pt Salt bridge

12 Pt Reaction: 2I - + 2Fe 3+  I 2 + 2Fe 2+ Electrons! Other Cells 2 I-I- Fe 3+ I2I2 Fe 2+ e-e- e-e- Salt bridge

13 PtC Reaction: Cl - + MnO 4 -  Cl 2 + Mn 2+ 10Cl - + 2MnO 4 - + 16H +  5Cl 2 + 2Mn 2+ + 8H 2 O Electrons! Other Cells 3 Cl - MnO 4 - /H + Cl 2 Mn 2+ e-e- e-e- Salt bridge

14 Cell Diagrams Since it would suck to draw cells every time you wanted to talk about them, scientists were kind enough to come up with a shorthand Zn 2+ (aq) Cu 2+ (aq) Cu (s) //// Zn (s) Oxidation on leftReduction on right Salt bridge electrode electrolyte(s) electrolyte(s) Electrolytes: If there’s more than one, What you start with on the left, end with on the right

15 Zn 2+ H2H2 Zn Reaction: Zn + 2H +  H 2 + Zn 2+ Electrons! Other Cells 1 H+H+ Pt Salt bridge Zn (s) /Zn 2+ (aq) //H + (aq), H 2(g) /Pt (s)

16 Pt Reaction: 2I - + 2Fe 3+  I 2 + 2Fe 2+ Electrons! Other Cells 2 I-I- Fe 3+ e-e- e-e- Salt bridge Fe 2+ I2I2 Pt (s) /I - (aq), I 2(aq) //Fe 3+ (aq), Fe 2+ (aq) /Pt (s)

17 PtC Reaction: 10Cl - + 2MnO 4 - + 16H +  5Cl 2 + 2Mn 2+ + 8H 2 O Electrons! Other Cells 3 Cl - MnO 4 - /H + Salt bridge Cl 2 Mn 2+ C (s) /Cl - (aq), Cl 2(aq) //MnO 4 - (aq), Mn 2+ (aq) /Pt (s)

18 Try it! Draw a cell schematic to show the reaction: Cu  Cu 2+ + 2e - I 2 + 2e -  2I - (platinum electrode) Cu + I 2  Cu 2+ + 2I - Label the anode, cathode, and direction of electron flow Calculate the voltage

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20 PtCu Cu + I 2  Cu 2+ + 2I - Electrons! Try It! Answer Salt bridge Cu 2+ I2I2 Anode- oxidation Cathode- reduction Voltage = -0.34 + 0.54 = 0.20V

21 Videos Galvanic cells (2 min)Galvanic cells Animation (24 sec)Animation Li Batteries (5.5 min)Li Batteries Potentials (2 min)Potentials Song (3.5 min, This guy is seriously prolific)Song

22 QUESTION FOUR: THE ELECTROCHEMICAL CELL An electrochemical cell is set up as shown below E°(Ag + /Ag) = +0.80 VE°(Cu 2+ /Cu) = +0.34 V (a)What would be the reading on the voltmeter in the cell? (b)Give the standard cell diagram that represents this cell. 2004 NZQA Exam Q4 E o (red)- E o (ox)= E o (Ag)- E o (Cu)= 0.80- 0.34 = 0.46V Cu (s) /Cu 2+ //Ag + /Ag (s)

23 2004 NZQA Exam Q4 The voltmeter is replaced by a piece of wire that allows the flow of electrons. (c)Discuss the flow of charge in this cell, both in terms of direction of movement and the species involved. Include the role of the salt bridge in your answer. (You may draw on the above diagram if it helps to clarify the description.) Electrons flow through the circuit from Cu to Ag electrode Cations from the salt bridge move toward the Ag half-cell Anions from the salt bridge move toward the Cu half cell The Cu electrode turns into Cu 2+ ions The Ag + ions plate in the electrode as Ag

24 2004 NZQA Exam Q4 (d)(i)Write a balanced equation for the reaction occurring in the electrochemical cell shown at the start of this Question 4. (ii)Describe what would be observed in each half-cell as the cell discharges and explain these observations in terms of the chemical reactions occurring. Ag + + e  Ag Cu  Cu 2+ + 2e 2Ag + + Cu  Cu 2+ + 2Ag In the Cu half cell, the blue colour would darken and the electrode would dissolve. In the Ag cell, the electrode would increase mass and have a dark deposit on the outside.

25 2004 NZQA Exam Q4 (e)The cell is allowed to discharge for a period of time during which the mass of the copper electrode changes by 3.20 g. Calculate the mass change that would be expected on the silver electrode. (Assume that the change in mass is only due to the oxidation-reduction reaction occurring as the cell discharges.) 2Ag + + Cu  Cu 2+ + 2Ag nCu = m/M = 3.2g/63.5g mol -1 = 0.05mol nAg = 2(nCu) = 0.10mol mAg = nM = 0.10mol(108gmol -1 ) = 10.8g Ag created The electrode mass would increase by 10.8g


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