2. Unit 5: Everything You Wanted to Know About Electrochemical Cells, But Were Too Afraid to Ask By : Michael “Chuy el Chulo” Bilow And “H”Elliot Pinkus

3. Redox Reactions A Redox Reaction features the transfer of electrons between ions. X + Y  X n+ + Y n- Oxidation Half-Reaction: X  X n+ + e - Reduction Half-Reaction: Y + e -  Y n-

4. Redox Reactions Determine Oxidation Numbers Atoms in a pure element have oxidation number of zero A monatomic ion has oxidation number equal to its charge. Sum of oxidation numbers equals overall charge of compound. Fluorine is always –1 with other elements H is +1 and O is –2 in most compounds Cl, Br, and I are –1 except with Oxygen or Fluorine.

5. Redox Reactions H 2 SO 4 –1 H = +1 –1 O = -2 –-8 + 2 + S = 0 –1 S = +6 Cr 2 O 7 2- –1 O = -2 –-14 + 2(Cr) = -2 –1 Cr = +6

6. Balancing Redox Equations Balance: Cr 2 O 7 2- + Cl -  Cr 3+ + Cl 2 Split Equation into half-reactions –Cr 2 O 7 2-  2Cr 3+ –Cl -  Cl 2 Add H +, then H 2 O, then e - to balance. –6e - + 14H + + Cr 2 O 7  2Cr 3+ + 7H 2 O –2Cl -  Cl 2 + 2e - Combine into overall reaction –6Cl - + 14H + + Cr 2 O 7  2Cr 3+ + 7H 2 O + 3Cl 2

7. Balancing Redox Equations To balance in a BASIC solution: Take final answer for Acidic Solution: –6Cl - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O + 3Cl 2 Add OH - to cancel H + and add H 2 O –6Cl - + 7H 2 O + Cr 2 O 7 2-  2Cr 3+ + 3Cl 2 +14OH -

8. What are Electrochemical (EC) Cells? An Electrochemical Cell converts chemical energy into electrical energy by reducing one substance and oxidizing another. For example: Cu+F 2  Cu 2+ +2F - The copper is oxidized and the fluorine is reduced because of a transfer of electrons, thus creating a current.

9. What are EC Cells? There are two types of EC cells: Galvanic cells spontaneously produce energy Electrolytic cells must have work done on them to go to completion, and are thus nonspontaneous

10. Electrolytic and Galvanic Cells In both electrolytic and galvanic cells, oxidation takes place at the anode and reduction takes place at the cathode But, galvanic cells have positively charged cathodes and negatively charged anodes And electrolytic cells have negative cathodes and positive anodes

11. WHY?

12. Because reduction is forced in electrolytic cells, electrons collect there, giving a negative charge. And because the oxidation is not favored, the anode develops a positive charge

13. How do You Make a Galvanic Cell? Many EC cells are made with two metals in a solution of one of their sulfate or nitrate The two metal bars are connected by a salt bridge. The salt bridge allows anions to pass through to the oxidized side to restore charge For example, take zinc and copper in solutions of CuSO 4 and ZnSO 4.

14. How a Galvanic Cell is made In this reaction, Zn (s) would be oxidized to Zn 2+ (aq) and Cu 2+ (aq) would be reduced to Cu (s) The zinc-copper galvanic cell would look like this: http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/electrochem.html

15. How do Galvanic Cells Produce Electricity? The electron flow from cathode to anode produces a current, and thus electricity. Over time, the Zn anode will deteriorate as it is oxidized to Zn 2+, and Cu 2+ ions will be reduced to Cu and leave the solution, plating the Cu cathode

16. How are Electrolytic Cells Made? There are many ways to make electrolytic cells, but all require an outside source of energy to force the reaction towards the products This shows the electrolysis of NaCl (l) to Na (l) and Cl 2(g)

17. How Much Electricity?

18. Cell Potentials To find how much electricity is produced or needed, you must use the oxidation and reduction potentials of each of the half-reactions that take place in the system. Reduction Potentials show how much energy is either released or needed to cause a reduction half-reaction to occur Since oxidation is the opposite of reduction, reduction potentials are the opposite of oxidation potentials.

19. Cell Potentials To find a cell’s potential difference (voltage), first find its standard oxidation and reduction potentials of its half- reactions, usually listed as E o. Then, subtract the standard reduction potential for the oxidized species from the standard reduction potential of the reduced species to get: E o cell =E o red -E o ox

20. Cell Potentials Let’s go back to the zinc-copper cell Make two half-reactions: Zn (s)  Zn 2+ (aq) +2e - And Cu 2+ (aq) +2e -  Cu (s) The reduction potential for the copper (II) ion to copper metal is +0.34 V The reduction potential for the zinc (II) ion to zinc metal is -0.76 V

21. Cell Potentials So, E o cell =E o Cu 2+ -E o Zn Or E o cell =.34 V- (-.76)V Therefore E o cell =1.10 V Remember that oxidation and reduction potentials change, and most are only listed for 1M concentrations of electrolytes at 25 o C and 1 atm of pressure. Changes to this will result in changes in potentials.

22. Cell Potentials Determine the spontaneous cell reaction and the cell potential of a cell that has these two half reactions Al 3+ +3e -  Al(s) E o Al 3+ =-1.66V Cu 2+ +2e -  Cu(s) E o Cu 2+ =0.34V First determine which species is to be oxidized and which to be reduced The oxidized substance in a spontaneous cell will always have the lesser potential

23. Cell Potentials Remember to reverse the equation of the oxidized species and balance the total ionic equations so that no electrons are left over. 3Cu 2+ +2Al  2Al 3+ +3Cu Now, find the E o cell E o cell =E o red -E o ox E o cell =E o Cu 2+ -E o Al 3+

24. Cell Potentials E o cell =.34-(-1.66) E o cell = 2.00 V Note that the reduction potentials are not multiplied by the coefficients in the equation.

25. What else? The SI unit of electric current is the ampere (A) and the SI unit of charge is the coulomb (C). 1A= 1 coulomb per second It has been determined that the charge of one mole of electrons is 9.65x10 4 C, which is referred to as Faraday’s constant and symbolized F

26. Faraday’s Constant From this, we can determine how much anode material is used up or how much is produced at the cathode For example, how many grams of copper will be deposited on the cathode of an electrolytic cell if a current of 4.00 A is run through a solution of CuSO 4 for 10.0 min?

27. Faraday’s Constant First, convert the minutes to seconds to coulombs 10.0 min*60.0sec*min -1 *4.00A=2.40x10 3 C Then coulombs to moles of electrons 2.40x10 3 C*1mol e - /9.65x10 4 C=.0249 mol e - To grams of copper. Remember that it takes 2 mol e - to reduce 1 mol Cu 2+.0249 mol e - *63.55 g Cu/2 mol e - =.791 g Cu

28. Gibbs Free Energy The maximum amount of work that can be done is the opposite of ΔG, the change in Gibbs Free Energy Since 1 Volt= 1 joule/1coulomb, and the joule is the SI unit of work, we get ΔG=-nFE cell Where n is the moles of electrons transferred and F is Faraday’s constant

29. Equilibrium Constants and Cell Potentials To find the equilibrium constant of an equation from its E o cell, the equation is: E o cell =RTln(K c )/nF E o cell =RTln(K c )/nF Where R=8.314 J/molK, T is temperature in Kelvin, ln(K c ) is the natural logarithm (log base e) of the equilibrium constant, n is the number of moles of electrons transferred, and F is faraday’s constant.

30. The Nernst Equation The Nernst Equation relates the calculated potential of a cell to its potential at a certain time. E cell =E o cell -RT/nFln(Q) Where R,T, n, and F are the same as above and Q is the mass-action constant of the equation, which equals the concentrations of products that can change concentration to their coefficient’s power, divided by reactants that act similarly.

31. The End