Unit 5: Everything You Wanted to Know About Electrochemical Cells, But Were Too Afraid to Ask

1. 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

2. Redox Reactions • A Redox Reaction features the transfer of electrons between ions. • X + Y  Xn+ + Yn- • Oxidation Half-Reaction: X  Xn+ + e- • Reduction Half-Reaction: Y + e-  Yn-

3. 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.

4. Redox Reactions • H2SO4 • 1 H = +1 • 1 O = -2 • -8 + 2 + S = 0 • 1 S = +6 • Cr2O72- • 1 O = -2 • -14 + 2(Cr) = -2 • 1 Cr = +6

5. Balancing Redox Equations • Balance: Cr2O72- + Cl- Cr3+ + Cl2 • Split Equation into half-reactions • Cr2O72-  2Cr3+ • Cl-  Cl2 • Add H+, then H2O, then e- to balance. • 6e- + 14H+ + Cr2O7 2Cr3+ + 7H2O • 2Cl-  Cl2 + 2e- • Combine into overall reaction • 6Cl- + 14H+ + Cr2O7 2Cr3+ + 7H2O + 3Cl2

6. Balancing Redox Equations • To balance in a BASIC solution: • Take final answer for Acidic Solution: • 6Cl- + 14H+ + Cr2O72-  2Cr3+ + 7H2O + 3Cl2 • Add OH- to cancel H+ and add H2O • 6Cl- + 7H2O + Cr2O72-  2Cr3+ + 3Cl2 +14OH-

7. What are Electrochemical (EC) Cells? • An Electrochemical Cell converts chemical energy into electrical energy by reducing one substance and oxidizing another. • For example: • Cu+F2Cu2++2F- • The copper is oxidized and the fluorine is reduced because of a transfer of electrons, thus creating a current.

8. 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

9. 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

10. WHY?

11. 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

12. 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 CuSO4 and ZnSO4.

13. How a Galvanic Cell is made • In this reaction, Zn(s) would be oxidized to Zn2+(aq) and Cu2+(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

14. 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 Zn2+, and Cu2+ ions will be reduced to Cu and leave the solution, plating the Cu cathode

15. 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 Cl2(g)

16. How Much Electricity?

17. 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.

18. 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 Eo. • Then, subtract the standard reduction potential for the oxidized species from the standard reduction potential of the reduced species to get: • Eocell=Eored-Eoox

19. Cell Potentials • Let’s go back to the zinc-copper cell • Make two half-reactions: Zn(s) Zn2+(aq) +2e- • And Cu2+(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

20. Cell Potentials • So, Eocell=EoCu2+-EoZn • Or Eocell= .34 V- (-.76)V • Therefore Eocell=1.10 V • Remember that oxidation and reduction potentials change, and most are only listed for 1M concentrations of electrolytes at 25oC and 1 atm of pressure. Changes to this will result in changes in potentials.

21. Cell Potentials • Determine the spontaneous cell reaction and the cell potential of a cell that has these two half reactions • Al3++3e-Al(s) EoAl3+=-1.66V • Cu2++2e-Cu(s) EoCu2+=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

22. Cell Potentials • Remember to reverse the equation of the oxidized species and balance the total ionic equations so that no electrons are left over. • 3Cu2+ +2Al2Al3++3Cu • Now, find the Eocell • Eocell=Eored-Eoox • Eocell=EoCu2+-EoAl3+

23. Cell Potentials • Eocell=.34-(-1.66) • Eocell= 2.00 V • Note that the reduction potentials are not multiplied by the coefficients in the equation.

24. 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 9.65x104 C, which is referred to as Faraday’s constant and symbolized F

25. 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 CuSO4 for 10.0 min?

26. Faraday’s Constant • First, convert the minutes to seconds to coulombs • 10.0 min*60.0sec*min-1*4.00A=2.40x103 C • Then coulombs to moles of electrons • 2.40x103 C*1mol e-/9.65x104 C=.0249 mol e- • To grams of copper. Remember that it takes 2 mol e- to reduce 1 mol Cu2+ • .0249 mol e-*63.55 g Cu/2 mol e-=.791 g Cu

27. 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=-nFEcell • Where n is the moles of electrons transferred and F is Faraday’s constant

28. Equilibrium Constants and Cell Potentials • To find the equilibrium constant of an equation from its Eocell, the equation is: • Eocell=RTln(Kc)/nF • Where R=8.314 J/molK, T is temperature in Kelvin, ln(Kc) 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.

29. The Nernst Equation • The Nernst Equation relates the calculated potential of a cell to its potential at a certain time. • Ecell=Eocell-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.

30. The End