1. Electrochemistry IV The galvanic cell continued

2. Consider the galvanic cell based On the unbalanced redox reaction: FFe 3+ (aq) + Cu(s)  Cu 2+ (aq) + Fe 2+ (aq) Find the half reactions as reductions to find their cell potential. A.Fe 3+ (aq) +1e -  Fe 2+ (aq) E ˚= 0.77V B.Cu 2+ (aq) + 2 e -  Cu (s) E ˚= 0.34V

3. To balance the cell reaction and calculate E ˚cell  Equation B must be reversed A.Cu 2+ (aq) + 2 e -  Cu (s) E ˚= 0.34V B.Cu (s)  Cu 2+ (aq) + 2 e - E ˚= - 0.34V When you reverse the equation, you reverse the sign of te cell potential. -----------------------------------------

4. And one must double the A equation to balance the e - ’s A.Fe 3+ (aq) +1e -  Fe 2+ (aq) E ˚= 0.77V  2Fe 3+ (aq) +2e -  2Fe 2+ (aq) E ˚= 0.77V Note that the cell potential is NOT multiplied -----------------------------------------

5. Sum up the two half reactions  Cu (s)  Cu 2+ (aq) + 2 e - E ˚= - 0.34V  2Fe 3+ (aq) +2e -  2Fe 2+ (aq) E ˚= 0.77V  ________________________________  2Fe 3+ (aq)+Cu(s)  2Fe 2+ (aq)+Cu 2+ (aq) E ˚= 0.77V / /

6. Class Examples:  Consider a galvanic cell based on the unbalanced equation:  Al 3+ (aq) + Mg(s)  Al(s) + Mg 2+ (aq)

7. Solution: start with half reactions  Al 3+ (aq)+ 3e -  Al(s) E ˚= - 1.66V  Mg 2+ (aq) + 2e -  Mg(s) E ˚= - 2.37V  Reverse the equation that will give you an overall positive cell potential and balance the electrons going in with the electrons going out.  Find the cell potential.

8. Solution continued  3( Mg(s)  Mg 2+ (aq) + 2e - ) E ˚= 2.37V  2( Al 3+ (aq)+ 3e -  Al(s)) E ˚= - 1.66V  ________________________________ 2Al 3+ (aq)+ 3Mg(s)  3Mg 2+ (aq) +2Al(s) E =0.71V / /

9. Second example:  Set up a galvanic cell using the following two half reactions:  MnO 4 - +5e - +8H +  Mn 2+ +4H 2 O E ˚=1.51V  ClO 4 - +2H + +2e -  ClO 3 - +H 2 O E ˚=1.19V  Find the balanced cell reaction and the cell potential

10. Solution  First: Find which reduction reaction should switch to oxidation. It must result in a positive cell potential.  The second reaction has the smaller absolute value and will be switched.  ClO 4 - +2H + +2e -  ClO 3 - +H 2 O E ˚=1.19V  ClO 3 - +H 2 O  ClO 4 - +2H + +2e - E ˚= - 1.19V --------------------------------------------------

11. Balance the electrons:  2(MnO 4 - +5e - +8H +  Mn 2+ +4H 2 O) E ˚=1.51V  5(ClO 3 - +H 2 O  ClO 4 - +2H + +2e - ) E ˚=-1.19V  _________________________________  Again, note the cell potential does not change with the multiplication of the reaction. It is an INTENSIVE PROPERTY.  2MnO 4 - +6H + +5ClO 3 -  5ClO 4 - +2Mn 2+ +3H 2 O  E ˚=0.32V // // / /

12. Line Notation  Is a useful short hand description of a galvanic cell.  In this notation, anode (oxidation) components are listed from the left, starting with the electrode and cathode (reduction) to the right, ending in the electrode, separated by a double line which represents the salt bridge or porous disk.

13. Line notation continued  Al 3+ (aq) + Mg(s)  Al(s) + Mg 2+ (aq)  Mg(s)|Mg 2+ (aq)|| Al 3+ (aq)|Al(s)  In this notation the single line is an indication of the phase difference between the electrode Mg(s) and the oxidized Mg 2+ (aq).  On the cathode side, the line between Al 3+ (aq) and Al(s) shows before and after reduction in the cathode compartment. Electrode

14. What about the last example in slide #11?  2MnO 4 - +6H + +5ClO 3 -  5ClO 4 - +2Mn 2+ +3H 2 O  All the components are aqueous and there is no solid metal for an electrode.  Platinum, a fairly non reactive metal, is normally used as an elecrode in just such an event. Thus:  Pt (s) |ClO 3 - (aq),ClO 4 - (aq),H + (aq) ||H + (aq),MnO 4 - (aq),Mn 2+ (aq) |Pt (s)

15. A Complete description of a Galvanic Cell, from ½ reactions.  Fe 2+ + 2e -  Fe E ˚= -0.44V  MnO 4 - +5e -  Mn 2+ +4H 2 O E ˚= 1.51V  Describe the cell, complete the reaction and find the cell potential.

16. Which rxn to reverse?  Fe since that will result in a positive cell potential.  How to balance the electrons? Lowest common denominator.  Don’t forget to balance the half rxn!  5(Fe (s)  Fe 2+ (aq) + 2e - ) E ˚=-0.44V  2(MnO 4 - (aq) +5e - +8H + (aq)  Mn 2+ (aq) +4H 2 O (l) ) E ˚= 1.51V  5Fe (s) + 2MnO 4 - (aq) + 16H + (aq)  5Fe 2+ (aq) + Mn 2+ (aq) + 4H 2 O (l)  E ˚=1.95V

17. Line notation  Fe (s) | Fe 2+ (aq) ||H + (aq),MnO 4 - (aq),Mn 2+ (aq) | Pt (s)

18. Diagram

19. Summary: A complete description usually has 4 things. 1.Cell potential is positive (+) and has a balenced cell reaction. 2.Direction of the electron flow which is found from the ½ reactions using the direction that gives a + E ˚ 3.Designation of Anode (oxidation) & Cathode (reduction) 4.Nature of the electrodes (may be Pt) and the ions present in each compartment.