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Electrochemistry IV

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

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Electrochemistry IV

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  1. Electrochemistry IV The galvanic cell continued

  2. Consider the galvanic cell based On the unbalanced redox reaction: • Fe3+(aq) + Cu(s)Cu2+(aq) + Fe 2+(aq) Find the half reactions as reductions to find their cell potential. • Fe3+(aq) +1e- Fe2+(aq) E ˚= 0.77V • Cu2+(aq) + 2e- Cu (s) E ˚= 0.34V

  3. To balance the cell reaction and calculate E ˚cell • Equation B must be reversed • Cu2+(aq) + 2e- Cu (s) E ˚= 0.34V • Cu (s)  Cu2+(aq) + 2e-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 • Fe3+(aq) +1e- Fe2+(aq) E ˚= 0.77V • 2Fe3+(aq) +2e- 2Fe2+(aq)E ˚= 0.77V Note that the cell potential is NOT multiplied -----------------------------------------

  5. Sum up the two half reactions • Cu (s)  Cu2+(aq) + 2e-E ˚= -0.34V • 2Fe3+(aq) +2e- 2Fe2+(aq)E ˚= 0.77V • ________________________________ • 2Fe3+(aq)+Cu(s) 2Fe2+(aq)+Cu2+(aq)E ˚= 0.77V / /

  6. Class Examples: • Consider a galvanic cell based on the unbalanced equation: • Al3+(aq) + Mg(s) Al(s) + Mg2+(aq)

  7. Solution: start with half reactions • Al3+(aq)+ 3e- Al(s) E ˚= -1.66V • Mg2+(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) Mg2+(aq) + 2e-) E ˚= 2.37V • 2(Al3+(aq)+ 3e- Al(s)) E ˚= -1.66V • ________________________________ 2Al3+(aq)+ 3Mg(s) 3Mg2+(aq) +2Al(s) E=0.71V /

  9. Second example: • Set up a galvanic cell using the following two half reactions: • MnO4-+5e-+8H+Mn2++4H2O E˚=1.51V • ClO4-+2H+ +2e-ClO3-+H2O 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. • ClO4-+2H+ +2e-ClO3-+H2O E˚=1.19V • ClO3-+H2O ClO4-+2H+ +2e-E˚=-1.19V --------------------------------------------------

  11. Balance the electrons: / / / • 2(MnO4-+5e-+8H+Mn2++4H2O) E˚=1.51V • 5(ClO3-+H2O ClO4-+2H+ +2e-) E˚=-1.19V • _________________________________ • Again, note the cell potential does not change with the multiplication of the reaction. It is an INTENSIVE PROPERTY. • 2MnO4-+6H++5ClO3-5ClO4- +2Mn2++3H2O • 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 • Al3+(aq) + Mg(s) Al(s) + Mg2+(aq) • Mg(s)|Mg2+(aq)|| Al3+(aq)|Al(s) • In this notation the single line is an indication of the phase difference between the electrode Mg(s) and the oxidized Mg2+ (aq). • On the cathode side, the line between Al3+(aq) and Al(s) shows before and after reduction in the cathode compartment. Electrode Electrode

  14. What about the last example in slide #11? • 2MnO4-+6H++5ClO3-5ClO4- +2Mn2++3H2O • 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)|ClO3-(aq),ClO4-(aq),H+(aq)||H+(aq),MnO4- (aq),Mn2+(aq)|Pt(s)

  15. A Complete description of a Galvanic Cell, from ½ reactions. • Fe2+ + 2e- Fe E˚= -0.44V • MnO4-+5e-Mn2++4H2O 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)Fe2+(aq) + 2e-)E˚=-0.44V • 2(MnO4-(aq)+5e-+8H +(aq)Mn2+(aq)+4H2O (l) )E˚= 1.51V • 5Fe(s)+2MnO4-(aq)+ 16H +(aq) 5Fe2+(aq)+Mn2+(aq)+4H2O (l) • E˚=1.95V

  17. Line notation • Fe(s)|Fe2+(aq) ||H +(aq),MnO4-(aq) ,Mn2+(aq)|Pt(s)

  18. Diagram

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

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