Electrochemistry IV The galvanic cell continued
Consider the galvanic cell based On the unbalanced redox reaction: FFe 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
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 ˚= V When you reverse the equation, you reverse the sign of te cell potential
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
Sum up the two half reactions Cu (s) Cu 2+ (aq) + 2 e - E ˚= V 2Fe 3+ (aq) +2e - 2Fe 2+ (aq) E ˚= 0.77V ________________________________ 2Fe 3+ (aq)+Cu(s) 2Fe 2+ (aq)+Cu 2+ (aq) E ˚= 0.77V / /
Class Examples: Consider a galvanic cell based on the unbalanced equation: Al 3+ (aq) + Mg(s) Al(s) + Mg 2+ (aq)
Solution: start with half reactions Al 3+ (aq)+ 3e - Al(s) E ˚= V Mg 2+ (aq) + 2e - Mg(s) E ˚= V 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.
Solution continued 3( Mg(s) Mg 2+ (aq) + 2e - ) E ˚= 2.37V 2( Al 3+ (aq)+ 3e - Al(s)) E ˚= V ________________________________ 2Al 3+ (aq)+ 3Mg(s) 3Mg 2+ (aq) +2Al(s) E =0.71V / /
Second example: Set up a galvanic cell using the following two half reactions: MnO e - +8H + Mn 2+ +4H 2 O E ˚=1.51V ClO H + +2e - ClO 3 - +H 2 O E ˚=1.19V Find the balanced cell reaction and the cell potential
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 H + +2e - ClO 3 - +H 2 O E ˚=1.19V ClO 3 - +H 2 O ClO H + +2e - E ˚= V
Balance the electrons: 2(MnO e - +8H + Mn 2+ +4H 2 O) E ˚=1.51V 5(ClO 3 - +H 2 O ClO H + +2e - ) E ˚=-1.19V _________________________________ Again, note the cell potential does not change with the multiplication of the reaction. It is an INTENSIVE PROPERTY. 2MnO H + +5ClO 3 - 5ClO Mn 2+ +3H 2 O E ˚=0.32V // // / /
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.
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
What about the last example in slide #11? 2MnO H + +5ClO 3 - 5ClO Mn 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)
A Complete description of a Galvanic Cell, from ½ reactions. Fe e - Fe E ˚= -0.44V MnO e - Mn 2+ +4H 2 O E ˚= 1.51V Describe the cell, complete the reaction and find the cell potential.
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
Line notation Fe (s) | Fe 2+ (aq) ||H + (aq),MnO 4 - (aq),Mn 2+ (aq) | Pt (s)
Diagram
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.