Lecture 27 Chapter 19, Sections 3-4 Galvanic Cells Electrochemical Potential Galvanic Cells Defined Standard Hydrogen Electrode Standard Reduction Potentials
Redox Balancing One More Example This time we ll do it a little differently than text When balancing hydrogen, we ll add H + (regardless of whether conditions are acidic or basic) This is simpler, because it doesn t muck up the O balance Then when we are all done If acidic add H 2 O to both sides to turn H + into H 3 O + If basic add OH to both sides to turn H + into H 2 O OK, then here is the example Se + Cr(OH) 3 Cr + SeO 2 3 (basic conditions) Solution: 3Se + 4Cr(OH) 3 + 6OH 4Cr + 3SeO 3 2 + 9H 2 O
Redox Shmedox, who cares? All of the reactions we have described so far have involved direct transfer of electrons from reductant to oxidant In many cases we can separate the two species and connect them with a wire Electrons must travel through wire to get from site of oxidation reaction to site of reduction reaction This is a Galvanic Cell Batteries are most common galvanic cells
The Essential Components of a Galvanic Cell 1. Spontaneous redox rxn 2. Barrier to direct electron transfer 3. Contact b/n chemical and electrical parts of cell electrodes 4. External electric circuit 5. Means to complete circuit salt bridge 6. Ions present in solution ANODE CATHODE Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s)
Electrodes Oxidation occurs at anode Reduction occurs at cathode Previous scene depicted active electrodes Same materials participating in redox reaction Often use passive electrodes (often platinum or gold) Simply pick up and drop off electrons
How much electricity can we get? What determines how useful our galvanic cell can be? Thermodynamics of course! If the G for our reaction is large and negative Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) Then there is lots of energy available to make the electrons do work The electrons have lots of potential energy There is a large potential difference between the anode and the cathode
Electrons flow from high potential to low Cathode (+) Anode (-)
Standard Potential E We use electrochemical potential (voltage) to quantify the potential energy difference. Just as G, H, and S are tabulated, we can tabulate E If we have a table, we must have standard conditions 1 M 1 bar (for gaseous things) 298 K
E for copper/zinc under standard conditions
Relative Potential Electrons want to leave zinc more badly than they want to leave copper So, zinc is oxidized and copper reduced when they are put together. This difference is our cell potential, E = 1.10 V But if we put copper together with silver, copper would be oxidized and silver reduced. All that matters is the potential difference b/n the two half reactions
Standard Hydrogen Electrode The standard commonly used is called the standard hydrogen electrode. (SHE) This is defined to have a reduction potential of zero. Any half-reaction can now be compared to this standard.
Standard Hydrogen Electrode When H 3 O + is reduced: 2 H 3 O + (aq, 1 M) + 2 e - H 2 (g, 1 bar) + 2 H 2 O(l) E = 0.00 V And when H 3 O + is oxidized: H 2 (g, 1 bar) + 2 H 2 O(l) 2 H 3 O + (aq, 1 M) + 2 e - E = 0.00 V
Representative Standard Reduction Potentials See also Appendix F Note that the thing that is more negative will tend to be oxidized, while the more positive will be reduced.
Potentials and Redox The half-reaction with the more negative reduction potential occurs at the anode as oxidation. The half-reaction with the more positive reduction potential occurs at the cathode as reduction The book suggests: E cell = E E cathode anode Zn 2+ + 2e Zn 0.7618 V (oxidation) Cu 2+ + 2e Cu 0.3419 V (reduction) E cell = E red E ox = 0.3419 ( 0.7618) = 1.1037 V
Potentials and Redox I suggest remembering that the cell potential will tend to be positive. So, the reaction with the more negative reduction potential will be the site of oxidation This reaction will run the other way Standard Potentials: Zn 2+ + 2e Zn 0.7618 V (oxidation) Cu 2+ + 2e Cu 0.3419 V (reduction) What actually will happen: Zn Zn 2+ + 2e 0.7618 V (oxidation) Cu 2+ + 2e Cu 0.3419 V (reduction) E cell = sum of E ½ = 0.3419 + 0.7618 = 1.1037 V Just like overall rxn is sum of ½ reactions: Zn + Cu 2+ Zn 2+ + Cu
Example A galvanic cell can be constructed from a zinc electrode immersed in a solution of zinc sulfate and an iron electrode immersed in a solution of iron (II) sulfate. What is the standard potential of this cell, and what is its spontaneous direction under standard conditions? Write down the two reduction reactions Zn 2+ + 2e Zn Fe 2+ + 2e Fe Look up standard reduction potentials Zn 2+ + 2e Zn 0.7618 V Fe 2+ + 2e Fe 0.447 V
So, if the standard reduction potentials are: Zn 2+ + 2e Zn 0.7618 V Fe 2+ + 2e Fe 0.447 V Which species will get oxidized at the anode? 25% 1. Zinc 25% 2. Iron 25% 25% 3. Neither 4. Both 1 2 3 4 5
What will be the net cell potential? Zn 2+ + 2e Zn 0.7618 V Fe 2+ + 2e Fe 0.447 V 25% 1. 1.209 V 25% 2. 0.315 V 25% 3. 0.315 V 25% 4. 1.209 V 1 2 3 4 5
Another Example What about a cell made from the following two reactions? Ag + + e Ag 0.7996 V Pb 2+ + 2 e Pb 0.1262 V Lead is more negative, so it will be reversed to oxidation Silver reaction will happen twice for each lead reaction BUT, each electron added to silver sees a potential of 0.7996 V Twice as many e move, but each sees the same potential. 2Ag + + 2e 2Ag Pb Pb 2+ + 2 e 0.7996 V 0.1262 V E cell = 0.7996 + 0.1262 = 0.9258 V
Another Example The following half-reactions occur in the rechargeable nickel-cadmium battery: Cd(OH) 2 + 2 e Cd + 2 OH NiO(OH) + H 2 O + e Ni(OH) 2 + OH This battery has a potential of 1.35 V under standard conditions, with nickel as the cathode. Determine the net reaction, and use the tabulated standard reduction potential for the cadmium half-reaction to find Eº for the nickel half-reaction. Look up book value for Cd rxn: 0.860 V Ni is the cathode, so Ni reaction must occur as reduction as written. Cd must occur as oxidation reversed. Cd + 2 OH Cd(OH) 2 + 2 e Eº = 0.860 V E cell = E Cd + E Ni 1.35 V = 0.860 V + E Ni E Ni = 0.49 V Again, note that the factor of two needed to balance the Ni reaction does NOT affect the cell potential.
Today Go to Seminar 3:00 write report! Next week s seminar will be Thursday 4:00 Monday Finish CAPA #16 Keep reviewing