Mr. Grodski’s voltaic cell design : 

I chose Al as my anode because it was most reactive reducing agent (most negative reduction potential = best at oxidizing)

                                                Cu⁺² +  2e-  →   Cu               +.34v

                                                Sn⁺²  +  2e-  →   Sn             -0.15 v

                                                Fe⁺²  +  2e-  →    Fe             -0.44 v

                                                Zn⁺²  +  2e-  →   Zn             -0.76 v

                                                Al⁺³  +  3e-   →    Al           -1.66 v                                                          

I chose  Cu as my cathode because it will not react with the water in the solutions due the Net potential with water being overall negative.  The other metals listed will oxidize in the presence of water!.  If the cathode oxidizes and produces electrons those electrons will lower the potential difference between the two half cells.  Electrons flow because of a place of high electrons to a place with lowered amount of electrons.  2 half cell oxidizing will cause a significant drop in voltage.

                                                Cu⁺² +  2e-  →   Cu                      +.34v

                                             2 H₂O  +  2e-  →  H₂(g)  +  2OH^–     -0.83 v

So the anode will react spontaneously with water in the solutions based on the above reduction potential of water.

This means that Al and water will react spontaneously (just like the magnesium in water in lab 1) so the trade off is that I could get more voltage out of the my voltaic cell with Al instead of Zn (second best oxidizing metal) because the although Zn would react less with water its oxidation potential =.76 is more than half that of Al = 1.66.  So the limiting factor I thought was the metal ability to oxidize and its ability to act as the strongest reducing agent. Even though a more reactive metal like Al will mean more reactivity with water compromising the voltage from the Al, I still think Al is  the best choice because its oxidation potential is so much higher than Zn.  With Cu⁺² Al will achieve a theoretical voltage of 2.00 v (.36 + 1.66) while Zn and Cu⁺² will achieve a theoretical voltage of 1.10 v (.36 + .76).

Now Cu was a great choice of a cathode because it was the least reactive metal, which would prevent it from being oxidized from ions of the other solutions if you mixed and matched metals (electrodes) with different solutions. When students mixed and matched they wound up with the metals being oxidized in the presence of certain solution based on the standard reduction potentials. If a complete redox reaction occurs at the anode or cathode and not being stretched between the wire then there will be very low voltage. If a cathode is oxidizing, electrons will not flow to it!

Now another problem is the leaking or the diffusion of the Cu⁺² into the oxidation half cell due to entropy and the dispersion of energy as Cu+2 goes from a concentrated source (reduction half cell) to a dilute source (the oxidation half cell via the salt bridge).  Once Cu+2 enters oxidation half cell it will react directly with Al and the voltage drops quickly as the Al has no reason to push electrons across the wire if it can do so in the same half cell. One way to combat this is to put a solution that has ions that will precipitate with Cu⁺² in the salt bridge preventing the leaking as the Cu⁺²reacts with the OH- to become Cu(OH)2.  A precipitate will build up in the salt bridge. A potassium hydroxide (KOH) solution that I used prevented Cu⁺² from reaching the oxidation cell.

                                                     Cu⁺² (aq)  +  2 OH^–   (aq)  →  Cu(OH)₂ (s)

One other adaptation is to not put Al+3 in the oxidation cell because that would lower the ions in the product and by Le Chateliers principle the reaction should become more spontaneous in the forward increasing the voltage. 

                                                       2Al    +    3Cu⁺²  →  2Al⁺³  +   3Cu                                                                                                                                                          ↓           

                                                            Shift ———————————> ↑ volts

So I put water but that would lower the concentration of OH- as it would diffuse out in the salt bridge so I put KOH. This was helpful because of the passivity of Al that creates oxides or aluminum hydroxide on the surface of the Al lowering the Al to further oxidize.  *In an acidic or basic environment Aluminum hydroxide (OH-) will dissolve and thus KOH will keep the oxide layer from forming. *I know this is something that you did not know but something that you could of investigated.

Only issue is that the OH- will move into the reduction half-cell by diffusion, precipitating Cu+2 and thus lowering the concentration of the reactant and thus lowering the spontaneity of the forward direction and Voltage by Le Chateliers Principle. This was a necessary evil and a compromise I had to make given the materials I had but that would take some time and the task was to make a battery with the highest sustained voltage for 30 seconds.

                                                    Cu⁺² (aq)  +  2 OH^–   (aq)  →  Cu(OH)₂ (s)

                                                         2Al    +    3Cu⁺²  →  2Al⁺³  +   3Cu

                                                         ↓

                                                         ↓ volts   <—————————————— Shift

What I could try to do differently is use FeCl₃ or Fe⁺³ ions as the oxidizing agent instead of Cu⁺² because it has the greater reduction potential than Cu⁺², (.77 to .34).

So I would get a (.77 + 1.66) 2.43 volts theoretically. Now this is not clear cut but generally the ion complexes that Fe⁺³make with water produce H+ ions are very acidic and react with Fe making the cathode who is Fe start reacting and making Hydrogen which means that it is oxidizing!!!!! How many of the students had their anodes and cathodes different from what they expected?  Also these complexes create very insoluble compounds quickly and that is why it is cloudy!! So the free Fe⁺³ does NOT really exist with Iron (III) chloride.

                                                      Fe+3  + 1e-  →  Fe               +.77 v

                                                      Cu⁺² +  2e-  →   Cu             +.34 v

                                                      2 H+  +  2e-  →  H₂(g)         0.00 v

                                                     Sn⁺²  +  2e-  →   Sn             -0.15 v

                                                     Fe⁺²  +  2e-  →    Fe             -0.44 v

                                                     Zn⁺²  +  2e-  →   Zn             -0.76 v

                                                     Al⁺³  +  3e-   →    Al             -1.66 v 

Electrolytic cell Assignment:

2: Electrolytic Cell intro lecture : 

2: Electrolytic Cell Lecture : 

Electroplating Demo : For Today’s Notes 

Electroplating Demo : For Today’s Notes

Stoichiometry 8 redox worksheet review: Review of Calculation of amount of mass of metal deposited at the cathode.

Today’s lesson : Actual Faraday’s findings into electrolytic calculations.

1: Lecture on Electrolytic cell – Molarity calculations worksheet : 

Electrolysis of water review video : 

Electrolytic Problems Expanded Review video: