Wednesday, May 25

The majority of our day was put into page 12 as well as our only lab for this unit. Of course we asked Mr. H about six flags first and apparently iron wolf is not every mans bestfriend.... hmmm... ANYWAY before flipping to page 12 we spoke about 11 and Mr. H explained that the difference is that there is no picture. So basically for the problems you should first separate the two material into half equations and then using the chart find which one is lower on the chart and is more easily oxidized. That equation will be your cathode. Generally our cathode equations did not have to be flipped. The other half equation is your anode and is usually flipped and the value of the oxidation potential is switched from negative to positive or vice versa. You find the voltage by adding the two potentials together and then you write the standard line notation with the anode farthest to the write and the ions in the middle. The answers are as follows:

a. Cathode half-equation: [Zn]2+ + 2[e]- > Zn

Anode half-equation: Al > [Al]3+ + 3[e]-

Volts: .90

Standard line notation: Al|[Al]3+||[Zn]2+|Zn

b. Cathode half-equation: [Fe]2+ + 2[e]- > Fe

Anode half-equation: Mg > [Mg]2+ + 2[e]-

Volts: 1.93

Standard line notation: Mg|[Mg]2+||[Fe]2+|Fe

c. Cathode half-equation: [Au]3+ + 3[e]- > Au

Anode half-equation: Al > [Ag]+ + [e]-

Volts: .70

Standard line notation: Ag|[Ag]+||[Au]3+|Au

d. Cathode half-equation: [Au]3+ + 3[e]- > Au

Anode half-equation: Mg > [Mg]2+ + 2[e]-

Volts: 3.87

Standard line notation: Mg|[Mg]2+||[Au]3+|Au

e. Cathode half-equation: [Cd]2+ + 2[e]- > Cd

Anode half-equation: Li > [Li]+ + [e]-

Volts: 2.65

Standard line notation: Li|[Li]+||[Cd]2+|Cd

Then we did our lab using juicy fruits to make a battery. You should have ended up seeing Mg most frequently in your negative column and it should have had the highest voltages. The order is Mg, Zn, Al, Pb, Fe, Cu. There is a WebAssign due Friday and a quest tomorrow.