This lab will require the student to create solutions of iron and aluminium by using electrolysis. Students will then observe the voltage created in a number of galvanic cells and create a reactivity order. Video: A galvanic cell is a spontaneous production of a redox reaction. This is an electrochemical cell that will be demonstrated in the lab. A galvanic cell has an anode (oxidation reaction) and a cathode (reduction reaction). The electrons flow through a salt bridge which does not react with other ions in solution. The salt bridge could be a paper towel or filter paper soaked in an ionic solution or it can be a porous disk between the anode and cathode chambers. Each half reaction has a standard reduction potential that can be looked up on a table. The standard reduction potentials can be used to calculate the cell potential, in Volts, of a galvanic cell. Eocell = Eocathode – Eoanode For a cell that is made up of Zinc and Cobalt, the half reactions and their standard reduction potentials are: Zn2+(aq) + 2e- → Zn(s) Eo = – 0.76V Co2+(aq) + 2e- → Co(s) Eo = – 0.28V Since cobalt has a more positive reduction potential, it will be the reduction reaction at the cathode. Zinc will be the oxidation reaction at the anode. The Eocell can then be calculated as: Eocell = Eocathode – Eoanode Eocell = EoCo – EoZn Eocell = (-0.28) – (-0.76) Eocell = +0.48V Video: Calculating redox potentials and determining spontaneity in non-standard conditions requires the use of the Nernst equation. (at 25 oC the following equation applies) E = Eo – (0.0592V/n) logQ Recall that Q is determined by the concentration of the produc
ts over the concentration of reactants. For this course, you should have only one species for each of the products and the reactants. For example, What is the potential of a cell made up of Zn/Zn2+ and Cu/Cu2+ half-cells at 25oC if [Zn2+] = 2.5M and the [Cu2+]= 0.25M? First step is to determine the reaction equation by looking at the standard reduction potentials to determine which species is reducing and which is oxidation. Zn2+ + 2e- → Zn Eo = -0.76 Cu2+ + 2e- → Cu Eo = +0.34 Copper is the cathode because it is more positive. Now calculate the Eo. Eocell = Eocathode – Eoanode Eocell = EoCu – EoZn Eocell = (0.34) – (-0.76) Eocell = 1.10V There are 2 electrons being transferred, n=2. Complete the Nernst Equation. E = Eo – (0.0592V/n) log([Zn2+]/[ Cu2+]) E = 1.10 – (0.0592V/2) log(2.5M/0.25M) E = 1.10 – 0.0296V E = 1.07 V A concentration cell is a galvanic cell that is composed of the same species, but in different concentrations. For example, a cell composed of 2 zinc electrodes with one cell holding 0.10M zinc sulfate and the other holding 1.0M zinc sulfate can be calculated using a form of the Nernst equation as well. E = Eo – (0.0592V/n) log([Zn2+]dilute/[ Zn2+]concentrated) E = 0V – (0.0592V/2) log(0.1/1.0) E = -0.030V will be used to create solutions of iron and aluminum. Electrolysis is the use of electricity to force a reaction that otherwise is nonspontaneous. will then be set up and studied. A of CuSO4 will also be studied. When making these solutions, be sure to record your observations for the Discussion section. Fe2+ solution Al3+ solution CuSO4 solution Fe electrode Al electrode Cu electrode