Calculate the cell potentials for the following voltaic cells:

2Al (s) + 3Cu 2+ (aq) → 2Al 3+ (aq) + 3Cu (s)

3Mg (s) + 2Al 3+ (aq) → 3Mg 2+ (aq) + 2Al (s)

Mg (s) + Cu 2+ (aq) → Mg 2+ (aq) + Cu (s)

2Al (s) + 3Cu 2+ (aq) → 2Al 3+ (aq) + 3Cu (s)

Al(s) ==> Al^3+ + 3e Look up Eo oxidation
Cu^2+ + 2e ==> Cu(s) Look up Eo reduction.
For Al you will find it in the standard reduction potential table at something like -3 volts but you need to look it up and use the correct value. I don't remember all of these tables. Then Eo oxidation will be the same number but opposite sign. The cell voltage will be the sum of Eo ox + Eo red = Ecell.
Do the same for the others. Post your work if you get stuck.

To calculate the cell potentials for the given voltaic cells, we need to use the standard reduction potentials of the half-reactions involved.

Step 1: Write down the balanced half-reactions for each cell.

Cell 1:
Cathode: 3Cu 2+ (aq) + 6e- → 3Cu (s)
Anode: 2Al (s) → 2Al 3+ (aq) + 6e-

Cell 2:
Cathode: 2Al 3+ (aq) + 6e- → 2Al (s)
Anode: 3Mg (s) → 3Mg 2+ (aq) + 6e-

Cell 3:
Cathode: Cu 2+ (aq) + 2e- → Cu (s)
Anode: Mg (s) → Mg 2+ (aq) + 2e-

Step 2: Look up the standard reduction potentials for each half-reaction. These values can be found in a standard reduction potential table.

Cell 1:
E°cell = E°cathode - E°anode
E°cell = E°Cu2+/Cu - E°Al3+/Al

Cell 2:
E°cell = E°cathode - E°anode
E°cell = E°Al3+/Al - E°Mg2+/Mg

Cell 3:
E°cell = E°cathode - E°anode
E°cell = E°Cu2+/Cu - E°Mg2+/Mg

Step 3: Substitute the standard reduction potential values into the respective equations and calculate the cell potentials.

For Cell 1:
E°cell = (+0.34 V) - (-1.66 V) = +2.00 V

For Cell 2:
E°cell = (-1.66 V) - (-2.37 V) = +0.71 V

For Cell 3:
E°cell = (+0.34 V) - (-2.37 V) = +2.71 V

Therefore, the cell potentials for the given voltaic cells are:
Cell 1: +2.00 V
Cell 2: +0.71 V
Cell 3: +2.71 V