Question

You have 5 metals A, B, C, D, E in 1.00M solutions of A^(3+), B^(3+), C^(3+), D^(3+), and E^(3+) respectively. When some of the half-cells are connected, the following results are found.

Cathode /Anode /Cell Voltage
C E 0.10V
D B 0.08V
A C 1.41V
D C 0.55V

Arrange the ions in decreasign strength as oxidizing agents.

How do you find the cell potential when they're all unknown?

Answer 1

You must compare them.
If C is the anode and A is the cathode, you have 1.41 volts generated; therefore, C is a better reducing agent than A (or A is a better oxidizing agent than C). So on a chart, place
A --stronger oxidizing agent.

C --weaker oxidizing agent.
Then do the others and fill in the gaps.

Answer 2

After comparing them I came up with A^(3+)>D^(3+)> ? > ? > E^(3+)

I can't figure out whether B is the stronger oxidizing agent or C.