Which of the metals (A, B, C, and D) has the highest ionization tendency (highest tendency to form an ions or to be oxidized) and lowest ionization tendency? Explain your answer.

Question

RECORD:

Electrodes: Positive (P) and Negative (N)

A & B: Positive: A, Negative: B

A & C: Positive: A, Negative: C

A & D: Positive: D, Negative: A

B & C: Positive: C, Negative: B

B & D: Positive: D, Negative: B

C & D: Positive: D, Negative: C

Also: Arrange the metals in decreasing order of their ionization tendencies.

Answer 1

I think this question could be worded better since ionization tendency could mean a tendency towards either gaining or losing electrons. However, going off of the assumption that ionization tendency = loss of electrons/formation of cations = oxidation and presuming that these metals are in a galvanic cell (discharging):

Remember that the positive cell is the cathode (+) and the negative cell is the anode (-).

Reduction occurs at the cathode and oxidation occurs at the anode; use mnemonic RED-CAT AN-OX.

You can imagine that these metals A-D can be ordered by oxidation potential (tendency to lose electrons). A species that is more likely to lose electrons (be oxidized) should be a "more negative" electrode. You can start by looking at each of the electrodes in pairs and from there can figure out the order of oxidation potential (the negative electrode should have the higher oxidation potential of the pair)

B - more negative than A,C, and D so it has the highest oxidation potential

C - more negative than A and D but less than B (second highest)

A - more negative than D, but less than B and C (third highest)

D - not more negative than any of the other electrodes (lowest reduction potential)

Therefore, the order of ionization tendencies (oxidation potential) is as follows: B>C>A>D