Periodic Trend - Standard reduction potentials

Dear colleagues,

I would like to ask your thoughts about a question that I am trying to explain to a group of students whom I am helping to prepare for the Chemistry Olympiad.

Right answer: B

My explanation (I am trying to link it to periodic trends):

Reduction is the inverse of ionization energy.

In ionization energy, we have M(g) → M⁺(g) + e; the reduction runs backward: M⁺(g) + e⁻ → M(g).

The half-reaction in this question is M²⁺(aq) + 2e⁻ → M(s). Strip away the “aq” and “s” and look at the electrons: the cation M²⁺ gains two electrons to go from M²⁺ to neutral M (reduction process). It is the reverse of the element M losing two electrons to go from neutral M to M²⁺ (ionization energy).

Therefore, an element with a high ionization energy (IE) is hard to pull electrons off; this same element has a high reduction potential (E°), meaning electrons want to come back on. So, the element with the highest ionization energy has the higher reduction potential.

Now, let's analyze how ionization energy varies among Ni, Pd, and Pt.

In general, ionization energy increases from bottom to top; however, because we are dealing with transition metals, we must account for the anomalous behavior of d and f orbitals.

The reason why ionization energy increases going up a group is because each added shell brings more shielding while Zeff stays roughly the same. By that logic, we’d expect Ni to have the highest IE of the three (fewest shells, least shielding) and Pt to have the lowest (most shells, most shielding), with Pd in between.

That’s not what happens. Ni is the lowest; Pd and Pt are both higher than Ni. The reason is that between each pair, a subshell fills in that shields unusually poorly (d and f orbitals).

Between Ni and Pd, Ni’s 3d subshell is filling right before Pd. d electrons already shield worse than s or p electrons, so Pd’s valence electrons feel more of the nuclear charge.

Between Pd and Pt, the effect repeats, but worse. Before you reach Pt, all 14 lanthanides have to fill their 4f subshell, and f electrons shield even more poorly than d electrons. This lanthanide contraction raises Pt’s Zeff further still, which is why Pt’s IE stays elevated instead of dropping the way you’d expect from adding an entire extra shell.

Since IE and E° are linked, Ni’s low IE becomes Ni’s low E° (least willing to be reduced), and Pd and Pt’s anomalously high IE values become their anomalously high E° values.