Video: Applying Knowledge of the Reducing Power and Atomic Radii of Potassium and Sodium

For statements I and II, state for each if they are true or false. (I) Potassium is a stronger reducing agent than sodium. (II) Potassium has a larger atomic radius than sodium. If both are true, state if (II) is a correct explanation for (I).

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Video Transcript

For statements I and II, state for each if they are true or false. (I) Potassium is a stronger reducing agent than sodium. (II) Potassium has a larger atomic radius than sodium. If both are true, state if (II) is a correct explanation for (I).

In statement (I), we see reference to the phrase “reducing agent”. In chemistry, a loss of electrons is called oxidation. Gain of electrons is called reduction. When these two processes happen together in a chemical reaction, this is called redox. These two definitions are most easily remembered by using the mnemonic OILRIG. From this definition, it follows that to have a good reducing agent, you need a species that can donate electrons. And therefore it will get oxidized. Conversely, an oxidizing agent accepts electrons and gets reduced in the process.

Oxidation and reduction always happen together in a chemical reaction, as in the example here of sodium reacting with chlorine. In this reaction, sodium is behaving as the reducing agent. It donates electrons and therefore gets oxidized. Sodium’s oxidation state changes from zero to positive one in this reaction. Chlorine is the oxidizing agent here. It accepts electrons and gets reduced. Its oxidation state changes from zero to negative one.

To decide if potassium is a stronger reducing agent than sodium, we need to compare some of the trends found within group one. That is the alkali metals. That’s where potassium and sodium belong. In the diagram, we see the first three elements in group one, as you descend the group. The sizes of the atoms are presented to scale here. And we see that as we descend the group, the atomic radius gets larger. This increase in size is caused quite simply by the effect of adding extra electron shells to the atoms. This has an impact on the relative ease with which these alkali metals can lose their valence-shell electron.

We can quantify how easily this happens by looking at the first ionization energies of these elements. First ionization energy of the element is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms to produce one mole of gaseous ions with a single positive charge. Notice that we are dealing with atoms in the gas phase here and a generic equation for this process is shown in the left-hand side. First ionization energies are influenced strongly by the effect of the nuclear charge, the distance of the valence shell from the nucleus, and the amount of shielding of the valence-shell electron from the nucleus by inner shells.

As we move from lithium to potassium within group one, the valence-shell electron gets further from the nucleus. It gets easier to remove this electron. And this is borne out by the trend shown in the first ionization energies. So now that we have examined some of the trends within group one, we can decide if statements (I) and (II) are true or false.

Potassium in the gas phase will certainly lose electrons more easily than sodium. We would therefore expect potassium to be a stronger reducing agent than sodium. And statement (I) is, in fact, true. As the potassium atom has an additional shell compared to a sodium atom, we have already seen that the atomic radius gets larger. The valence electron in potassium is more shielded or screened from the nucleus than it is in sodium. And therefore statement (II) is true. Since statement (II) is part of the argument for establishing the reducing power of potassium relative to sodium, it is a correct explanation for statement (I).

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