# Question #6e450

Jan 18, 2017

See explanation.

#### Explanation:

As you know, ionization energy is defined as the energy needed to remove one mole of electrons from one mole of isolated neutral atoms in the gaseous state and form one mole of cations.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{neutral atom"_ ((g)) + color(blue)("energy") -> "cation"_ ((g))^(+) + "e}}^{-}}}}$

The first ionization of a neutral atom always produces a $1 +$ cation.

Now, an element's reducing power is nothing more than a measure of how effective it is at losing electrons. In a redox reaction, you know that

• $\text{one atom is being oxidized " -> " it loses electrons}$
• $\text{another atom is being reduced " -> " it gains electrons}$

More specifically, an atom that is being oxidized gives electrons to the atom that is being reduced, i.e. it acts as a reducing agent.

In order for an atom to be an effective reducing agent, it must be very good at giving up electrons. You can thus say that atoms that lose electrons easily, i.e. are very good at giving up electrons, have a higher reducing power.

keep in mind that for most atoms, this is a relative term, meaning that their reducing power depends on the actual reaction in which they take part.

So, ionization energy tells you how much energy is needed in order to remove electrons from an atom and reducing power tells you how easily an atom gives up electrons.

You can thus say that atoms that have a low ionization energy will be relatively good at reducing other atoms, i.e. they will have a high reducing power.

On the other hand, atoms that have a high ionization energy will be relatively weak reducing agents, i.e. they will have a low reducing power.

Think of it like this -- the less energy is needed in order to remove an electron from an atom, the easier it will be for the atom to give up that electron, which in turn will make it a relatively strong reducing agent.

Similarly, the more energy is needed in order to remove an electron from an atom, the harder it will be for the atom to give up that electron, which in turn will make it a relatively weak reducing agent.

As a conclusion, you have

• $\text{high ionization energy " -> " relatively weak reducing agent}$
• $\text{low ionization energy " -> " relatively strong reducing agent}$