# Question #04d43

Jan 14, 2016

${\text{X}}^{-}$

#### Explanation:

Before jumping in, it's worth taking a moment to make sure that you understand what you're dealing with here.

The problem wants you to pick the strongest reducing agent by looking at what should actually be four redox equilibria.

So, what does it mean for a chemical species to be a reducing agent?

As you know, oxidation and reduction reactions involve the transfer of electrons from one element to another. More specifically, you know that

• the element that donates electrons is being oxidized
• the element that accepts electrons is being reduced

So, in order for a chemical species to be a reducing agent, it must reduce another chemical species, i.e. it must donate electrons to that species.

The strongest reducing agent will thus be the chemical species that donates electrons with the most ease.

Now, take a look at the four redox equilibria given to you

$\text{M"_text((aq])^(2+) + 2"e"^(-) rightleftharpoons "M"_text((s])" " " "E^@ = +"1.21 V}$

$\text{Q"_text((aq[)^(+) + "e"^(-) rightleftharpoons "Q"_text((s])" " " " color(white)(a) E^@ = +"1.03 V}$

$\text{Z"_text((aq])^(3+) + 3"e"^(-) rightleftharpoons "Z"_text((s])" " " " color(white)(a)E^@ = - "0.21 V}$

$\text{X"_text(2(aq]) + 2"e"^(-) rightleftharpoons 2"X"_text((s])^(-) " " " "E^@ = -"1.23 V}$

The idea here is that the ${E}^{\circ}$ value for each redox equilibria tells you where will the equilibrium lie compared with a reference electrode, usually a hydrogen electrode.

Simply put, a positive ${E}^{\circ}$ value tells you that your chemical species forms ions less readily than hydrogen does. When this is the case, the redox equilibrium will lie further to the right, since this side is relatively more positive than the left side.

Likewise, a negative ${E}^{\circ}$ value tells you that the chemical species forms ions more readily than hydrogen does. When this is the case, the redox equilibrium will lie further to the left, since that side is relatively more negative than the right side.

You're looking for the chemical species that is most willing to donate electrons. This means that its redox equilibrium must lie furthest to the left.

This in turn means that you're looking for more negative ${E}^{\circ}$ values. The strongest reducing agent will thus have the most negative ${E}^{\circ}$ value.

This means that ${\text{X}}^{-}$, which has an ${E}^{\circ}$ value of $- \text{1.23 V}$, will be the strongest reducing agent of the group.

$\text{ " " " "X"_text(2(aq]) + 2"e"^(-) rightleftharpoons 2"X"_text((s])^(-) " " " "E^@ = -"1.23 V}$