# Question d9281

Jul 10, 2016

$\text{Pb}$, $\text{Zn}$, and $\text{Cu}$.

#### Explanation:

The idea here is that you need to use the standard reduction potentials, ${E}^{\circ}$, given to you to determine the identity of the three metals.

You know that you have

$\text{Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Zn"_ ((s))" "E^@ = -"0.76 V}$

$\text{Pb"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V}$

$\text{Cu"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V}$

Now, the standard reduction potentials measure the tendency of a chemical species to release electrons and form cations when compared to that of hydrogen.

A negative ${E}^{\circ}$ means that the chemical species loses electrons more readily than hydrogen, and that its reduction equilibrium lies to the left.

A positive ${E}^{\circ}$ means that the chemical species loses electrons less readily than hydrogen, and that its reduction equilibrium lies to the right.

Now, chemical species that lose electrons more readily are stronger reducing agents than chemical species that tend to lose electrons less readily.

When you compare the ${E}^{\circ}$ values for two reduction equilibria, you can say that

• the equilibrium with the less positive / more negative ${E}^{\circ}$ value will lie further to the left
• the equilibrium with the less negative / more positive ${E}^{\circ}$ value will lie further to the right

Let's take the first two reduction equilibria. You have

E^@("Zn"^(2+), "Zn") = -"0.76 V"

E^@("Pb"^(2+), "Pb") = -"0.13 V"

Here ${E}^{\circ} = - \text{0.76 V}$ is more negative than ${E}^{\circ} = - \text{0.13 V}$, which means that the first equilibrium lies further to the left than the second one.

You can thus say that zinc metal, $\text{Zn}$, will reduce ${\text{Pb}}^{2 +}$ because zinc loses electrons more readily than lead metal does.

Consequently, zinc will also reduce ${\text{Cu}}^{2 +}$, since ${E}^{\circ} = - \text{0.76 V}$ is more negative than ${E}^{\circ} = + \text{0.34 V}$.

In other words, when zinc metal is placed in a solution that contains ${\text{Pb}}^{2 +}$ and ${\text{Cu}}^{2 +}$ cations, the cations will be reduced to lead metal and copper metal, respectively, and zinc will be oxidized to ${\text{Zn}}^{2 +}$.

This implies that zinc is metal $\text{B}$.

Finally, notice that metal $\text{C}$ cannot reduce ${\text{A}}^{2 +}$, since it will not be oxidized when placed in a solution that contains ${\text{A}}^{2 +}$. This means that the ${E}^{\circ}$ of metal $\text{C}$ is more positive than ${E}^{\circ}$ for metal $\text{A}$. As a result, metal $\text{C}$ will be copper and metal $\text{A}$ will be lead.

$\text{Metal A "-> " Lead}$

$\text{Metal B " -> " Zinc}$

$\text{Metal C " -> " Copper}$

So, to answer such problems quickly, list the reduction half-reactions in order of increasing ${E}^{\circ}$ values

$\text{Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons color(blue)("Zn"_ ((s)))" "E^@ = -"0.76 V}$

color(red)("Pb"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V"

color(red)("Cu"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V"#

and keep in mind that the chemical species listed in the top right can reduce the chemical species listed in the bottom left.

For example, zinc metal is $\textcolor{b l u e}{\text{top right}}$, so it can reduce ${\text{Pb}}^{2 +}$ and ${\text{Cu}}^{2 +}$ because they are located in the $\textcolor{red}{\text{bottom left}}$.

Similarly, you can say that the chemical species located in the bottom left can oxidize the chemical species located to the top right.

Here ${\text{Cu}}^{2 +}$ and ${\text{Pb}}^{2 +}$ are $\textcolor{red}{\text{bottom left}}$, so they can oxidize zinc metal, which is located in the $\textcolor{b l u e}{\text{top right}}$.

$\left\{\left(\textcolor{red}{\text{Bottom left")color(white)(a)"oxidizes" color(white)(a)color(blue)("top right ")), (color(blue)("Top right")color(white)(a)"reduces" color(white)(a)color(red)("bottom left}}\right)\right.$