Q21.140 CP - Chemistry: Molecular Nature Of Matter And Change

Question

From the skeleton equations below, create a list of balanced half-reactions in which the strongest oxidizing agent is on top and the weakest is on the bottom:

$$\begin{gathered} {\mathbf{\text{U}}}^{\mathbf{\text{3 + }}}{\mathbf{\text{(aq) + Cr}}}^{\mathbf{\text{3 + }}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{Cr}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + U}}}^{\mathbf{\text{4 + }}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{Fe(s) + Sn}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{Sn(s) + Fe}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{Fe(s) + U}}}^{\mathbf{\text{4 + }}}\mathbf{\text{(aq)}}\mathbf{\to }\mathbf{\text{ no reaction }} \\ {\mathbf{\text{Cr}}}^{\mathbf{\text{3 + }}}\mathbf{\text{(aq) + Fe(s)}}\mathbf{\to }{\mathbf{\text{Cr}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + Fe}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{Cr}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + Sn}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{Sn(s) + Cr}}}^{\mathbf{\text{3 + }}}\mathbf{\text{(aq)}} \end{gathered}$$

Step-by-Step Solution

Answer

The list of balanced half-reactions from top to bottom is arranged as:

$$\begin{gathered} {\text{Sn}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Sn}}_{\text{(s)}} \\ {\text{Cr}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Cr}}_{\text{(s)}} \\ {\text{Fe}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Fe}}_{\text{(s)}} \\ {\text{U}}_{\text{(aq)}}^{\text{3 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{U}}_{\text{(s)}} \end{gathered}$$

Step 1: Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Step 2: Creating a list of balanced half-reactions in order of top to bottom

The reactions are given as:

 $$\begin{gathered} {\text{U}}^{\text{3 + }}{\text{(aq) + Cr}}^{\text{3 + }}\text{(aq)}\to {\text{Cr}}^{\text{2 + }}{\text{(aq) + U}}^{\text{4 + }}\text{(aq)}.............\text{(1)} \\ {\text{Fe(s) + Sn}}^{\text{2 + }}\text{(aq)}\to {\text{Sn(s) + Fe}}^{\text{2 + }}\text{(aq)}...............\text{(2)} \\ {\text{Fe(s) + U}}^{\text{4 + }}\text{(aq)}\to \text{no reaction }...................\text{(3)} \\ {\text{Cr}}^{\text{3 + }}\text{(aq) + Fe(s)}\to {\text{Cr}}^{\text{2 + }}{\text{(aq) + Fe}}^{\text{2 + }}\text{(aq)}...........\text{(4)} \\ {\text{Cr}}^{\text{2 + }}{\text{(aq) + Sn}}^{\text{2 + }}\text{(aq)}\to {\text{Sn(s) + Cr}}^{\text{3 + }}\text{(aq)}................\text{(5)} \end{gathered}$$

The following may be deduced from the information provided:

The ${\text{U}}^{\text{3 + }}$  is oxidized in the first reaction, ${\text{Cr}}^{\text{3 + }}$  is a more powerful reducing agent.

The Fe is oxidized in the second reaction, ${\text{Sn}}^{\text{2 + }}$  is a more powerful oxidizing agent.

The ${\text{U}}^{\text{3 + }}$   cannot oxidise Fe, there is no reaction in the third reaction; hence, ${\text{Fe}}^{\text{2 + }}$  is a stronger oxidizing agent than ${\text{U}}^{\text{3 + }}$ .

It can oxidize Fe, ${\text{Cr}}^{\text{3 + }}$  is a stronger oxidizing agent than Fe in the fourth reaction.

${\text{Sn}}^{\text{2 + }}$  is a more powerful oxidizing agent in the fifth reaction since it oxidizedCr(Il) ions.

Therefore,

 ${\text{Sn}}^{\text{2 + }}$ is the strongest oxidizing agent because it cannot be oxidized in any of the reactions.

${\text{Cr}}^{\text{2 + }}$  has the ability to oxidize  ${\text{U}}^{\text{3 + }}$ and ${\text{Fe}}^{\text{2 + }}$ , making it more powerful than two other ions.

As  ${\text{U}}^{\text{3 + }}$ may be oxidized by all three ions, it is the weakest oxidizing agent, whereasis the strongest.

Then, the list of ions will be:

$$\begin{gathered} {\text{Sn}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Sn}}_{\text{(s)}} \\ {\text{Cr}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Cr}}_{\text{(s)}} \\ {\text{Fe}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Fe}}_{\text{(s)}} \\ {\text{U}}_{\text{(aq)}}^{\text{3 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{U}}_{\text{(s)}} \end{gathered}$$

Therefore, the list is arranged in order is:

$$\begin{gathered} {\text{Sn}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Sn}}_{\text{(s)}} \\ {\text{Cr}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Cr}}_{\text{(s)}} \\ {\text{Fe}}_{\text{(aq)}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{Fe}}_{\text{(s)}} \\ {\text{U}}_{\text{(aq)}}^{\text{3 + }}{\text{ + 2e}}^{\text{ - }}\to {\text{U}}_{\text{(s)}} \end{gathered}$$