Q21.40P - Chemistry: Molecular Nature Of Matter And Change

Question

Use the emf series (Appendix D) to arrange the species.

(a) In order of decreasing strength as oxidizing agents:  ${\mathbf{\text{Fe}}}^{\mathbf{\text{3+}}}{\mathbf{\text{,Br}}}_{\mathbf{\text{2}}}{\mathbf{\text{,Cu}}}^{\mathbf{\text{2+}}}$

(b) In order of increasing strength as oxidizing agents : ${\mathbf{\text{Ca}}}^{\mathbf{\text{2+}}}\mathbf{\text{,}}$${\mathbf{\text{Cr}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{7}}}^{\mathbf{\text{2-}}}$,${\mathbf{\text{Ag}}}^{\mathbf{\text{+}}}$

Step-by-Step Solution

Answer

The required work use the emf series increasing and decreasing order strength as oxidizing agents given the highest and smallest value.

 

Step 1: Find largest and smallest reduction in decreasing order

(a) Based on the table of standard reduction potentials, the substance with the largest standard reduction potential is the strongest oxidizing agent :${\text{Br}}_{\text{2}}$,${\text{Fe}}^{\text{3+}}$,${\text{Cu}}^{\text{2+}}$. The larger the standard reduction potential, the more likely the substance is to be reduced. The stronger the reduction of a substance, the stronger it is as an oxidizing agent. Given the following standard reduction potentials, the highest value is the strongest oxidizing agent, while the smallest/ most negative value is the weakest oxidizing agent. They have to be written in decreasing order, hence the order, ${\text{Br}}_{\text{2}}{\text{,Fe}}^{\text{3}}{\text{,Cu}}^{\text{2}}$

$\begin{array}{ccc}\text{ given }& {\text{E}}^{\text{°}}\text{(V)}& \text{ rank }\\ {\text{Br}}_{\text{2}}& \text{1.07}& \text{1}\\ {\text{Fe}}^{\text{3+}}& \text{0.77}& \text{2}\\ {\text{Cu}}^{\text{2+}}& \text{0.34}& \text{3}\end{array}$

Step 2: Find largest and smallest reduction in increasing order

(b) Based on the table of standard reduction potentials, the substance with the largest standard reduction potential is the strongest oxidizing agent: ${\text{Ca}}^{\text{2}}{\text{,Ag}}^{\text{†}}\text{,}$ ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}$The larger the standard reduction potential, the more likely the substance is to be reduced. The stronger the reduction of a substance, the stronger it is as an oxidizing agent. Given the following standard reduction potentials, the highest value is the strongest oxidizing agent, while the smallest/ most negative value is the weakest oxidizing agent. They have to be written in increasing order, hence the order, ${\text{Ca}}^{\text{2}}{\text{,Ag}}^{\text{†}}\text{,}$${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}$

 $\begin{array}{ccc}\text{ given }& {\text{E}}^{\text{°}}\text{(V)}& \text{ rank }\\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}& \text{1.33}& \text{1}\\ {\text{Ag}}^{\text{+}}& \text{0.88}& \text{2}\\ {\text{Ca}}^{\text{2+}}& \text{-2.87}& \text{3}\end{array}$

 Therefore, the work done is

        ${\text{(a)Br}}_{\text{2}}{\text{,Fe}}^{\text{3+}}{\text{,Cu}}^{\text{2+}}{\text{(b)Ca}}^{\text{2+}}{\text{,Ag}}^{\text{+}}{\text{,Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}\text{.}$