Q21.152CP - Chemistry: Molecular Nature Of Matter And Change

Question

For the reaction:

${\mathbf{S}}_{\mathbf{4}}{{\mathbf{O}}_{\mathbf{6}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{I}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{S}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\Delta }{\mathbf{G}}^{\mathbf{o}}\mathbf{=}\mathbf{87}\mathbf{.}\mathbf{8}\mathbf{kJ}\mathbf{/}\mathbf{mol}$

(a) Identify the oxidizing and reducing agents.

(b) Calculate ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{cell}}$.

(c) For the reduction half-reaction, write a balanced equation, give the oxidation number of each element, and calculate ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{half}\mathbf{-}\mathbf{cell}}$

Step-by-Step Solution

Answer

(a) The oxidising agent is ${\mathrm{I}}_2$  and reducing agent is ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$ .

(b)  The value of ${E^o}_{cell}$  is obtained as: $-0.4546 \mathrm{V}$ .

(c)  The balanced equation is: ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ and the value of ${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}$  is obtained as:  $-0.985\mathrm{V}$.

Step 1: Chemical reaction

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

Step 2: Subpart (a)

To identify the oxidizing and reducing agents, divide the reaction into half-reactions.

 $\begin{array}{c}{\mathrm{Oxidation}:2\mathrm{I}}_{\left(\mathrm{aq}\right)}^{-}\to 2{\mathrm{e}}^{-}+{\mathrm{I}}_{2\left(\mathrm{s}\right)}\\ \mathrm{Reduction}:{\mathrm{S}}_4{{\mathrm{O}}_6}_{\left(\mathrm{aq}\right)}^{2-}+2{\mathrm{e}}^{-}\to {\mathrm{S}}_2{{\mathrm{O}}_3}_{\left(\mathrm{aq}\right)}^{2-}\end{array}$

${\mathrm{I}}^{-}$  is the reducing agent because it is oxidised, whereas ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$  is the oxidising agent because it is reduced.

Step 3: Subpart (b)

The cell potential may be calculated using the ${\mathrm{\Delta G}}^{\mathrm{o}}$ .

$\begin{array}{c}{\mathrm{\Delta G}}^{\mathrm{o}}{=-\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}\\ 87,800\mathrm{J}=-6{\mathrm{e}}^{-}\times 96,500\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}{\times \mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=\frac{87,800 \mathrm{J}}{-2{\mathrm{e}}^{-}\times 96500\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}}\\ =-0.4546 \mathrm{V}\end{array}$

Therefore, the value is: $-0.4546 \mathrm{V}$ .

Step 4: Subpart (c)

Non-zero atoms and electrons must be balanced.

${\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}$

The reduction potential for ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$  may be calculated if  ${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}$ for the reaction  ${\mathrm{I}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{I}}^{-}$ is $0.53\mathrm{V}$ and the cell potential is $-0.4546\mathrm{V}$.

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}^{\mathrm{reduction}}{-\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}\\ {=\mathrm{E}}_{{\mathrm{S}}_4{\mathrm{O}}_6^{2-}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}\\ -0{.4546 \mathrm{V}=\mathrm{E}}_{{\mathrm{S}}_4{\mathrm{O}}_6^{2-}}^{\mathrm{o}}-0.53 \mathrm{V}\\ {\mathrm{E}}_{{\mathrm{S}}_4{\mathrm{O}}_6^{2-}}^{\mathrm{o}}=-0.4546 \mathrm{V}-0.53 \mathrm{V}\\ =-0.985\mathrm{V}\end{array}$$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}^{\mathrm{reduction}}{-\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}\\ {=\mathrm{E}}_{{\mathrm{S}}_4{\mathrm{O}}_6^{2-}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}\\ -0{.4546 \mathrm{V}=\mathrm{E}}_{{\mathrm{S}}_4{\mathrm{O}}_6^{2-}}^{\mathrm{o}}-0.53 \mathrm{V}\\ {\mathrm{E}}_{{\mathrm{S}}_4{\mathrm{O}}_6^{2-}}^{\mathrm{o}}=-0.4546 \mathrm{V}-0.53 \mathrm{V}\\ =-0.985\mathrm{V}\end{array}$

Therefore, the equation is:  ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ and ${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}$  is: $-0.985\mathrm{V}$ .