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 ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{half}\mathbf{-}\mathbf{cell}}$

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.

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.

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: $-0.4546\mathrm{V}$.

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: $-0.985\mathrm{V}$.