You are given the following three half-reactions:

$$\begin{gathered} {\text{(1) Fe}}^{\text{3 +}}{\text{(aq) +e}}^{\text{-}}\rightleftharpoons {\text{Fe}}^{\text{2 +}}\text{(aq)} \\ {\text{(2) Fe}}^{\text{2 +}}{\text{(aq) + 2e}}^{\text{-}}\rightleftharpoons \text{Fe(s)} \\ {\text{(3) Fe}}^{\text{3 +}}{\text{(aq) + 3e}}^{\text{-}}\rightleftharpoons \text{Fe(s)} \end{gathered}$$

(a). Use ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$values for ( 1 ) and ( 2 ) to find ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$ for ( 3 ).

(b) Calculate$∆G^0$ for ( 1) and ( 2) from their ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$values.

(c) Calculate$∆G^0$ for (3) from (1) and ( 2 ).

(d) Calculate ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$for ( 3) from its $∆G^0$.

(e) What is the relationship between the ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$values for (1) and (2) and the${{\text{E}}^{\text{o}}}_{\text{half - cell}}$value for (3 )?

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.

$$\begin{gathered} {\text{Fe}}^{\text{3 +}}{\text{(aq) +e}}^{\text{-}}\rightleftharpoons {\text{Fe}}^{\text{2 +}}\text{(aq)}{{\text{E}}^{\text{o}}}_{\text{half - cell}}\text{= 0.77 V (1)} \\ {\text{Fe}}^{\text{2 +}}{\text{(aq) + 2e}}^{\text{-}}\rightleftharpoons \text{Fe(s)}{{\text{E}}^{\text{o}}}_{\text{half - cell}}\text{= - 0.44 V (2)} \\ {\text{Fe}}^{\text{3 +}}{\text{(aq) + 3e}}^{\text{-}}\rightleftharpoons \text{Fe(s)}{{\text{E}}^{\text{o}}}_{\text{half - cell}}\text{=}{{\text{E}}^{\text{o}}}_{\text{1}}\text{-}{{\text{E}}^{\text{o}}}_{\text{2}} \end{gathered}$$

$\begin{array}{rcl}& & {{\text{E}}^{\text{o}}}_{\text{cell}}\text{=}{{\text{E}}^{\text{o}}}_{\text{1}}\text{-}{{\text{E}}^{\text{o}}}_{\text{2}}\\ & & \text{= - 0.44V - 0.77V}\\ & & \text{= 0.33V}\\ & & \end{array}$

Therefore, the value is:

$$\begin{gathered} {{\text{E}}^{\text{o}}}_{\text{cell 1}}\text{= 0.77 V} \\ {{\text{E}}^{\text{o}}}_{\text{cell 2}}\text{= - 0.44 V} \\ {{\text{E}}^{\text{o}}}_{\text{cell 3}}\text{= 0.33V} \end{gathered}$$

$∆{\text{G}}^{\text{o}}\text{= - nF}{{\text{E}}^{\text{o}}}_{\text{cell}}$

$$\begin{gathered} {\text{Fe}}^{\text{3 +}}{\text{(aq) +e}}^{\text{-}}\rightleftharpoons {\text{Fe}}^{\text{2 +}}\text{(aq)}{{\text{E}}^{\text{o}}}_{\text{half - cell}}\text{= 0.77 V} \\ \text{n = 1} \\ {\text{Fe}}^{\text{2 +}}{\text{(aq) + 2e}}^{\text{-}}\rightleftharpoons \text{Fe(s)}{{\text{E}}^{\text{o}}}_{\text{half - cell}}\text{= - 0.44 V} \\ \text{n = 2} \\ {\text{Fe}}^{\text{3 +}}{\text{(aq) + 3e}}^{\text{-}}\rightleftharpoons \text{Fe(s)}{{\text{E}}^{\text{o}}}_{\text{half - cell}}\text{= 0.33 V} \\ \text{n = 3} \end{gathered}$$

$$\begin{gathered} ∆{\text{G}}_{\text{1}}^{\text{o}}\text{= - nF}{{\text{E}}^{\text{o}}}_{\text{cell}} \\ {\text{= - 1e}}^{\text{-}}\text{×96500}\frac{\text{C}}{{\text{mole}}^{\text{-}}}\text{×0.77 V} \\ \text{= - 74,305 J} \end{gathered}$$

$$\begin{gathered} ∆{\text{G}}_{\text{2}}^{\text{o}}\text{= - nF}{{\text{E}}^{\text{o}}}_{\text{cell}} \\ {\text{= - 2e}}^{\text{-}}\text{×96500}\frac{\text{C}}{{\text{mole}}^{\text{-}}}\text{×- 0.44 V} \\ \text{= 84,920 J} \end{gathered}$$

Therefore, the values are: $$\begin{gathered} ∆{\text{G}}_{\text{1}}^{\text{o}}\text{= - 74,305 J} \\ ∆{\text{G}}_{\text{2}}^{\text{o}}\text{= 84,920 J} \end{gathered}$$.

$$\begin{gathered} {\text{∆G}}_{\text{3}}^{\text{o}}{\text{=∆G}}_{\text{1}}^{\text{o}}{\text{+∆G}}_{\text{2}}^{\text{o}} \\ \text{= - 74,305 J + 84,920 J} \\ \text{= 10,615 J} \end{gathered}$$

Therefore, the value is: $∆{\text{G}}_{\text{3}}^{\text{o}}\text{= 10,615 J}$.

$$\begin{gathered} {{\text{E}}^{\text{o}}}_{\text{cell}}\text{=}\frac{∆{\displaystyle {{\displaystyle \text{G}}}^{\text{o}}}}{{\displaystyle \text{nF}}} \\ \text{=}\frac{{\displaystyle \text{10,615J}}}{{\displaystyle {\text{3e}}^{\text{-}}\text{×96500}\frac{\text{C}}{{\text{mole}}^{\text{-}}}}} \\ \text{= - 0.037 V} \end{gathered}$$

Therefore, the value is: ${{\text{E}}^{\text{o}}}_{\text{cell}}\text{= - 0.037 V}$.

Therefore,Through the Gibbs Free Energy computation, the half-cell potentials for ( 1 ) and ( 2) may be utilised to calculate the half-cell potential for ( 3 ).