Q21.141 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 )?

Step-by-Step Solution

Answer

1. The values of ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$ are obtained as: $$\begin{gathered} {{\text{E}}^{\text{o}}}_{\text{cell 1}}\text{ = 0.77 V} \\ {{\text{E}}^{\text{o}}}_{\text{cell 2}}\text{ = - 0.44 V} \\ {{\text{E}}^{\text{o}}}_{\text{cell 3}}\text{ = 0.33V} \end{gathered}$$ .
2. The value of  $∆{\text{G}}^{\text{o}}$ are obtained as: $$\begin{gathered} ∆{\text{G}}_{\text{1}}^{\text{o}}\text{ = - 74,305 J} \\ ∆{\text{G}}_{\text{2}}^{\text{o}}\text{ = 84,920 J} \end{gathered}$$ .
3. The value of  $∆{\text{G}}^{\text{o}}$ is obtained as: $∆{\text{G}}_{\text{3}}^{\text{o}}\text{ = 10,615 J}$ .
4. The value of  ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$ from  $∆{\text{G}}^{\text{o}}$ for the third reaction is obtained as: ${{\text{E}}^{\text{o}}}_{\text{cell}}\text{ = - 0.037 V}$ .
5. The Gibbs Free Energy calculation may be utilised to derive the half-cell potential for (3 ) using the half-cell potentials for ( 1 ) and ( 2 ).

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: Evaluate the value of E o half - cell for the third reaction

a. The value of ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$ for the reaction is evaluated as:

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

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

 

Therefore, the value is:

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

Step 3: Evaluate the value of ∆ G o for first and second reaction

b. The value of  $∆{\text{G}}^{\text{o}}$ is obtained using the equation:

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

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

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

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

 

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

Step 4: Evaluate the value of ∆ G o for third reaction

c. The value of  $∆{\text{G}}^{\text{o}}$ is obtained by adding the free energy of reaction one and two:

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

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

Step 5: Evaluate the value of for third reaction from

d. The value of  ${{\text{E}}^{\text{o}}}_{\text{half - cell}}$ can be evaluated using $∆{\text{G}}^{\text{o}}\text{ = nF}{{\text{E}}^{\text{o}}}_{\text{cell}}$ :

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

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

Step 6: Explanation of relationship between all the three reactions

e. The Gibbs Free Energy calculation may be utilised to obtain the half-cell potential for ( 3) using the half-cell potentials for (1 ) and (2 ).

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 ).