Q21.62P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate $\mathbf{△}{\mathit{G}}^{\mathbf{\circ }}$for each of the reactions 

$$\begin{gathered} {\mathbf{\text{(a)Ag(s) and Mn}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{(b)Cl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) and Br}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq)}} \end{gathered}$$

Step-by-Step Solution

Answer

a. $△G^{\circ }=382.14KJ/mol$

b. $△G^{\circ }=55.97KJ/mol$

Standard electrode potential and △ G ∘

For a redox reaction taking place, the standard reduction potential is the difference between the respective cell potential.

${\text{E}}_{\text{cell}}°={\text{E}}_{\text{cathode}}°{\text{ - E}}_{\text{anode}}°$

The relation between and the standard electrode potential is given below.

$△G^{\circ }=-nF{E^{\circ }}_{cell}$

Where,

n = number of electrons involved in the redox reaction

$\text{F} \text{ = } \text{96500} \text{C/mol}$.

△ G ∘ for Ag(s) and M g 2 + ( a q )

The redox reaction taking place between $\text{Ag}\left(\text{s}\right) \text{and} {\text{Mn}}^{\text{ + 2}}\left(\text{aq}\right)$ is given below.

$2\text{Ag}\left(\text{s}\right)+{\text{Mn}}^{2+}\to 2{\text{Ag}}^{+}\left(\text{aq}\right)+\text{Mn}\left(\text{s}\right)$

Now, we have to write down the cathode and anode half-cell reaction along with their respective electrode potential.

$$\begin{gathered} \text{2Ag}\left(\text{s}\right)\to {\text{2Ag}}^{\text{ + }}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }} {{\text{E}}^{\text{0}}}_{\text{anode}}=0.80 \text{V} \\ {\text{Mn}}^{\text{ + 2}}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }}\to \text{Mn}\left(\text{s}\right) {{\text{E}}^{\text{0}}}_{\text{cathode}}=-1.18 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{Cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=-1.18V-0.80V \\ {E^{\circ }}_{cell}=-1.98V \end{gathered}$$

We know that,

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-2\times 96500\times -1.98 \\ △G^{\circ }=382.14KJ/mol \end{gathered}$$

Hence $△G^{\circ }=382.14KJ/mol$.

△ G ∘ for C l 2 ( g )   a n d   B r - ( a q )

The redox reaction taking place between ${\text{l}}_2\left(\text{s}\right) \text{and} {\text{Br}}^{\text{ - }}\left(\text{aq}\right)$ is given below.

${\text{Cl}}_2( \text{s})+2{\text{Br}}^{-}\left(\text{aq}\right)\to 2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Br}}_2\left(\text{l}\right)$

Now, we have to write down the cathode and anode half-cell reaction along with their respective electrode potential.

$$\begin{gathered} {\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{ + 2e}}^{\text{ - }}\to {\text{2Cl}}^{\text{ - }}\left(\text{aq}\right) {{\text{E}}^{\text{0}}}_{\text{cathode}}=1.36\text{V} \\ {\text{2Br}}^{\text{ - }}\left(\text{aq}\right) \to {\text{Br}}_{\text{2}}\left(\text{s}\right){\text{ + 2e}}^{\text{ - }} {{\text{E}}^{\text{0}}}_{\text{anode}}=1.07 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{reduction}-{E^{\circ }}_{oxidation} \\ {E^{\circ }}_{cell}=\left(1.36-1.07\right)V \\ {E^{\circ }}_{cell}=-0.29V \end{gathered}$$

We know that.

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-2.96500C/mol\times 0.29V \\ △G^{\circ }=-55.97KJ/mol \end{gathered}$$

hence $△G^{\circ }=-55.97KJ/mol$