Q21.61P - Chemistry: Molecular Nature Of Matter And Change

Question

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

$$\begin{gathered} {\mathbf{\text{(a)Al(s) and Cd}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{(b)I}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s) and Br}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq)}} \end{gathered}$$

Step-by-Step Solution

Answer

a. $△G^{\circ }=-729.54KJ/mol$

b. $△G^{\circ }=104.22KJ/mol$

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 $△G^{\circ }$ 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 Al s     and     Cd + 2 aq

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

$2Al\left(\text{s}\right)+3{\text{Cd}}^{2+}\text{(aq)}\to 2{\text{Al}}^{3+}\left(\text{aq}\right)+3\text{Cd}\left(\text{s}\right)$

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

$$\begin{gathered} \text{2Al}\left(\text{s}\right)\to 2{\text{Al}}^{\text{ + 3}}\left(\text{aq}\right){\text{ + 6e}}^{\text{ - }} {\text{E}°}_{\text{anode}}=-1.66 V \\ {\text{3Cd}}^{\text{ + 2}}\left(\text{aq}\right){\text{ + 6e}}^{\text{ - }}\to 2\text{Cd}\left(\text{s}\right) {{\text{E}}^{\circ \pm }}_{\text{cathode}}=-0.40 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=-0.40V-\left(-1.66\right)V \\ {E^{\circ }}_{cell}=1.26 V \end{gathered}$$

 

We know that,

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-6\times 96500C/mol\times 1.26V \\ △G^{\circ }=-729.54KJ/mol \end{gathered}$$

Hence $△G^{\circ }=-729.54KJ/mol$

△ G ∘ for l 2 s     and     Br - aq

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.

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

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

$$\begin{gathered} {\text{l}}_2\left(\text{s}\right){\text{ + 2e}}^{\text{ - }}\to {\text{2I}}^{\text{ - }}\left(\text{aq}\right) {\text{E}°}_{\text{cathode}}=0.53\text{V} \\ {\text{2Br}}^{\text{ - }}\left(\text{aq}\right) \to {\text{Br}}_{\text{2}}\left(\text{s}\right){\text{ + 2e}}^{\text{ - }} {\text{E}°}_{\text{anode}}=1.07 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=\left(0.53-1.07\right)V \\ {E^{\circ }}_{cell}=-0.54V \end{gathered}$$

We know that,

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-2\times 96500C/mol\times -0.54 V \\ △G^{\circ }=104.22 KJ/mol \end{gathered}$$

Hence $△G^{\circ }=104.22 KJ/mol$.