Q21.60P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate for each of the reactions 

$$\begin{gathered} {\mathbf{\text{(a)Ni(s)andAg}}}^{\mathbf{\text{ + }}}\mathbf{\text{(aq)}} \\ {\mathbf{\text{(b)Fe(s)andCr}}}^{\mathbf{\text{3 + }}}\mathbf{\text{(aq)}} \end{gathered}$$

Step-by-Step Solution

Answer

a. $△G^{\circ }=-202.65KJ/mol$

b. $△G^{\circ }=173.7KJ/mol$

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.

$△G^{\circ }=-nF{E_{cell}}^{\circ }$

Where,

n = number of electrons involved in the redox reaction

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

△ G ∘ for Ni(s) and A g + ( a q )

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

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

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

$$\begin{gathered} \text{Ni}\left(\text{s}\right)\to {\text{Ni}}^{\text{ + 2}}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }} {\text{E}°}_{\text{anode}}=-0.25 \text{V} \\ {\text{2Ag}}^{\text{ + }}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }}\to 2\text{Ag}\left(\text{s}\right) {\text{E}°}_{\text{cathode}}=0.80 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=0.80V-\left(-0.25V\right) \\ {E^{\circ }}_{cell}=1.05 V \end{gathered}$$

Since two electrons are involved in this redox reaction, n=2.

$$\begin{gathered} △G^{\circ }=-nF{E^{\circ }}_{cell} \\ △G^{\circ }=-2\times 96500C/mol\times 1.05V \\ △G^{\circ }=-202.65KJ/mol \end{gathered}$$

Hence $△G^{\circ }=-202.65KJ/mol$

△ G ∘ for Fe(s) and C r 3 + ( a q )

The redox reaction taking place between $\text{Fe}\left(\text{s}\right) \text{and} {\text{Cr}}^{\text{ + 3}}\left(\text{aq}\right)$ is given below.

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

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

$$\begin{gathered} \text{3Fe}\left(\text{s}\right)\to 3{\text{Fe}}^{\text{ + 2}}\left(\text{aq}\right){\text{ + 6e}}^{\text{ - }} {\text{E}°}_{\text{anode}}=-0.44 \text{V} \\ {\text{2Cr}}^{\text{ + 3}}\left(\text{aq}\right){\text{ + 6e}}^{\text{ - }}\to 2\text{Cr}\left(\text{s}\right) {\text{E}°}_{\text{cathode}}=-0.74 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{cathode}-{E^{\circ }}_{anode} \\ {E^{\circ }}_{cell}=-0.74 V-\left(-0.44 V\right) \\ {E^{\circ }}_{cell}=-0.30 V \end{gathered}$$

Since six electrons are involved in this redox reaction, n=6

$$\begin{gathered} △G^{\circ }=-6\times 96500\times \left(-0.30\right) \\ △G^{\circ }=173.7KJ/mol \end{gathered}$$

Hence$△G^{\circ }=173.7KJ/mol$