Q21.56P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the value of the equilibrium constant for the reaction between each pair at 25°C?

 $\begin{array}{l}{\mathbf{\text{(a) Ni(s) and Ag}}}^{\mathbf{\text{+}}}\mathbf{\text{(aq)}}\\ {\mathbf{\text{(b) Fe(s) and Cr}}}^{\mathbf{\text{3+}}}\mathbf{\text{(aq)}}\end{array}$

Step-by-Step Solution

Answer

(a) Equilibrium constant for the reaction between   ${\text{Ni(s) and Ag}}^{\text{+}}{\text{(aq)at 25}}^{\text{°}}{\text{C is 2.95×10}}^{\text{35}}$ 

(b) Equilibrium constant for the reaction between  ${\text{Fe(s) and Cr}}^{\text{3+}}{\text{(aq)at 25}}^{\text{°}}{\text{Cis 3.935×10}}^{\text{-31}}$ 

Step 1: Standard electrode potential and equilibrium constant

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

  ${\text{E}}_{\text{cell}}^{\text{0}}={\text{E}}_{\text{cathode}}^{\text{0}}{\text{-E}}_{\text{anode}}^{\text{0}}$

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

 ${\text{E}}_{\text{cell}}^{\text{0}}=\frac{\text{RT}}{\text{nF}}\text{lnK}$

Step 2: Equilibrium constant between Ni ( s )    and    Ag + ( aq )

The redox reaction taking place between  $\text{Ni}\left(\text{s}\right){\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}Ag}}^{\text{+}}\left(\text{aq}\right)$ 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.

$\begin{array}{c}\text{Ni}\left(\text{s}\right)\to {\text{Ni}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{anode}}=-0.25\text{\hspace{0.17em}V}\\ {\text{2Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to 2\text{Ag}\left(\text{s}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{cathode}}=0.80\text{\hspace{0.17em}V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell }}^{\text{°}}{\text{=E}}_{\text{reduction }}^{\text{°}}{\text{-E}}_{\text{oxidation }}^{\text{°}}\\ {\text{E}}_{\text{cell }}^{\text{°}}\text{=0.80-(-0.25)}\\ {\text{E}}_{\text{cell }}^{\text{°}}\text{=1.05 V}\end{array}$

At  ${\text{25}}^{\text{o}}\text{C}$ ,

 $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=\frac{\text{8.314 J/Kmol×298\hspace{0.17em}K}\times \text{2.303}}{\text{n}\times \text{96485\hspace{0.17em}C}}\text{logK}\\ \text{= }\frac{\text{0.0592}}{\text{n}}\text{logK}\end{array}$

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

$\begin{array}{c}\text{logK=}\frac{\text{1.05 V×2}}{\text{0.0592 V}}\\ \text{ =35.47}\\ {\text{K =10}}^{\text{35.47}}\\ {\text{ =2.95×10}}^{\text{35}}\end{array}$

Equilibrium constant for the reaction between  ${\text{Ni(s) and Ag}}^{\text{+}}{\text{(aq)at 25}}^{\text{°}}{\text{C is 2.95×10}}^{\text{35}}$

Step 3: Equilibrium constant between Fe ( s )    and    Cr + 3 ( aq )

The redox reaction taking place between $\text{Fe}\left(\text{s}\right){\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}Cr}}^{\text{+3}}\left(\text{aq}\right)$ 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.

$\begin{array}{l}\text{3Fe}\left(\text{s}\right)\to 3{\text{Fe}}^{\text{+2}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{ anode}}=-0.44\text{\hspace{0.17em}V}\\ {\text{2Cr}}^{\text{+3}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}\to 2\text{Cr}\left(\text{s}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{ cathode}}=-0.74\text{\hspace{0.17em}V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell }}^{\text{°}}{\text{=E}}_{\text{cathode }}^{\text{°}}{\text{-E}}_{\text{anode }}^{\text{°}}\\ {\text{E}}_{\text{cell }}^{\text{°}}\text{=-0.74-(-0.44)}\\ {\text{E}}_{\text{cell }}^{\text{°}}\text{=-0.30 V}\end{array}$

At  ${\text{25}}^{\text{o}}\text{C}$ ,

 $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=\frac{\text{8.314 J/Kmol×298\hspace{0.17em}K}\times \text{2.303}}{\text{n}\times \text{96485\hspace{0.17em}C}}\text{logK}\\ \text{= }\frac{\text{0.0592}}{\text{n}}\text{logK}\end{array}$

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

$\begin{array}{c}\text{logK=}\frac{\text{-0.30 V×6}}{\text{0.0592 V}}\\ \text{ = -30.405 }\\ {\text{K=100}}^{\text{-30.405}}\\ {\text{ =3.935×10}}^{-31}\end{array}$

Equilibrium constant for the reaction between  ${\text{Fe(s) and Cr}}^{\text{3+}}{\text{(aq)at 25}}^{\text{°}}{\text{Cis 3.935×10}}^{\text{-31}}$.