Question

What is the value of the equilibrium constant for the reaction between each pair at $\mathbf{25}\mathbf{°}\mathbf{C}$?

$\begin{array}{l}{\mathbf{\text{(a)Cr(s)andCu}}}^{\mathbf{\text{2+}}}\mathbf{\text{(aq)}}\\ {\mathbf{\text{(b)Sn(s)andPb}}}^{\mathbf{\text{2+}}}\mathbf{\text{(aq)}}\end{array}$

Short answer

(a) Equilibrium constant for the reaction between${\text{Cr(s)andCu}}^{\text{2+}}{\text{(aq)at25}}^{\text{°}}{\text{Cis5.37×10}}^{\text{79}}$

(b) Equilibrium constant for the reaction between${\text{Sn(s)andPb}}^{\text{2+}}{\text{(aq)at25}}^{\text{°}}\text{Cis2.19.}$

Step-by-step solution

Standard electrode potential and equilibrium constant

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

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

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

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

Equilibrium constant between Cr(s)  and  Cu+2(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.

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

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

$\begin{array}{l}\text{2Cr}\left(\text{s}\right)\to {\text{2Cr}}^{\text{+3}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{anode}}=-0.73\text{\hspace{0.17em}V}\\ {\text{3Cu}}^{\text{+2}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}\to 3\text{Cu}\left(\text{s}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{cathode}}=0.34\text{\hspace{0.17em}V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=0.34V-(-0.73\hspace{0.17em}V)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=1.07V}\end{array}$

At $25°\mathrm{C}$,

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

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

$\begin{array}{l}\text{logK\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1.07V×6}}{\text{0.0592V}}\\ \text{=\hspace{0.17em}\hspace{0.17em}79.73}\\ {\text{K\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}10}}^{\text{79.73}}\\ {\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}5.37×10}}^{\text{79}}\end{array}$

Equilibrium constant for the reaction between${\text{Cr(s)andCu}}^{\text{2+}}{\text{(aq)at25}}^{\text{°}}{\text{Cis5.37×10}}^{\text{79}}$.

Equilibrium constant between Sn(s)  and  Pb+2(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.

$\mathrm{Sn}\left(\mathrm{s}\right)+{\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Pb}\left(\mathrm{s}\right)$

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

$\begin{array}{l}\text{Sn}\left(\text{s}\right)\to {\text{Sn}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{anode}}\text{=\hspace{0.17em}-0.14\hspace{0.17em}V}\\ {\text{Pb}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{Pb}\left(\text{s}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{cathode}}=-0.13\text{\hspace{0.17em}V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.13V-(-0.14\hspace{0.17em}V)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=0.01V}\end{array}$

At $25°\mathrm{C}$,

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

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

$\begin{array}{c}\text{logK=}\frac{\text{0.01V×2}}{\text{0.0592V}}\\ \text{=0.34}\\ {\text{K=10}}^{\text{0.34}}\\ \text{=2.19}\end{array}$

Equilibrium constant for the reaction between_{${\text{Sn(s)andPb}}^{\text{2+}}{\text{(aq)at25}}^{\text{°}}\text{Cis2.19.}$}