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

$\begin{array}{l}{\mathbf{\text{(a)Al(s)andCd}}}^{\mathbf{\text{2+}}}\mathbf{\text{(aq)}}\\ {\mathbf{\text{(b)I}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s)andBr}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}\end{array}$

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.

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

The redox reaction taking place between $\text{Al}\left(\text{s}\right){\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}Cd}}^{\text{+2}}\left(\text{aq}\right)$is given below.

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

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

$\begin{array}{l}\text{2Al}\left(\text{s}\right)\to 2{\text{Al}}^{\text{+3}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{anode}}=-1.66\text{\hspace{0.17em}V}\\ {\text{3Cd}}^{\text{+2}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}\to 2\text{Cd}\left(\text{s}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ \pm }}_{\text{cathode}}=-0.40\text{\hspace{0.17em}V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.40-(-1.66)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=1.26V}\end{array}$

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.

$\begin{array}{l}\text{logK=}\frac{\text{1.26V×6}}{\text{0.0592V}}\\ \text{=127.70}\\ {\text{K=10}}^{127.70}\\ =5.01\times {10}^{127}\end{array}$

The redox reaction taking place between ${\text{l}}_2\left(\text{s}\right){\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}Br}}^{\text{-}}\left(\text{aq}\right)$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.

$\begin{array}{l}{\text{l}}_2\left(\text{s}\right){\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}\left(\text{aq}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{cathode}}=0.53\text{V}\\ {\text{2Br}}^{\text{-}}\left(\text{aq}\right)\to {\text{Br}}_{\text{2}}\left(\text{s}\right){\text{+2e}}^{\text{-}}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\circ }}_{\text{anode}}=1.07\text{\hspace{0.17em}V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=}\left(\text{0.53-1.07}\right)\text{V}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.54V}\end{array}$

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.

$\begin{array}{c}\text{logK=}\frac{\text{-0.54V×2}}{\text{0.0592V}}\\ \text{=-18.24}\\ {\text{K=10}}^{\text{-18.24}}\\ {\text{=5.75×10}}^{\text{-19}}\end{array}$