91影视

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

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

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

${\text{E}}_{\text{cell}}^{鈭�}=\text{鈥夆赌�}\frac{\text{RT}}{\text{nF}}\text{lnK}$

The redox reaction taking place between $\text{Al}\left(\text{s}\right){\text{鈥夆赌塧nd鈥夆赌塁d}}^{\text{+2}}\left(\text{aq}\right)$is given below.

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

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

$\begin{array}{l}\text{2Al}\left(\text{s}\right)鈫�2{\text{Al}}^{\text{+3}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}{{\text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌塃}}^{鈭�}}_{\text{anode}}=鈭�1.66\text{鈥塚}\\ {\text{3Cd}}^{\text{+2}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}鈫�2\text{Cd}\left(\text{s}\right){{\text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌塃}}^{鈭�卤}}_{\text{cathode}}=鈭�0.40\text{鈥塚}\end{array}$

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

At$25掳\mathrm{C}$ ,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=\text{鈥夆赌�}\frac{\text{8.314闯/碍尘辞濒脳298鈥块}脳\text{2.303}}{\text{n}脳\text{96485鈥塁}}\text{logK}\\ \text{=}\frac{\text{0.0592}}{\text{n}}\text{logK}\end{array}$

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

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

The redox reaction taking place between ${\text{l}}_2\left(\text{s}\right){\text{鈥夆赌塧nd鈥夆赌塀r}}^{\text{-}}\left(\text{aq}\right)$is given below.

${\text{I}}_2(\text{s})+2{\text{Br}}^{鈭�}\left(\text{aq}\right)鈫�2{\text{I}}^{鈭�}\left(\text{aq}\right)+{\text{Br}}_2\left(\text{l}\right)$

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

$\begin{array}{l}{\text{l}}_2\left(\text{s}\right){\text{+2e}}^{\text{-}}鈫�{\text{2I}}^{\text{-}}\left(\text{aq}\right){{\text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌塃}}^{鈭�}}_{\text{cathode}}=0.53\text{V}\\ {\text{2Br}}^{\text{-}}\left(\text{aq}\right)\text{鈥夆赌夆�夆赌�}鈫�{\text{Br}}_{\text{2}}\left(\text{s}\right){\text{+2e}}^{\text{-}}{{\text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌塃}}^{鈭�}}_{\text{anode}}=1.07\text{鈥塚}\end{array}$

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

At $25掳\mathrm{C}$,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=\text{鈥夆赌�}\frac{\text{8.314闯/碍尘辞濒脳298鈥块}脳\text{2.303}}{\text{n}脳\text{96485鈥塁}}\text{logK}\\ \text{=}\frac{\text{0.0592}}{\text{n}}\text{logK}\end{array}$

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

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