91Ó°ÊÓ

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.

${\text{E}}_{\text{cell}}^{\text{0}}=\text{ â¶Ä‰}\frac{\text{RT}}{\text{nF}}\text{lnK}$

The redox reaction taking place between $\text{Ag}\left(\text{s}\right){\text{ â¶Ä‰and â¶Ä‰Mn}}^{\text{+2}}\left(\text{aq}\right)$is given below.

$2\text{Ag}\left(\text{s}\right)+{\text{Mn}}^{2+}→2{\text{Ag}}^{+}\left(\text{aq}\right)+\text{Mn}\left(\text{s}\right)$

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

$\begin{array}{l}\text{2Ag}\left(\text{s}\right)→{\text{2Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}{{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{\text{0}}}_{\text{anode}}=0.80\text{ V}\\ {\text{Mn}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}→\text{Mn}\left(\text{s}\right){{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰E}}^{\text{0}}}_{\text{cathode}}=−1.18\text{ V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{Cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-1.18 V-0.80³Õ}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-1.98V}\end{array}$

At $25°\mathrm{C}$,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=\text{ â¶Ä‰}\frac{\text{8.314´³/°­³¾´Ç±ô×298‿é}×\text{2.303}}{\text{n}×\text{96485 C}}\text{logK}\\ \text{=}\frac{\text{0.0592}}{\text{n}}\text{logK}\end{array}$

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

$\begin{array}{l}\text{logK=}\frac{\text{-1.98³Õ×2}}{\text{0.0592V}}\\ \text{=-66.89}\\ {\text{K=10}}^{\text{-66.89}}\\ {\text{=1.29×100}}^{\text{-67}}\end{array}$

Equilibrium constant for the reaction between ${\text{Ag(s)andMn}}^{\text{2+}}{\text{(aq)at25}}^{\text{°}}{\text{°ä¾±²õ 1.29×10}}^{\text{-67}}$

The redox reaction taking place between${\text{l}}_2\left(\text{s}\right){\text{ â¶Ä‰and â¶Ä‰Br}}^{\text{-}}\left(\text{aq}\right)$is given below.

${\text{Cl}}_2(\text{s})+2{\text{Br}}^{−}\left(\text{aq}\right)→2{\text{Cl}}^{−}\left(\text{aq}\right)+{\text{Br}}_2\left(\text{l}\right)$

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

$\begin{array}{l}{\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{+2e}}^{\text{-}}→{\text{2Cl}}^{\text{-}}\left(\text{aq}\right){{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{\text{0}}}_{\text{cathode}}=1.36\text{V}\\ {\text{2Br}}^{\text{-}}\left(\text{aq}\right)\text{ â¶Ä‰â€‰â¶Ä‰}→{\text{Br}}_{\text{2}}\left(\text{s}\right){\text{+2e}}^{\text{-}}{{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰E}}^{\text{0}}}_{\text{anode}}=1.07\text{ V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{reduction}}^{\text{°}}{\text{-E}}_{\text{oxidation}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=}\left(\text{1.36-1.07}\right)\text{V}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.29V}\end{array}$

At$25°\mathrm{C}$

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=\text{ â¶Ä‰}\frac{\text{8.314´³/°­³¾´Ç±ô×298‿é}×\text{2.303}}{\text{n}×\text{96485 C}}\text{logK}\\ \text{=}\frac{\text{0.0592}}{\text{n}}\text{logK}\end{array}$,

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

$\begin{array}{l}\text{logK=}\frac{\text{0.29³Õ×2}}{\text{0.0592V}}\\ \text{=9.80}\\ {\text{K=100}}^{\text{9.80}}\\ {\text{=6.27×10}}^{\text{9}}\end{array}$

Equilibrium constant for the reaction between _{${\text{Cl}}_{\text{2}}{\text{(s)andBr}}^{\text{-}}{\text{(aq)at25}}^{\text{°}}{\text{°ä¾±²õ6.27×10}}^{\text{9}}$.}