91Ó°ÊÓ

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

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

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

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

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

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

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

$\begin{array}{l}\text{2Cr}\left(\text{s}\right)→{\text{2Cr}}^{\text{+3}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}{{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{∘}}_{\text{anode}}=−0.73\text{ V}\\ {\text{3Cu}}^{\text{+2}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}→3\text{Cu}\left(\text{s}\right){{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰E}}^{∘}}_{\text{cathode}}=0.34\text{ V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=0.34³Õ-(-0.73 V)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=1.07V}\end{array}$

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=6.

$\begin{array}{l}\text{±ô´Ç²µ°­â€‰= }\frac{\text{1.07³Õ×6}}{\text{0.0592V}}\\ \text{= â¶Ä‰79.73}\\ {\text{K â¶Ä‰= â¶Ä‰10}}^{\text{79.73}}\\ {\text{ â¶Ä‰= â¶Ä‰5.37×10}}^{\text{79}}\end{array}$

Equilibrium constant for the reaction between${\text{Cr(s)andCu}}^{\text{2+}}{\text{(aq)at25}}^{\text{°}}{\text{°ä¾±²õ5.37×10}}^{\text{79}}$.

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

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

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

$\begin{array}{l}\text{Sn}\left(\text{s}\right)→{\text{Sn}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}{{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{∘}}_{\text{anode}}\text{= -0.14 V}\\ {\text{Pb}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}→\text{Pb}\left(\text{s}\right){{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰E}}^{∘}}_{\text{cathode}}=−0.13\text{ V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.13³Õ-(-0.14 V)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=0.01V}\end{array}$

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.

$\begin{array}{c}\text{logK=}\frac{\text{0.01³Õ×2}}{\text{0.0592V}}\\ \text{=0.34}\\ {\text{K=10}}^{\text{0.34}}\\ \text{=2.19}\end{array}$

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