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

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

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

$\begin{array}{c}\text{Ni}\left(\text{s}\right)→{\text{Ni}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}{{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{∘}}_{\text{anode}}=−0.25\text{ V}\\ {\text{2Ag}}^{\text{+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}→2\text{Ag}\left(\text{s}\right){{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{∘}}_{\text{cathode}}=0.80\text{ V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{reduction}}^{\text{°}}{\text{-E}}_{\text{oxidation}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=0.80-(-0.25)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=1.05V}\end{array}$

At ${\text{25}}^{\text{o}}\text{C}$,

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

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

$\begin{array}{c}\text{logK=}\frac{\text{1.05³Õ×2}}{\text{0.0592V}}\\ \text{=35.47}\\ {\text{K=10}}^{\text{35.47}}\\ {\text{=2.95×10}}^{\text{35}}\end{array}$

Equilibrium constant for the reaction between${\text{Ni(s)andAg}}^{\text{+}}{\text{(aq)at25}}^{\text{°}}{\text{°ä¾±²õ2.95×10}}^{\text{35}}$

The redox reaction taking place between $\text{Fe}\left(\text{s}\right){\text{ â¶Ä‰and â¶Ä‰Cr}}^{\text{+3}}\left(\text{aq}\right)$is given below.

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

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

$\begin{array}{l}\text{3Fe}\left(\text{s}\right)→3{\text{Fe}}^{\text{+2}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}{{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{∘}}_{\text{anode}}=−0.44\text{ V}\\ {\text{2Cr}}^{\text{+3}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}→2\text{Cr}\left(\text{s}\right){{\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰E}}^{∘}}_{\text{cathode}}=−0.74\text{ V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.74-(-0.44)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.30V}\end{array}$

At ${\text{25}}^{\text{o}}\text{C}$,

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

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

$\begin{array}{c}\text{logK=}\frac{\text{-0.30³Õ×6}}{\text{0.0592V}}\\ \text{=-30.405}\\ {\text{K=100}}^{\text{-30.405}}\\ {\text{=3.935×10}}^{−31}\end{array}$

Equilibrium constant for the reaction between${\text{Fe(s)andCr}}^{\text{3+}}{\text{(aq)at25}}^{\text{°}}{\text{°ä¾±²õ3.935×10}}^{\text{-31}}$.