Question

Consider the following electrode process of a cell, $ {Cl^{-1} -> \frac{1}{2} Cl_2 + e^{-}}$ $ {[MCl + e^{-} -> M + Cl^{-} ] }$ If EMF of this cell is $-1.140\, V$ and $E^0$ value of the cell is $-0.55\,V$ at $298\,K$, the value of the equilibrium constant of the sparingly soluble salt MCl is in the order of

• A. $10^{-10}$
• B. $10^{-8}$
• C. $10^{-7}$
• D. $10^{-11}$

Answer 1

The Correct Option is A

\(M Cl + e ^{-} \longrightarrow M+ Cl ^{-}\) cathode (reduction)
\(Cl ^{-} \longrightarrow \frac{1}{2} Cl _{2}+e^{-}\) anode (oxidation)
\(M Cl \rightarrow M+\frac{1}{2} Cl _{2}\)
The \(K_{C}\) of the cell reaction is calculated from 
Nernst equation \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log K_{c}\)
\(-1.140=-0.55-\frac{0.059}{1} \,\log _{C}\)
\(-0.59=-0.059 \log K_{C}\)
\(\log K_{C}=\frac{0.59}{0.059}=10\)
\(\therefore K_{C}=10^{10}\)
\(K_{S p}\) is for \(M+\frac{1}{2} Cl _{2} \longrightarrow M C l \longrightarrow M^{+}+ Cl ^{-}\)
\(\therefore K_{S p}=\frac{1}{K_{C}}\)
\(=\frac{1}{10^{10}}=10^{-10}\)

The given electrode process is:

Cl^-(aq) -> 1/2 Cl2(g) + e^-

The standard electrode potential for this process can be obtained from the reduction potential of chlorine gas and the standard reduction potential of the Cl^-/Cl^2- couple:

E°(Cl2/Cl^-) = 1.36 V (from standard reduction potential tables)

E°(Cl^-/Cl2-) = -E°(Cl2/Cl^-) = -1.36 V

The given EMF of the cell, Ecell, is -1.140 V, which is less negative than the standard reduction potential of the Cl^-/Cl2- couple. This means that the reaction is not at standard conditions and the reaction quotient Q is less than the equilibrium constant K.

The relationship between Ecell, E° and the reaction quotient Q is given by the Nernst equation:

Ecell = E° - (RT/nF) * ln Q

where R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant, and ln is the natural logarithm.

Substituting the values given in the problem, we get:

-1.140 V = -0.55 V - (0.0257 V/K) * ln Q

Solving for ln Q, we get:

ln Q = -22.48

Taking the exponential of both sides, we get:

Q = e^(-22.48) = 4.19 x 10^(-10)

The equilibrium constant K is related to the reaction quotient Q by the equation:

K = Q/[Cl^-]

where [Cl^-] is the concentration of chloride ions in the solution. Since the problem states that MCl is a sparingly soluble salt, we can assume that its concentration is much less than the concentration of chloride ions in the solution. Therefore, we can approximate [Cl^-] to be the same as the initial concentration of Cl^- in the solution, which is usually 1 M.

Thus, the equilibrium constant K is:

K = Q/[Cl^-] = (4.19 x 10^(-10))/1 = 4.19 x 10^(-10)

Therefore, the value of the equilibrium constant of the sparingly soluble salt MCl is in the order of 10^(-10).