The metal-insoluble salt electrode consists of a metal M coated with a porous insoluble salt MX In a solution of `X^(-),` A good example is Ihe silver, silver-chloride electrode for which the half-cell reaction is `AgCI(s)+e Leftrightarrow Ag(s) + CI (aq),` where, the reduction of sold silver chloride produces solid silver and releases chloride ion into solute for the cell
`M(s)|M^(2+) (aq)||Ag^(+) (aq)|Ag(s)" "E_(M//M^(2+))^(@)=+2.37V and E_(Ag//Ag)^(@)=+0.80V`
If cell reaction is in equilibrium stage, value of logarithms of equilibrium constant is

Options

A. 1.0727
B. 107.27
C. 1072.7
D. 10.727

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the half-cell reactions
The half-cell reactions for the given electrochemical cell are:
- At the anode (oxidation):
\[ M(s) \rightarrow M^{2+}(aq) + 2e^- \]
- At the cathode (reduction):
\[ Ag^+(aq) + e^- \rightarrow Ag(s) \]

### Step 2: Balance the overall cell reaction
To balance the overall cell reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. Since the cathode reaction involves 1 electron, we need to multiply it by 2 to match the 2 electrons lost at the anode:
\[ M(s) + 2Ag^+(aq) \rightarrow M^{2+}(aq) + 2Ag(s) \]

### Step 3: Determine the standard cell potential (E°cell)
The standard cell potential can be calculated by adding the standard reduction potentials of the half-reactions:
- For \(M/M^{2+}\): \(E^\circ_{M/M^{2+}} = +2.37 \, V\)
- For \(Ag/Ag^+\): \(E^\circ_{Ag/Ag^+} = +0.80 \, V\)

Thus, the overall standard cell potential is:
\[ E^\circ_{cell} = E^\circ_{M/M^{2+}} - E^\circ_{Ag/Ag^+} = 2.37 \, V - 0.80 \, V = 1.57 \, V \]

### Step 4: Apply the Nernst equation at equilibrium
At equilibrium, the cell potential \(E_{cell} = 0\). The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log K \]
Where \(n\) is the number of moles of electrons transferred in the balanced equation. Here, \(n = 2\).

Setting \(E_{cell} = 0\) gives:
\[ 0 = E^\circ_{cell} - \frac{0.0591}{2} \log K \]

### Step 5: Solve for log K
Rearranging the equation to solve for \(\log K\):
\[ \log K = \frac{2 \cdot E^\circ_{cell}}{0.0591} \]

Substituting \(E^\circ_{cell} = 1.57 \, V\):
\[ \log K = \frac{2 \cdot 1.57}{0.0591} = \frac{3.14}{0.0591} \approx 53.1 \]

### Conclusion
Thus, the logarithm of the equilibrium constant \(K\) is approximately \(53.1\).