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`
Emf of given cell, when concentration of `M^(2+)` ion and concentration of `Ag^(+)` ion both are 0.01 M

Options

A. 3.17
B. 3.11
C. 3.01
D. 3.21

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the half-cell reactions
The half-cell reactions for the metal-insoluble salt electrode are:
- **Anode (oxidation)**: \( M(s) \rightarrow M^{2+}(aq) + 2e^- \)
- **Cathode (reduction)**: \( Ag^+(aq) + e^- \rightarrow Ag(s) \)

### Step 2: Write the overall cell reaction
To combine the half-cell reactions, we need to balance the number of electrons. Since the cathode reaction involves one electron, we will multiply the cathode reaction by 2:
- Overall reaction:
\[ M(s) + 2Ag^+(aq) \rightarrow Ag(s) + M^{2+}(aq) \]

### Step 3: Calculate the standard cell potential \( E^\circ_{cell} \)
The standard cell potential can be calculated using the standard reduction potentials given:
- \( E^\circ_{M/M^{2+}} = +2.37 \, V \) (oxidation potential)
- \( E^\circ_{Ag/Ag^+} = +0.80 \, V \) (reduction potential)

The standard cell potential is given by:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 0.80 \, V - 2.37 \, V = -1.57 \, V \]

### Step 4: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) is defined as:
\[ Q = \frac{[M^{2+}]}{[Ag^+]^2} \]
Given that both concentrations are \( 0.01 \, M \):
\[ Q = \frac{0.01}{(0.01)^2} = \frac{0.01}{0.0001} = 100 \]

### Step 5: Use the Nernst equation to find the cell potential
The Nernst equation is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q \]
Where \( n \) is the number of moles of electrons transferred in the balanced equation. Here, \( n = 2 \) (from the overall reaction).

Substituting the values:
\[ E_{cell} = -1.57 \, V - \frac{0.0592}{2} \log(100) \]
Calculating \( \log(100) = 2 \):
\[ E_{cell} = -1.57 \, V - \frac{0.0592}{2} \cdot 2 \]
\[ E_{cell} = -1.57 \, V - 0.0592 \]
\[ E_{cell} = -1.57 \, V - 0.0592 = -1.6292 \, V \]

### Step 6: Final Result
The EMF of the given cell is approximately \( -1.63 \, V \).