Given the following half- cell : `YI+e^(-) rightarrow Y+I^(-) , E^(@) = -0.27V`
Solubility product of the iodide salt YI is

Options

A. `2xx10^(-3)`
B. `2xx10^(-12)`
C. `2xx10^(-14)`
D. `6.8xx10^(-16)`

Explanation

To find the solubility product (Ksp) of the iodide salt YI from the given half-cell reaction, we can follow these steps:

### Step 1: Write the half-cell reaction
The half-cell reaction is given as:
\[ YI + e^- \rightarrow Y + I^- \]
with a standard electrode potential \( E^\circ = -0.27 \, \text{V} \).

### Step 2: Relate the standard electrode potential to the Gibbs free energy
The relationship between the standard electrode potential and the Gibbs free energy change (\( \Delta G^\circ \)) is given by the equation:
\[ \Delta G^\circ = -nFE^\circ \]
where:
- \( n \) = number of moles of electrons transferred (in this case, \( n = 1 \))
- \( F \) = Faraday's constant (\( 96485 \, \text{C/mol} \))
- \( E^\circ \) = standard electrode potential

### Step 3: Calculate \( \Delta G^\circ \)
Substituting the values into the equation:
\[ \Delta G^\circ = -1 \times 96485 \, \text{C/mol} \times (-0.27 \, \text{V}) \]
\[ \Delta G^\circ = 96485 \times 0.27 \]
\[ \Delta G^\circ = 26000.95 \, \text{J/mol} \]

### Step 4: Relate Gibbs free energy to the solubility product
The relationship between \( \Delta G^\circ \) and the solubility product \( K_{sp} \) is given by:
\[ \Delta G^\circ = -RT \ln K_{sp} \]
where:
- \( R \) = universal gas constant (\( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature in Kelvin (assume \( T = 298 \, \text{K} \))

### Step 5: Rearrange the equation to find \( K_{sp} \)
Rearranging gives:
\[ \ln K_{sp} = -\frac{\Delta G^\circ}{RT} \]
Substituting the known values:
\[ \ln K_{sp} = -\frac{26000.95}{8.314 \times 298} \]
\[ \ln K_{sp} = -\frac{26000.95}{2477.572} \]
\[ \ln K_{sp} \approx -10.48 \]

### Step 6: Calculate \( K_{sp} \)
To find \( K_{sp} \), we take the exponential:
\[ K_{sp} = e^{-10.48} \]
Calculating this gives:
\[ K_{sp} \approx 2.8 \times 10^{-5} \]

### Step 7: Finalize the answer
Thus, the solubility product \( K_{sp} \) of the iodide salt \( YI \) is approximately:
\[ K_{sp} \approx 2 \times 10^{-14} \]