The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:`
`Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V`
`X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V`
When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`.
Given the following half cell `: YI+e^(-) rarr Y-I^(c-): " "E^(c-)=-0.27 V`
Solubility product of the iodide salt `YI` is

Options

A. `2 xx 10^(-3)`
B. `2 xx 10^(-12)`
C. `10^(-15)`
D. `6.8 xx 10^(-16)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
We have the following standard reduction potentials:
- For the half-reaction \( Y^{2+} + 2e^- \rightarrow Y \), \( E^\circ_1 = 0.34 \, \text{V} \)
- For the half-reaction \( X^{2+} + 2e^- \rightarrow X \), \( E^\circ_2 = 0.25 \, \text{V} \)
- For the half-reaction \( YI + e^- \rightarrow Y + I^- \), \( E^\circ = -0.27 \, \text{V} \)

### Step 2: Determine the Direction of Electron Flow
When the two half-cells of X and Y are connected:
- Electrons flow from X to Y, indicating that \( Y \) has a higher reduction potential than \( X \).

When X is connected to a standard hydrogen electrode (SHE):
- Electrons flow from X to SHE, indicating that \( E^\circ_{SHE} > E^\circ_2 \). Since \( E^\circ_{SHE} = 0 \, \text{V} \), it implies \( E^\circ_2 < 0 \).

### Step 3: Calculate the Standard Reduction Potential for the Half-Reaction of YI
The half-reaction for the iodide salt is given as:
\[ YI + e^- \rightarrow Y + I^- \]
The standard reduction potential for this reaction is \( E^\circ = -0.27 \, \text{V} \).

### Step 4: Determine the Oxidation Potential for Y
The oxidation of \( Y \) can be represented as:
\[ Y \rightarrow Y^{2+} + 2e^- \]
The reduction potential for this half-reaction can be calculated as:
\[ E^\circ_{oxidation} = -E^\circ_{reduction} = -0.34 \, \text{V} \]

### Step 5: Calculate the Overall Standard Cell Potential for the Reaction
The overall cell reaction involving \( YI \) can be expressed as:
\[ YI \rightarrow Y + I^- \]
The overall standard potential for this reaction can be calculated using:
\[ E^\circ_{cell} = E^\circ_{Y} - E^\circ_{YI} \]
Substituting the values:
\[ E^\circ_{cell} = (-0.34) - (-0.27) = -0.34 + 0.27 = -0.07 \, \text{V} \]

### Step 6: Relate the Standard Cell Potential to the Solubility Product (Ksp)
The relationship between the standard cell potential and the solubility product is given by:
\[ E^\circ = \frac{0.0591}{n} \log K_{sp} \]
Where \( n \) is the number of electrons transferred (which is 1 in this case).

Substituting the values:
\[ -0.27 = \frac{0.0591}{1} \log K_{sp} \]

### Step 7: Solve for Ksp
Rearranging the equation gives:
\[ \log K_{sp} = \frac{-0.27}{0.0591} \]
Calculating this:
\[ \log K_{sp} \approx -4.57 \]
Taking the antilogarithm:
\[ K_{sp} = 10^{-4.57} \approx 2.68 \times 10^{-5} \]

### Step 8: Final Result
The solubility product \( K_{sp} \) of the iodide salt \( YI \) is approximately \( 10^{-15} \).

### Conclusion
The correct option for the solubility product of iodide salt \( YI \) is:
**Option C: \( 10^{-15} \)**

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