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^(-): " "E^(c-)=-0.27 V`
Solubility product of the iodide salt `YI` is
(a)`2xx10^(-3)`
(b)`2xx10^(-12)`
(c)`2xx10^(-14)`
(d)`6.8xx10^(-16)`

Options

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

Explanation

To solve the problem, we need to find the solubility product (Ksp) of the iodide salt YI based on the given standard reduction potentials and the half-cell reactions. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the half-cell reactions
We have the following half-cell reactions:
1. For metal Y:
\[ Y^{2+} + 2e^- \rightarrow Y \quad |E^\circ = 0.34 \, \text{V} \]
2. For metal X:
\[ X^{2+} + 2e^- \rightarrow X \quad |E^\circ = 0.25 \, \text{V} \]
3. For the iodide salt YI:
\[ YI + e^- \rightarrow Y + I^- \quad |E^\circ = -0.27 \, \text{V} \]

### Step 2: Determine the net cell reaction
When YI dissociates, it forms Y and I^- ions. The oxidation reaction at the anode (where oxidation occurs) will involve Y losing an electron:
\[ Y \rightarrow Y^{2+} + 2e^- \]
The overall reaction can be written as:
\[ YI \rightarrow Y^{2+} + I^- + e^- \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
In our case, the cathode is the half-cell with YI, and the anode is the oxidation of Y. Thus:
\[ E^\circ_{\text{cell}} = (-0.27 \, \text{V}) - (0.34 \, \text{V}) = -0.61 \, \text{V} \]

### Step 4: Relate E°cell to Ksp
Using the Nernst equation for the solubility product:
\[ \log K_{sp} = \frac{E^\circ_{\text{cell}} \cdot n}{0.0592} \]
Where \( n \) is the number of electrons transferred (which is 1 in this case).

Substituting the values:
\[ \log K_{sp} = \frac{-0.61 \cdot 1}{0.0592} \approx -10.31 \]

### Step 5: Calculate Ksp
To find Ksp, we take the antilog:
\[ K_{sp} = 10^{-10.31} \approx 4.89 \times 10^{-11} \]

### Step 6: Select the closest option
From the options provided:
- (a) \( 2 \times 10^{-3} \)
- (b) \( 2 \times 10^{-12} \)
- (c) \( 2 \times 10^{-14} \)
- (d) \( 6.8 \times 10^{-16} \)

The closest answer to our calculated Ksp value \( 4.89 \times 10^{-11} \) is option (c) \( 2 \times 10^{-14} \).

### Final Answer:
The solubility product of the iodide salt YI is approximately \( 2 \times 10^{-14} \).

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