Calculate the value of solubility product for solid AgI by using following data :
`Ag^+(aq)+e^(-)toAg(s)E^(@)=0.800` volt
`AgI(s)+e^(-)toAg(s)+I^(-)(aq)E^(@)=-0.164` volt

Options

A. `5xx10^(-17)`
B. `6xx10^(-15)`
C. `2xx10^(-14)`
D. `4xx10^(-12)`

Answer

To calculate the solubility product (Ksp) for solid AgI, we will follow these steps:

### Step 1: Write the Dissociation Reaction
When AgI dissolves in water, it dissociates into silver ions (Ag⁺) and iodide ions (I⁻):
\[ \text{AgI(s)} \rightleftharpoons \text{Ag}^+(aq) + \text{I}^-(aq) \]

### Step 2: Identify the Half-Reactions
We have two half-reactions provided:
1. \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \) with \( E^\circ = 0.800 \, \text{V} \)
2. \( \text{AgI}(s) + e^- \rightarrow \text{Ag}(s) + \text{I}^-(aq) \) with \( E^\circ = -0.164 \, \text{V} \)

### Step 3: Determine the Overall Reaction
To find the overall cell reaction, we will subtract the first half-reaction from the second:
\[ \text{AgI}(s) + e^- \rightarrow \text{Ag}(s) + \text{I}^-(aq) \\ - \left( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \right) \]
This simplifies to:
\[ \text{AgI}(s) \rightarrow \text{Ag}^+(aq) + \text{I}^-(aq) \]

### Step 4: Calculate the Cell Potential
The cell potential (E°cell) can be calculated by subtracting the potentials of the half-reactions:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.164) - (0.800) = -0.964 \, \text{V} \]

### Step 5: Use the Nernst Equation
According to the Nernst equation:
\[ E^\circ_{\text{cell}} = \frac{0.059}{n} \log K_{sp} \]
where \( n \) is the number of electrons transferred. In this case, \( n = 1 \).

### Step 6: Substitute and Solve for Ksp
Substituting the values into the Nernst equation:
\[ -0.964 = \frac{0.059}{1} \log K_{sp} \]
Rearranging gives:
\[ \log K_{sp} = \frac{-0.964}{0.059} \approx -16.13 \]
Now, converting from logarithmic form to Ksp:
\[ K_{sp} = 10^{-16.13} \approx 5 \times 10^{-17} \]

### Final Answer
The solubility product \( K_{sp} \) for AgI is:
\[ K_{sp} \approx 5 \times 10^{-17} \]

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