Given the data `25^(@) C`,
`Ag+ I^(-) to AgI+e^(-)` , `E^(@) = 0.152V`
`Ag to Ag^(+) + E(-)`, `E^(@) = - 0.800V`
What is the value of` K_(sp) ` for AgI?

Options

A. Antilog `(-8.12)`
B. antilog `(+8.612)`
C. antilog `(-37.83)`
D. antilog `(-16.1)`

Explanation

To find the value of \( K_{sp} \) for \( AgI \), we will follow these steps:

### Step 1: Write the half-reactions and their standard electrode potentials.

We have the following half-reactions and their standard electrode potentials:

1. \( Ag^+ + e^- \rightarrow Ag \) with \( E^\circ = 0.152 \, V \)
2. \( Ag \rightarrow Ag^+ + e^- \) with \( E^\circ = -0.800 \, V \)

### Step 2: Write the overall reaction for the dissolution of \( AgI \).

The dissolution of silver iodide \( AgI \) can be represented as:
\[ AgI (s) \rightleftharpoons Ag^+ (aq) + I^- (aq) \]

### Step 3: Combine the half-reactions to find the overall cell potential.

To find the overall cell potential for the dissolution of \( AgI \), we need to manipulate the half-reactions. We can reverse the second half-reaction (which will change the sign of its potential) and add it to the first half-reaction:

- Reverse the second reaction:
\[ Ag^+ + e^- \rightarrow Ag \quad (E^\circ = 0.800 \, V) \]

- Now, add the first reaction and the reversed second reaction:
\[ Ag^+ + e^- \rightarrow Ag \quad (E^\circ = 0.152 \, V) \\ Ag \rightarrow Ag^+ + e^- \quad (E^\circ = 0.800 \, V) \]

The overall reaction becomes:
\[ AgI (s) \rightleftharpoons Ag^+ (aq) + I^- (aq) \]

### Step 4: Calculate the cell potential \( E^\circ_{cell} \).

Now, we can calculate the overall cell potential:
\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} = 0.152 \, V - (-0.800 \, V) = 0.152 \, V + 0.800 \, V = 0.952 \, V \]

### Step 5: Use the Nernst equation to find \( K_{sp} \).

The relationship between the standard cell potential and the solubility product constant \( K_{sp} \) is given by:
\[ E^\circ_{cell} = \frac{0.059}{n} \log K_{sp} \]
where \( n \) is the number of electrons transferred. In this case, \( n = 1 \).

Substituting the values:
\[ 0.952 = \frac{0.059}{1} \log K_{sp} \]

### Step 6: Solve for \( \log K_{sp} \).

Rearranging the equation gives:
\[ \log K_{sp} = \frac{0.952}{0.059} \approx 16.1 \]

### Step 7: Calculate \( K_{sp} \).

Now, we can find \( K_{sp} \) by taking the antilogarithm:
\[ K_{sp} = 10^{16.1} \approx 1.26 \times 10^{16} \]

### Final Answer:
Thus, the value of \( K_{sp} \) for \( AgI \) is approximately \( 1.26 \times 10^{16} \).

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