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 solubility product constant (Ksp) for AgI, we can follow these steps:

### Step 1: Write the half-reactions and their standard electrode potentials (E°)

We have two half-reactions:
1. \( \text{Ag}^+ + e^- \rightarrow \text{AgI} \) with \( E^\circ = 0.152 \, \text{V} \)
2. \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \) with \( E^\circ = -0.800 \, \text{V} \)

### Step 2: Determine the overall cell reaction

The overall reaction for the formation of AgI from Ag+ and I- can be written as:
\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} + e^- \]

### Step 3: Calculate the standard cell potential (E°cell)

The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Here, Ag+ is reduced (cathode) and Ag is oxidized (anode):
- \( E^\circ_{\text{cathode}} = 0.152 \, \text{V} \)
- \( E^\circ_{\text{anode}} = -0.800 \, \text{V} \)

Thus,
\[ E^\circ_{\text{cell}} = 0.152 - (-0.800) = 0.152 + 0.800 = 0.952 \, \text{V} \]

### Step 4: Relate E°cell to Ksp using the Nernst equation

The Nernst equation relates the standard cell potential to the equilibrium constant (Ksp):
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_{sp} \]

Where \( n \) is the number of moles of electrons transferred in the balanced equation. Here, \( n = 1 \).

### Step 5: Rearranging the equation to solve for Ksp

Rearranging the Nernst equation gives:
\[ \log K_{sp} = \frac{n \cdot E^\circ_{\text{cell}}}{0.0591} \]

Substituting \( E^\circ_{\text{cell}} = 0.952 \, \text{V} \) and \( n = 1 \):
\[ \log K_{sp} = \frac{1 \cdot 0.952}{0.0591} \]

Calculating this gives:
\[ \log K_{sp} \approx 16.094 \]

### Step 6: Convert from log Ksp to Ksp

To find Ksp, we take the antilog:
\[ K_{sp} = 10^{16.094} \]

Calculating this gives:
\[ K_{sp} \approx 1.23 \times 10^{16} \]

### Final Answer

Thus, the value of \( K_{sp} \) for AgI is approximately \( 1.23 \times 10^{16} \).

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