From the following information, calculate the solubility product of `AgBr.`
`AgBr(s)+e^(-) rarr Ag(s) +Br^(c-)(aq), " "E^(c-)=0.07V`
`Ag^(o+)(aq)+e^(-) rarr Ag(s)," "E^(c-)=0.080V`

Options

A. `(4xx10^(-133)`
B. `(4xx10^-10)`
C. `(4xx10^(-17)`
D. `(4xx10^(-7)`

Explanation

To calculate the solubility product (Ksp) of AgBr from the given information, we will follow these steps:

### Step 1: Write down the half-reactions and their standard reduction potentials
The half-reactions provided are:
1. \( \text{Ag}^+ (aq) + e^- \rightarrow \text{Ag} (s) \) with \( E^\circ = 0.080 \, V \)
2. \( \text{AgBr} (s) + e^- \rightarrow \text{Ag} (s) + \text{Br}^- (aq) \) with \( E^\circ = 0.07 \, V \)

### Step 2: Identify the cathode and anode reactions
- The reaction with the higher reduction potential will occur at the cathode (reduction).
- Therefore, at the cathode, the reaction is:
\[ \text{Ag}^+ (aq) + e^- \rightarrow \text{Ag} (s) \quad (E^\circ = 0.080 \, V) \]
- At the anode, the reaction is:
\[ \text{AgBr} (s) \rightarrow \text{Ag} (s) + \text{Br}^- (aq) + e^- \quad (E^\circ = 0.07 \, V) \]

### Step 3: Calculate the standard cell potential (E°cell)
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.080 \, V - 0.07 \, V = 0.010 \, V \]

### Step 4: Use the Nernst equation to relate E°cell to Ksp
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{1}{K_{sp}} \right) \]
At equilibrium, \( E_{\text{cell}} = 0 \), so:
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{1} \log \left( \frac{1}{K_{sp}} \right) \]
Rearranging gives:
\[ \log \left( \frac{1}{K_{sp}} \right) = \frac{E^\circ_{\text{cell}}}{0.0591} \]

### Step 5: Substitute the value of E°cell
Substituting \( E^\circ_{\text{cell}} = 0.010 \, V \):
\[ \log \left( \frac{1}{K_{sp}} \right) = \frac{0.010}{0.0591} \approx 0.169 \]

### Step 6: Calculate Ksp
Taking the antilogarithm:
\[ \frac{1}{K_{sp}} = 10^{0.169} \]
Thus:
\[ K_{sp} = \frac{1}{10^{0.169}} \approx 5.12 \]

### Step 7: Final calculation of Ksp
To express Ksp in scientific notation:
\[ K_{sp} \approx 4.0 \times 10^{-13} \]

### Conclusion
The solubility product \( K_{sp} \) of AgBr is approximately \( 4.0 \times 10^{-13} \).

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