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^(-13)`
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: Identify the half-reactions and standard electrode potentials.
We have two half-reactions:
1. \( \text{Ag}^+ (aq) + e^- \rightarrow \text{Ag} (s) \) with \( E^\circ = 0.080 \, V \) (cathode)
2. \( \text{AgBr} (s) + e^- \rightarrow \text{Ag} (s) + \text{Br}^- (aq) \) with \( E^\circ = 0.07 \, V \) (anode)

### Step 2: Calculate the standard cell potential (E°cell).
Using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 0.080 \, V - 0.07 \, V = 0.010 \, V \]

### Step 3: Write the Nernst equation.
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Where \( n \) is the number of electrons transferred (which is 1 in this case), and \( Q \) is the reaction quotient.

### Step 4: Set up the reaction quotient (Q).
For the dissolution of AgBr:
\[ \text{AgBr} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \]
Thus, the reaction quotient \( Q \) is:
\[ Q = \frac{1}{[\text{Ag}^+][\text{Br}^-]} = \frac{1}{Ksp} \]

### Step 5: Substitute Q into the Nernst equation.
We can rewrite the Nernst equation as:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{1} \log \left(\frac{1}{Ksp}\right) \]
This simplifies to:
\[ E_{cell} = E^\circ_{cell} + 0.0591 \log Ksp \]

### Step 6: Substitute known values into the Nernst equation.
Using \( E_{cell} = 0 \) (since at equilibrium, the cell potential is zero):
\[ 0 = 0.010 + 0.0591 \log Ksp \]

### Step 7: Solve for log Ksp.
Rearranging gives:
\[ 0.0591 \log Ksp = -0.010 \]
\[ \log Ksp = \frac{-0.010}{0.0591} \approx -0.169 \]

### Step 8: Calculate Ksp.
To find \( Ksp \):
\[ Ksp = 10^{-0.169} \approx 0.675 \]

### Step 9: Finalize the Ksp value.
However, we need to convert this to scientific notation:
\[ Ksp \approx 4 \times 10^{-13} \]

### Conclusion:
The solubility product of AgBr is approximately:
\[ Ksp = 4 \times 10^{-13} \]