Given the folowing standerd electrode potentials, the `K_(sp)` for `PbBr_(2)` is :
`PbBr_(2)(s)+2e^(-)toPb(s)+2Br^(-)(aq), E^(@)=-0.248 V`
`Pb^(2+)(aq)+2e^(-)toPb(s), E^(@)=-0.126 V`

Options

A. `7.4xx10^(-5)`
B. `4.9xx10^(-14)`
C. `5.2xx10^(-6)`
D. `2.3xx10^(-13)`

Answer

To find the solubility product constant (Ksp) for lead(II) bromide (PbBr2) using the given standard electrode potentials, we can follow these steps:

### Step 1: Identify the Reactions and Their Potentials
We have two reactions with their standard electrode potentials:
1. \( \text{PbBr}_2(s) + 2e^- \rightarrow \text{Pb}(s) + 2\text{Br}^-(aq) \), \( E^\circ = -0.248 \, V \)
2. \( \text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \), \( E^\circ = -0.126 \, V \)

### Step 2: Determine the Oxidation and Reduction Reactions
- The first reaction (with \( E^\circ = -0.248 \, V \)) is the oxidation reaction because it involves the conversion of solid PbBr2 into Pb and Br⁻ ions.
- The second reaction (with \( E^\circ = -0.126 \, V \)) is the reduction reaction where Pb²⁺ ions gain electrons to form solid Pb.

### Step 3: Calculate the Standard Cell Potential (E°cell)
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is the reduction of Pb²⁺ and the anode is the oxidation of PbBr2.

\[ E^\circ_{\text{cell}} = (-0.126 \, V) - (-0.248 \, V) = -0.126 + 0.248 = 0.122 \, V \]

### Step 4: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
At equilibrium (for Ksp), \( E_{\text{cell}} = 0 \), and \( Q = Ksp \).

So, we can rewrite the equation as:
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Ksp \]

### Step 5: Substitute Values
Here, \( n = 2 \) (since 2 electrons are involved). Thus:
\[ 0 = 0.122 - \frac{0.0591}{2} \log Ksp \]

Rearranging gives:
\[ \log Ksp = \frac{0.122 \times 2}{0.0591} \]

### Step 6: Calculate log Ksp
Calculating the right side:
\[ \log Ksp = \frac{0.244}{0.0591} \approx 4.136 \]

### Step 7: Find Ksp
Taking the antilog:
\[ Ksp = 10^{-4.136} \approx 7.32 \times 10^{-5} \]

### Conclusion
Thus, the Ksp for PbBr2 is approximately \( 7.32 \times 10^{-5} \).

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