Calculate the cell potential of following cell
`Pt(s)|H_(2)(g)"(0.1 bar)"|BOH(0.1M)||HA(0.1M)|H_(2)(g)("1 bar")|Pt`
Given
`K(a)(HA10^(-7),K_(b)(BOH)=10^(-5)`

Options

A. 0.39 V
B. 0.36 V
C. 0.93 V
D. 0.63 V

Answer

To calculate the cell potential of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the Components of the Cell
The cell is given as:
\[ \text{Pt(s)} | \text{H}_2(g)(0.1 \text{ bar}) | \text{BOH}(0.1M) || \text{HA}(0.1M) | \text{H}_2(g)(1 \text{ bar}) | \text{Pt} \]

Here, we have:
- Anode: \( \text{H}_2(g)(0.1 \text{ bar}) \) in the presence of \( \text{BOH}(0.1M) \)
- Cathode: \( \text{H}_2(g)(1 \text{ bar}) \) in the presence of \( \text{HA}(0.1M) \)

### Step 2: Calculate \( \text{[OH}^-] \) for \( \text{BOH} \)
Using the formula for the concentration of hydroxide ions from a weak base:
\[ [\text{OH}^-] = \sqrt{K_b \times C} \]
Where \( K_b = 10^{-5} \) and \( C = 0.1 \):
\[ [\text{OH}^-] = \sqrt{10^{-5} \times 0.1} = \sqrt{10^{-6}} = 10^{-3} \text{ M} \]

### Step 3: Calculate \( [\text{H}^+] \) for \( \text{BOH} \)
Using the water dissociation constant:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
Thus,
\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \text{ M} \]

### Step 4: Calculate \( [\text{H}^+] \) for \( \text{HA} \)
Using the formula for the concentration of hydrogen ions from a weak acid:
\[ [\text{H}^+] = \sqrt{K_a \times C} \]
Where \( K_a = 10^{-7} \) and \( C = 0.1 \):
\[ [\text{H}^+] = \sqrt{10^{-7} \times 0.1} = \sqrt{10^{-8}} = 10^{-4} \text{ M} \]

### Step 5: Write the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log Q \]
Where \( n = 2 \) (number of electrons transferred).

### Step 6: Calculate the Reaction Quotient \( Q \)
For the cell reaction:
\[ Q = \frac{[\text{H}^+]^2 \cdot P_{\text{H}_2}^{\text{anode}}}{[\text{H}^+]^2 \cdot P_{\text{H}_2}^{\text{cathode}}} = \frac{(10^{-11})^2 \cdot (0.1)}{(10^{-4})^2 \cdot (1)} \]
Calculating \( Q \):
\[ Q = \frac{10^{-22} \cdot 0.1}{10^{-8}} = 10^{-22} \cdot 10^{8} = 10^{-14} \]

### Step 7: Substitute into the Nernst Equation
Now substituting \( Q \) into the Nernst equation:
\[ E = 0 - \frac{0.059}{2} \log(10^{-14}) \]
Calculating:
\[ E = -0.0295 \cdot (-14) = 0.413 \text{ V} \]

### Final Answer
The cell potential \( E \) is approximately:
\[ E \approx 0.413 \text{ V} \]