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)(HA)=10^(-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 can follow these steps:

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

Here:
- The left side (anode) has hydrogen gas at 0.1 bar and BOH at 0.1 M.
- The right side (cathode) has hydrogen gas at 1 bar and HA at 0.1 M.

### Step 2: Calculate the concentration of \( \text{OH}^- \) ions from \( \text{BOH} \)
Given \( K_b(\text{BOH}) = 10^{-5} \):
\[ [\text{OH}^-] = \sqrt{K_b \cdot [\text{BOH}]} = \sqrt{10^{-5} \cdot 0.1} = \sqrt{10^{-6}} = 10^{-3} \text{ M} \]

### Step 3: Calculate the concentration of \( \text{H}^+ \) ions
Using the ion product of water:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
Substituting \( [\text{OH}^-] \):
\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \text{ M} \]

### Step 4: Calculate the concentration of \( \text{H}^+ \) ions from the weak acid \( \text{HA} \)
Given \( K_a(\text{HA}) = 10^{-7} \):
\[ [\text{H}^+] = \sqrt{K_a \cdot [\text{HA}]} = \sqrt{10^{-7} \cdot 0.1} = \sqrt{10^{-8}} = 10^{-4} \text{ M} \]

### Step 5: Write the half-reactions
For the anode (oxidation):
\[ \text{H}_2(g) \rightarrow 2 \text{H}^+ + 2e^- \]
For the cathode (reduction):
\[ 2 \text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \]

### Step 6: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{H}^+]^2_{\text{anode}} \cdot P_{\text{H}_2}^{\text{cathode}}}{[\text{H}^+]^2_{\text{cathode}} \cdot P_{\text{H}_2}^{\text{anode}}} \]
Substituting the values:
\[ Q = \frac{(10^{-11})^2 \cdot 1}{(10^{-4})^2 \cdot 0.1} = \frac{10^{-22}}{10^{-8} \cdot 0.1} = \frac{10^{-22}}{10^{-9}} = 10^{-13} \]

### Step 7: Use the Nernst equation to calculate \( E_{cell} \)
The Nernst equation is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \]
For the hydrogen electrode, \( E^\circ_{cell} = 0 \) and \( n = 2 \):
\[ E_{cell} = 0 - \frac{0.059}{2} \log(10^{-13}) = -0.0295 \cdot (-13) = 0.3835 \text{ V} \]

### Step 8: Final result
Thus, the cell potential \( E_{cell} \) is approximately:
\[ E_{cell} \approx 0.39 \text{ V} \]