calculate the e.m.f (in V) of the cell:
`Pt|H_2(g)|BOH(Aq)"||"HA(Aq)|H_2(g)|Pt`,
0.1bar 1M 0.1M 1bar
Given : `K_a(HA)=10^(-7), K_b(BOH)=10^(-6)`
(a)0.39V
(b)0.36V
(c)0.93V
(d)None of these

Options

A. 0.39V
B. 0.36V
C. 0.93V
D. None of these

Answer

To calculate the electromotive force (e.m.f) of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions
In the given cell notation:
\[ \text{Pt} | \text{H}_2(g) | \text{BOH}(aq) || \text{HA}(aq) | \text{H}_2(g) | \text{Pt} \]

- The left side (anode) involves the oxidation of hydrogen gas to protons:
\[ \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \]

- The right side (cathode) involves the reduction of protons to hydrogen gas:
\[ 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \]

### Step 2: Calculate the concentrations of \( \text{H}^+ \) ions
1. **For the right side (HA)**:
- Given \( K_a(HA) = 10^{-7} \) and concentration \( [HA] = 0.1 \, M \).
- Using the formula for \( K_a \):
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[HA]} \implies [\text{H}^+]^2 = K_a \cdot [HA] \implies [\text{H}^+]^2 = 10^{-7} \cdot 0.1 = 10^{-8} \]
\[ [\text{H}^+] = \sqrt{10^{-8}} = 10^{-4} \, M \]

2. **For the left side (BOH)**:
- Given \( K_b(BOH) = 10^{-6} \) and concentration \( [BOH] = 0.1 \, M \).
- Using the formula for \( K_b \):
\[ K_b = \frac{[\text{OH}^-][\text{B}^+]}{[BOH]} \implies [\text{OH}^-]^2 = K_b \cdot [BOH] \implies [\text{OH}^-]^2 = 10^{-6} \cdot 0.1 = 10^{-7} \]
\[ [\text{OH}^-] = \sqrt{10^{-7}} = 10^{-3} \, M \]

### Step 3: Calculate \( [\text{H}^+] \) from \( [\text{OH}^-] \)
Using the relation:
\[ [\text{H}^+][\text{OH}^-] = 10^{-14} \]
We can find \( [\text{H}^+] \) on the left side:
\[ [\text{H}^+] = \frac{10^{-14}}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \, M \]

### Step 4: Apply the Nernst Equation
The Nernst equation for the cell is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{H}^+]^2 \cdot P_{\text{H}_2}}{[\text{H}^+]^2} \right) \]
Where \( n = 2 \) (number of electrons transferred).

Given:
- \( E^\circ = 0 \) (since both half-reactions involve hydrogen)
- \( P_{\text{H}_2} = 0.1 \, \text{bar} \)

Substituting the values:
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{(10^{-4})^2 \cdot 0.1}{(10^{-11})^2} \right) \]
Calculating the log term:
\[ = \frac{(10^{-8}) \cdot 0.1}{10^{-22}} = 10^{14} \]
Thus:
\[ E = -\frac{0.0591}{2} \log(10^{14}) = -\frac{0.0591}{2} \cdot 14 \]
Calculating:
\[ E = -0.0591 \cdot 7 = -0.4137 \, \text{V} \]
Since we are calculating e.m.f, we take the absolute value:
\[ E = 0.4137 \, \text{V} \]

### Step 5: Match with the options
The closest option to our calculated e.m.f is:
(a) 0.39 V.

### Final Answer
The e.m.f of the cell is **0.39 V**.