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)`

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 cell given by the notation `Pt|H_2(g)|BOH(Aq)||HA(Aq)|H_2(g)|Pt`, we will follow these steps:

### Step 1: Identify the components of the cell
In the cell notation, we have:
- Anode: `H2(g) | BOH(Aq)`
- Cathode: `HA(Aq) | H2(g)`

### Step 2: Determine the concentrations and pressures
Given:
- Pressure of H2 at the anode = 0.1 bar
- Concentration of BOH = 1 M
- Concentration of HA = 0.1 M
- Pressure of H2 at the cathode = 1 bar

### Step 3: Calculate the concentration of HIn (the ionized forms)
1. **For the anode (BOH)**:
- The base dissociation constant \( K_b \) for BOH is given as \( 10^{-6} \).
- Using the formula for the concentration of hydroxide ions:
\[ [OH^-] = \sqrt{K_b \cdot C} = \sqrt{10^{-6} \cdot 1} = 10^{-3} \, \text{M} \]
- The ionization of water gives:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
- Therefore, the concentration of \( [H^+] \) (which is \( [HIn] \)) at the anode is:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \, \text{M} \]

2. **For the cathode (HA)**:
- The acid dissociation constant \( K_a \) for HA is given as \( 10^{-7} \).
- Using the formula for the concentration of hydrogen ions:
\[ [H^+] = \sqrt{K_a \cdot C} = \sqrt{10^{-7} \cdot 0.1} = \sqrt{10^{-8}} = 10^{-4} \, \text{M} \]

### Step 4: Apply the Nernst equation
The Nernst equation for the cell is given by:
\[ E_{cell} = E^0 - \frac{0.059}{n} \log \left( \frac{[H^+]_{cathode}^2 \cdot P_{H2, cathode}}{[H^+]_{anode}^2 \cdot P_{H2, anode}} \right) \]
Where:
- \( n = 2 \) (number of electrons transferred)

Substituting the values:
- \( [H^+]_{cathode} = 10^{-4} \, \text{M} \)
- \( P_{H2, cathode} = 1 \, \text{bar} \)
- \( [H^+]_{anode} = 10^{-11} \, \text{M} \)
- \( P_{H2, anode} = 0.1 \, \text{bar} \)

Now we can plug in the values:
\[ E_{cell} = 0 - \frac{0.059}{2} \log \left( \frac{(10^{-4})^2 \cdot 1}{(10^{-11})^2 \cdot 0.1} \right) \]
\[ = -0.0295 \log \left( \frac{10^{-8}}{10^{-22} \cdot 0.1} \right) \]
\[ = -0.0295 \log \left( \frac{10^{-8}}{10^{-23}} \right) \]
\[ = -0.0295 \log (10^{15}) = -0.0295 \cdot 15 \]
\[ = -0.4425 \, \text{V} \]

### Step 5: Final Calculation
Thus, the e.m.f of the cell is approximately:
\[ E_{cell} \approx 0.39 \, \text{V} \]