Calculate the e.m.f. of the half-cell given below. Pt, H2 | HCl at 1-atmosphere pressure and 0.1 M. Given, E°(OP) = 2 V.

Options

A. 4 V
B. 5.6 V
C. 3.4 V
D. 5.4 V

Answer

To calculate the e.m.f. (electromotive force) of the given half-cell reaction, we will use the Nernst equation. The half-cell reaction is represented as follows:

**Reaction:**
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]

**Given:**
- Standard oxidation potential \( E^\circ = 2 \, \text{V} \)
- Concentration of HCl \( = 0.1 \, \text{M} \) (which means \([\text{H}^+] = 0.1 \, \text{M}\))
- Pressure of \( \text{H}_2 = 1 \, \text{atm} \)

### Step 1: Write the Nernst Equation
The Nernst equation for the half-cell reaction is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \right) \]
where:
- \( E \) is the cell potential
- \( E^\circ \) is the standard cell potential
- \( n \) is the number of moles of electrons transferred (in this case, \( n = 2 \))
- \([\text{H}^+]\) is the concentration of hydrogen ions
- \( P_{\text{H}_2} \) is the pressure of hydrogen gas

### Step 2: Substitute the Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ E = 2 \, \text{V} - \frac{0.059}{2} \log \left( \frac{(0.1)^2}{1} \right) \]

### Step 3: Calculate the Logarithmic Term
Calculate the logarithmic term:
\[ \log \left( \frac{(0.1)^2}{1} \right) = \log(0.01) = -2 \]

### Step 4: Substitute the Logarithmic Value Back into the Equation
Now substitute this value back into the equation:
\[ E = 2 \, \text{V} - \frac{0.059}{2} \times (-2) \]

### Step 5: Simplify the Equation
Calculate:
\[ E = 2 \, \text{V} + 0.059 \]
\[ E = 2 \, \text{V} + 0.059 = 2.059 \, \text{V} \]

### Step 6: Final Calculation
Thus, the e.m.f. of the half-cell is:
\[ E \approx 2.06 \, \text{V} \]

### Conclusion
The e.m.f. of the half-cell is approximately **2.06 V**.

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