Calculate the electrode potential of the given electrode. Pt, `Cl_2`(2 bar)| 2`Cl^–(0.02 M)`; E°`(Cl_2 | 2Cl^–)` = 3.4 V

Options

A. 3.51 V
B. 3.55 V
C. 1.26 V
D. 2.95 V

Answer

To calculate the electrode potential of the given electrode, we will use the Nernst equation. The Nernst equation relates the standard electrode potential to the actual electrode potential under non-standard conditions.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The half-reaction for the given electrode is:
\[ \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^- \]

2. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log Q \]
where:
- \( E \) = electrode potential
- \( E^\circ \) = standard electrode potential
- \( n \) = number of moles of electrons transferred in the reaction
- \( Q \) = reaction quotient

3. **Substitute Known Values**:
From the problem, we know:
- \( E^\circ = 3.4 \, \text{V} \)
- \( n = 2 \) (since 2 electrons are involved)
- Concentration of \( \text{Cl}^- = 0.02 \, \text{M} \)
- Pressure of \( \text{Cl}_2 = 2 \, \text{bar} \)

The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Cl}^-]^2}{P_{\text{Cl}_2}} \]
Substituting the values:
\[ Q = \frac{(0.02)^2}{2} = \frac{0.0004}{2} = 0.0002 \]

4. **Calculate the Logarithm**:
Now, we calculate \( \log Q \):
\[ \log(0.0002) = \log(2 \times 10^{-4}) = \log(2) + \log(10^{-4}) \]
\[ \log(2) \approx 0.301 \quad \text{and} \quad \log(10^{-4}) = -4 \]
\[ \log(0.0002) \approx 0.301 - 4 = -3.699 \]

5. **Substitute into the Nernst Equation**:
Now, substitute \( \log Q \) back into the Nernst equation:
\[ E = 3.4 - \frac{0.059}{2} \times (-3.699) \]
\[ E = 3.4 + \frac{0.059 \times 3.699}{2} \]
\[ E = 3.4 + \frac{0.218 \, \text{V}}{2} \]
\[ E = 3.4 + 0.109 = 3.509 \, \text{V} \]

6. **Final Result**:
Rounding off, we get:
\[ E \approx 3.51 \, \text{V} \]