Find the electrode potentials of the following electrodes.
(a) Pt, `H_2`(1 atm)/HCl (0.1 M),
(b) Pt, `H_2`(2 atm)/`H_2SO_4` (0.01 M)

Answer

To find the electrode potentials of the given electrodes, we will use the Nernst equation, which relates the electrode potential to the concentrations of the reactants and products involved in the half-reaction.

### Part (a): Pt, H₂(1 atm)/HCl (0.1 M)

1. **Identify the half-reaction**:
The reduction half-reaction for hydrogen is:
\[ H_2 \rightarrow 2H^+ + 2e^- \]

2. **Determine the standard electrode potential (E°)**:
The standard electrode potential for the hydrogen electrode is defined as 0 V.

3. **Calculate the concentration of H⁺ ions**:
Given that the concentration of HCl is 0.1 M, the concentration of H⁺ ions is also 0.1 M.

4. **Use the Nernst equation**:
The Nernst equation is given by:
\[ E = E° - \frac{RT}{nF} \ln Q \]
Where:
- \(E° = 0 \, V\)
- \(R = 8.314 \, J/(mol \cdot K)\)
- \(T = 298 \, K\)
- \(n = 2\) (number of electrons transferred)
- \(F = 96500 \, C/mol\)
- \(Q = \frac{[H^+]^2}{P_{H_2}}\)

5. **Calculate Q**:
Given that \(P_{H_2} = 1 \, atm\) and \([H^+] = 0.1 \, M\):
\[ Q = \frac{(0.1)^2}{1} = 0.01 \]

6. **Substitute values into Nernst equation**:
\[ E = 0 - \frac{(8.314)(298)}{(2)(96500)} \ln(0.01) \]
\[ E = -0.0591 \cdot \log(0.01) \]
Since \(\log(0.01) = -2\):
\[ E = -0.0591 \cdot (-2) = 0.1182 \, V \]

### Part (b): Pt, H₂(2 atm)/H₂SO₄ (0.01 M)

1. **Identify the half-reaction**:
The reduction half-reaction is the same:
\[ H_2 \rightarrow 2H^+ + 2e^- \]

2. **Determine the standard electrode potential (E°)**:
Again, \(E° = 0 \, V\).

3. **Calculate the concentration of H⁺ ions**:
For H₂SO₄, the concentration of H⁺ ions will be double that of the H₂SO₄ concentration:
\[ [H^+] = 2 \times 0.01 = 0.02 \, M \]

4. **Use the Nernst equation**:
\[ E = E° - \frac{RT}{nF} \ln Q \]
Where:
- \(Q = \frac{[H^+]^2}{P_{H_2}}\)

5. **Calculate Q**:
Given that \(P_{H_2} = 2 \, atm\) and \([H^+] = 0.02 \, M\):
\[ Q = \frac{(0.02)^2}{2} = \frac{0.0004}{2} = 0.0002 \]

6. **Substitute values into Nernst equation**:
\[ E = 0 - \frac{(8.314)(298)}{(2)(96500)} \ln(0.0002) \]
\[ E = -0.0591 \cdot \log(0.0002) \]
Since \(\log(0.0002) \approx -3.699\):
\[ E = -0.0591 \cdot (-3.699) \approx 0.218 \, V \]

### Final Answers:
- (a) Electrode potential \(E = 0.1182 \, V\)
- (b) Electrode potential \(E = 0.218 \, V\)