Which has maximum potential for the half-cell reaction?
`2H^(+) + 2e^(-) rarr H_(2)`

Options

A. 1.0M HCl
B. A solution having pH 4
C. Pure water
D. 1.0 M NaOH

Answer

To determine which solution has the maximum potential for the half-cell reaction \(2H^+ + 2e^- \rightarrow H_2\), we will analyze the standard reduction potential and the concentration of hydrogen ions (\(H^+\)) in different solutions.

### Step-by-Step Solution:

1. **Identify the Half-Cell Reaction**:
The half-cell reaction is given as:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

2. **Standard Reduction Potential**:
The standard reduction potential (\(E^\circ\)) for the hydrogen electrode reaction is defined as 0 V. This means that under standard conditions (1 M concentration of \(H^+\), 1 atm pressure of \(H_2\)), the potential is 0 V.

3. **Nernst Equation**:
To calculate the electrode potential under non-standard conditions, we use the Nernst equation:
\[ E = E^\circ - \frac{0.059}{n} \log Q \]
where:
- \(E\) is the electrode potential,
- \(E^\circ\) is the standard electrode potential (0 V for hydrogen),
- \(n\) is the number of moles of electrons transferred (2 for this reaction),
- \(Q\) is the reaction quotient.

4. **Reaction Quotient (Q)**:
For the reaction \(2H^+ + 2e^- \rightarrow H_2\), the reaction quotient \(Q\) is given by:
\[ Q = \frac{P_{H_2}}{[H^+]^2} \]
where \(P_{H_2}\) is the partial pressure of hydrogen gas and \([H^+]\) is the concentration of hydrogen ions.

5. **Concentration of \(H^+\)**:
The potential \(E\) will depend on the concentration of \(H^+\). The higher the concentration of \(H^+\), the more positive the potential will be.

6. **Comparing Different Solutions**:
- **1 M HCl**: This is a strong acid that completely dissociates, providing 1 M \(H^+\).
- **1 M NaOH**: This is a strong base, and the concentration of \(H^+\) will be very low due to the neutralization with \(OH^-\). The concentration of \(H^+\) can be calculated using the ion product of water (\(K_w = 10^{-14}\)), leading to \([H^+] = 10^{-14} / 1 = 10^{-14} M\).
- **Pure Water**: The concentration of \(H^+\) in pure water is \(10^{-7} M\).

7. **Calculating Electrode Potential**:
- For **1 M HCl**:
\[ Q = \frac{1}{(1)^2} = 1 \]
\[ E = 0 - \frac{0.059}{2} \log(1) = 0 \text{ V} \]
- For **1 M NaOH**:
\[ Q = \frac{1}{(10^{-14})^2} = 10^{28} \]
\[ E = 0 - \frac{0.059}{2} \log(10^{28}) = 0 - 0.059 \times 14 = -0.826 \text{ V} \]
- For **Pure Water**:
\[ Q = \frac{1}{(10^{-7})^2} = 10^{14} \]
\[ E = 0 - \frac{0.059}{2} \log(10^{14}) = 0 - 0.059 \times 7 = -0.206 \text{ V} \]

8. **Conclusion**:
The solution with the maximum potential for the half-cell reaction \(2H^+ + 2e^- \rightarrow H_2\) is **1 M HCl**, as it provides the highest concentration of \(H^+\) ions.