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

Options

A. `1.0` M HCl
B. a solution having pH4
C. pure water
D. `1.0` M NaOH

Answer

To determine which half-cell reaction has the maximum potential for the reaction \(2H^+ + 2e^- \rightarrow H_2(g)\), we will analyze the provided options using the Nernst equation.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The half-cell reaction given is:
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
This reaction involves the reduction of hydrogen ions (\(H^+\)) to form hydrogen gas (\(H_2\)).

2. **Nernst Equation**:
The Nernst equation is used to calculate the electrode potential (\(E\)) under non-standard conditions:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where:
- \(E^\circ\) is the standard electrode potential,
- \(R\) is the gas constant (8.314 J/(mol·K)),
- \(T\) is the temperature in Kelvin,
- \(n\) is the number of moles of electrons exchanged,
- \(F\) is Faraday's constant (96500 C/mol),
- \(Q\) is the reaction quotient.

3. **Calculating Reaction Quotient \(Q\)**:
For the reaction:
\[ Q = \frac{1}{[H^+]^2} \]
This is because the concentration of \(H_2\) gas is taken as 1 atm in the standard state.

4. **Substituting Values**:
At room temperature (25°C or 298 K) and for the reaction involving 2 electrons (\(n = 2\)):
\[ E = E^\circ - \frac{(8.314)(298)}{(2)(96500)} \ln Q \]
Simplifying the constants gives:
\[ E = E^\circ - 0.0591 \log Q \]

5. **Evaluating Different Cases**:
We will evaluate the potential for different concentrations of \(H^+\) ions in the four options provided:

- **Option A: 1 M HCl**:
\[ [H^+] = 1 \, \text{M} \quad \Rightarrow \quad Q = \frac{1}{(1)^2} = 1 \]
\[ E = E^\circ - 0.0591 \log(1) = E^\circ \]

- **Option B: \(10^{-4}\) M HCl**:
\[ [H^+] = 10^{-4} \, \text{M} \quad \Rightarrow \quad Q = \frac{1}{(10^{-4})^2} = 10^8 \]
\[ E = E^\circ - 0.0591 \log(10^8) = E^\circ - 0.0591 \times 8 = E^\circ - 0.4728 \]

- **Option C: Pure Water**:
\[ [H^+] = 10^{-7} \, \text{M} \quad \Rightarrow \quad Q = \frac{1}{(10^{-7})^2} = 10^{14} \]
\[ E = E^\circ - 0.0591 \log(10^{14}) = E^\circ - 0.0591 \times 14 = E^\circ - 0.8274 \]

- **Option D: 1 M NaOH**:
\[ [H^+] = 10^{-14} \, \text{M} \quad \Rightarrow \quad Q = \frac{1}{(10^{-14})^2} = 10^{28} \]
\[ E = E^\circ - 0.0591 \log(10^{28}) = E^\circ - 0.0591 \times 28 = E^\circ - 1.6568 \]

6. **Comparing Values**:
- For 1 M HCl: \(E = E^\circ\)
- For \(10^{-4}\) M HCl: \(E = E^\circ - 0.4728\)
- For pure water: \(E = E^\circ - 0.8274\)
- For 1 M NaOH: \(E = E^\circ - 1.6568\)

7. **Conclusion**:
The maximum potential occurs in the case of 1 M HCl, where \(E = E^\circ\). Therefore, the half-cell reaction with the maximum potential is from the option with 1 M HCl.

### Final Answer:
**The half-cell reaction with the maximum potential is from the option with 1 M HCl.**