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 given by:

\[ 2H^+ + 2e^- \rightarrow H_2 \]

we will use the Nernst equation. The Nernst equation relates the cell potential (E) to the standard electrode potential (E°) and the concentrations of the reactants and products. The equation is given by:

\[ E = E° - \frac{0.0591}{n} \log \left( \frac{1}{[H^+]^2} \right) \]

where:
- \( E° \) is the standard electrode potential (which is 0 for the hydrogen half-cell),
- \( n \) is the number of electrons transferred in the half-reaction (which is 2 for this reaction),
- \([H^+]\) is the concentration of hydrogen ions.

### Step 1: Identify the Concentrations of \( H^+ \) for Each Option

1. **1 M HCl**:
- \([H^+] = 1 \, \text{M}\)

2. **Solution with pH = 4**:
- \([H^+] = 10^{-4} \, \text{M}\) (since \( pH = -\log[H^+] \))

3. **Pure Water**:
- \([H^+] = 10^{-7} \, \text{M}\) (neutral water)

4. **1 M NaOH**:
- \([OH^-] = 1 \, \text{M}\) implies \([H^+] = 10^{-14} \, \text{M}\) (using \( K_w = [H^+][OH^-] = 10^{-14} \))

### Step 2: Calculate the Cell Potential for Each Option

1. **For 1 M HCl**:
\[ E = 0 - \frac{0.0591}{2} \log(1^2) = 0 \, \text{V} \]

2. **For pH = 4**:
\[ E = 0 - \frac{0.0591}{2} \log(10^{-4}) = 0 - 0.0591 \times 2 \times (-4) = 0.236 \, \text{V} \]

3. **For Pure Water**:
\[ E = 0 - \frac{0.0591}{2} \log(10^{-7}) = 0 - 0.0591 \times 2 \times (-7) = 0.413 \, \text{V} \]

4. **For 1 M NaOH**:
\[ E = 0 - \frac{0.0591}{2} \log(10^{-14}) = 0 - 0.0591 \times 2 \times (-14) = 0.827 \, \text{V} \]

### Step 3: Compare the Potentials

- **1 M HCl**: \( E = 0 \, \text{V} \)
- **pH = 4**: \( E = 0.236 \, \text{V} \)
- **Pure Water**: \( E = 0.413 \, \text{V} \)
- **1 M NaOH**: \( E = 0.827 \, \text{V} \)

### Conclusion

The solution with the maximum potential for the half-cell reaction is **1 M NaOH** with a potential of \( 0.827 \, \text{V} \).