`Pt(H_(2))(1atm)|H_(2)O`, electrode potential at `298K` is

Options

A. `-0.2364V`
B. `-0.4137V`
C. `0.4137V`
D. `0.00V`

Explanation

To find the electrode potential of the cell `Pt(H2)(1 atm)|H2O` at 298 K, we can follow these steps:

### Step 1: Write the Half-Cell Reaction
The half-cell reaction for hydrogen gas can be written as:
\[ \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \]

### Step 2: Determine the Concentration of H⁺
At pH 7 (which is standard for biological conditions), the concentration of hydrogen ions \([\text{H}^+]\) is:
\[ [\text{H}^+] = 10^{-7} \, \text{mol/L} \]

### Step 3: Write the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Where:
- \( E^\circ_{\text{cell}} \) is the standard electrode potential (for hydrogen, it is 0 V).
- \( n \) is the number of electrons transferred (which is 2 for the hydrogen half-reaction).
- \( Q \) is the reaction quotient.

### Step 4: Calculate the Reaction Quotient (Q)
For the reaction, the reaction quotient \( Q \) can be expressed as:
\[ Q = \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \]
Given that \( P_{\text{H}_2} = 1 \, \text{atm} \) and \([\text{H}^+] = 10^{-7} \, \text{mol/L}\):
\[ Q = \frac{(10^{-7})^2}{1} = 10^{-14} \]

### Step 5: Substitute Values into the Nernst Equation
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log(10^{-14}) \]

### Step 6: Simplify the Logarithm
Using the property of logarithms:
\[ \log(10^{-14}) = -14 \]
Thus, substituting this into the equation gives:
\[ E_{\text{cell}} = -\frac{0.0591}{2} \times (-14) \]

### Step 7: Calculate the Final Value
Now, calculate:
\[ E_{\text{cell}} = \frac{0.0591 \times 14}{2} = \frac{0.8264}{2} = 0.4132 \, \text{V} \]

### Step 8: Final Answer
Since we are looking for the potential in volts, we can round this to:
\[ E_{\text{cell}} \approx 0.4137 \, \text{V} \]

### Conclusion
Thus, the electrode potential at 298 K for the cell `Pt(H2)(1 atm)|H2O` is approximately:
\[ E_{\text{cell}} = 0.4137 \, \text{V} \]