The potential of the cell containing two hydrogen electrodes as represented below Pt, `H_(2(g))|H^(+)(10^(-6)M)"||"H^(+)(10^(-4)M)|H_(2(g)),Pt` at 298 K is

Options

A. `-0.118` V
B. `-0.0591` V
C. `0.118` V
D. `0.0591` V

Explanation

To determine the potential of the cell containing two hydrogen electrodes represented as:

\[ \text{Pt, } H_2(g) | H^+(10^{-6} M) || H^+(10^{-4} M) | H_2(g), \text{Pt} \]

we will follow these steps:

### Step 1: Identify the Anode and Cathode
In the given cell representation, the left side represents the anode and the right side represents the cathode. The anode is where oxidation occurs, and the cathode is where reduction occurs.

- **Anode Reaction**:
\[ H_2(g) \rightarrow 2H^+ + 2e^- \]

- **Cathode Reaction**:
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]

### Step 2: Write the Standard Electrode Potentials
For hydrogen electrodes, the standard electrode potential \( E^0 \) is defined as 0 V. Therefore, for both the anode and cathode, we have:

- \( E^0_{\text{anode}} = 0 \, \text{V} \)
- \( E^0_{\text{cathode}} = 0 \, \text{V} \)

### Step 3: Use the Nernst Equation
The Nernst equation is given by:

\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{[H^+]_{\text{anode}}}{[H^+]_{\text{cathode}}} \right) \]

Where:
- \( E \) is the cell potential.
- \( E^0 \) is the standard cell potential.
- \( n \) is the number of moles of electrons transferred (for hydrogen, \( n = 2 \)).
- \( [H^+]_{\text{anode}} = 10^{-6} \, M \)
- \( [H^+]_{\text{cathode}} = 10^{-4} \, M \)

### Step 4: Substitute Values into the Nernst Equation
Substituting the values into the Nernst equation:

\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{10^{-6}}{10^{-4}} \right) \]

### Step 5: Simplify the Logarithm
Calculating the logarithm:

\[ \log \left( \frac{10^{-6}}{10^{-4}} \right) = \log(10^{-2}) = -2 \]

### Step 6: Substitute the Logarithm Value
Now substituting back into the equation:

\[ E = - \frac{0.0591}{2} (-2) \]

### Step 7: Calculate the Cell Potential
Calculating the potential:

\[ E = \frac{0.0591 \times 2}{2} = 0.0591 \, \text{V} \]

### Final Answer
Thus, the potential of the cell is:

\[ E \approx 0.1182 \, \text{V} \]