For the cell
`Pt||H_2(0.4atm)|H^+(pH=1)||(H^+(pH=2)|H_2(0.1atm)|Pt`
The measured potential at `25^(@)C` is

Options

A. `-0.322J//K`
B. `-0.635kJ//K`
C. `3.51kJ//K`
D. `-0.322kJ//K`

Answer

To find the measured potential of the given electrochemical cell at 25°C, we will follow these steps:

### Step 1: Identify the half-cell reactions
The cell notation indicates that we have two half-cells:
1. Anode (oxidation): \( H_2 \) at 0.4 atm converting to \( 2H^+ + 2e^- \)
2. Cathode (reduction): \( 2H^+ + 2e^- \) converting to \( H_2 \) at 0.1 atm

### Step 2: Determine the concentrations of \( H^+ \) ions
Using the pH values given:
- For the anode, \( \text{pH} = 1 \)
\[ [H^+] = 10^{-1} = 0.1 \, \text{M} \]
- For the cathode, \( \text{pH} = 2 \)
\[ [H^+] = 10^{-2} = 0.01 \, \text{M} \]

### Step 3: Calculate the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[H^+]_{\text{anode}} \cdot P_{H_2, \text{cathode}}}{[H^+]_{\text{cathode}} \cdot P_{H_2, \text{anode}}} \]
Substituting the values:
\[ K_c = \frac{(0.1) \cdot (0.1)}{(0.01) \cdot (0.4)} = \frac{0.01}{0.004} = 2.5 \]

### Step 4: Use the Nernst equation to calculate the cell potential
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K_c \]
For the hydrogen electrode, \( E^\circ_{\text{cell}} = 0 \) and \( n = 2 \) (since 2 electrons are involved). Thus:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log(2.5) \]

### Step 5: Calculate \( \log(2.5) \)
Using a calculator:
\[ \log(2.5) \approx 0.3979 \]

### Step 6: Substitute back into the Nernst equation
\[ E_{\text{cell}} = - \frac{0.0591}{2} \cdot 0.3979 \]
Calculating this gives:
\[ E_{\text{cell}} \approx -0.0118 \, \text{V} \]

### Step 7: Final result
The measured potential at 25°C is approximately:
\[ E_{\text{cell}} \approx -0.0118 \, \text{V} \]