`Pt(H_(2),"x atm")abs(0.0MH^(+))abs(0.1MH^(+))Pt(H_(2),"y atm").` if `E^(o) "" (cell) = 0.00V, ` then `x/y` is :

Options

A. 100
B. 10
C. `0.01`
D. `0.1`

Answer

To solve the problem, we need to find the ratio \( \frac{x}{y} \) where \( x \) is the pressure of hydrogen gas at the anode and \( y \) is the pressure of hydrogen gas at the cathode. We are given that the standard cell potential \( E^\circ_{cell} = 0.00 \, V \).

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- At the anode, hydrogen gas is oxidized:
\[ H_2 \rightarrow 2H^+ + 2e^- \]
- At the cathode, hydrogen ions are reduced:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

2. **Write the Nernst Equation**:
The Nernst equation for the cell can be expressed as:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[H^+]^2}{P_{H_2}} \right) \]
where \( n \) is the number of moles of electrons transferred (which is 2 in this case).

3. **Substituting Values**:
Given that \( E^\circ_{cell} = 0.00 \, V \), we can set up the equation:
\[ 0 = 0 - \frac{0.0591}{2} \log \left( \frac{(0.1)^2}{\frac{x}{y}} \right) \]
This simplifies to:
\[ 0 = -0.02955 \log \left( \frac{(0.1)^2}{\frac{x}{y}} \right) \]

4. **Solving the Logarithmic Equation**:
Since the left-hand side is zero, the right-hand side must also equal zero:
\[ \log \left( \frac{(0.1)^2}{\frac{x}{y}} \right) = 0 \]
This implies:
\[ \frac{(0.1)^2}{\frac{x}{y}} = 1 \]
Therefore:
\[ (0.1)^2 = \frac{x}{y} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ x = (0.1)^2 \cdot y \]
Thus:
\[ \frac{x}{y} = (0.1)^2 = 0.01 \]

### Final Answer:
The ratio \( \frac{x}{y} \) is:
\[ \frac{x}{y} = 0.01 \]