The following cell is found to have EMF equal to zero.
`Pt,H_(2)("x atm")|0.01MH^(+)||0.1MH^(+)|H_(2)(yatm),Pt`
The ratio x/y is,

Options

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

Explanation

To solve the problem, we need to analyze the given electrochemical cell and apply the Nernst equation. The cell has an EMF (electromotive force) of zero, indicating that the reaction is at equilibrium.

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- At the anode, hydrogen gas (H₂) is oxidized to hydrogen ions (H⁺):
\[ \text{Anode: } H_2(g) \rightarrow 2H^+(aq) + 2e^- \]
- At the cathode, hydrogen ions (H⁺) are reduced to hydrogen gas (H₂):
\[ \text{Cathode: } 2H^+(aq) + 2e^- \rightarrow H_2(g) \]

2. **Define Concentrations and Pressures**:
- For the anode:
- Pressure of H₂ = \( x \) atm
- Concentration of H⁺ = \( 0.01 \, M = 10^{-2} \, M \)
- For the cathode:
- Pressure of H₂ = \( y \) atm
- Concentration of H⁺ = \( 0.1 \, M = 10^{-1} \, M \)

3. **Apply the Nernst Equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[ \text{Products} ]}{[ \text{Reactants} ]} \right) \]
Since the cell EMF is zero, we have:
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{y \cdot (10^{-2})^2}{x \cdot (10^{-1})^2} \right) \]

4. **Substituting Values**:
Since this is a standard hydrogen electrode, \( E^\circ_{\text{cell}} = 0 \):
\[ 0 = - \frac{0.0591}{2} \log \left( \frac{y \cdot 10^{-4}}{x \cdot 10^{-2}} \right) \]
Simplifying gives:
\[ 0 = - \frac{0.0591}{2} \log \left( \frac{y}{x} \cdot 10^{-2} \right) \]

5. **Solving the Logarithmic Equation**:
Since the logarithm equals zero:
\[ \log \left( \frac{y}{x} \cdot 10^{-2} \right) = 0 \]
This implies:
\[ \frac{y}{x} \cdot 10^{-2} = 1 \quad \Rightarrow \quad \frac{y}{x} = 10^{2} = 100 \]

6. **Finding the Ratio \( \frac{x}{y} \)**:
To find the ratio \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{1}{100} = 10^{-2} \]

### Final Answer:
The ratio \( \frac{x}{y} \) is \( 10^{-2} \).