A gas X at 1 atm is bubbled through a solution containing a mixture of 1M `Y^(-)` and 1M `Z^(-)` at `25^(@)C`. If the reduction potential of `ZgtYgtX`, then

Options

A. Y will oxidise X and not Z
B. Y will oxidise Z and not X
C. Y will oxidise both X and Z.
D. Y will reduce both X and Z.

Explanation

To solve the problem, we need to analyze the information given about the gas X and the ions Y⁻ and Z⁻ in the solution. The key points to consider are the reduction potentials of the species involved.

### Step-by-Step Solution:

1. **Understanding Reduction Potentials**:
- The reduction potential indicates the tendency of a species to gain electrons (be reduced). A higher reduction potential means a greater ability to be reduced.
- The order given is: \( E^\circ(Z) > E^\circ(Y) > E^\circ(X) \). This means Z has the highest reduction potential, followed by Y, and X has the lowest.

2. **Identifying the Oxidizing and Reducing Agents**:
- Since Z has the highest reduction potential, it will be the best oxidizing agent among the three. It can reduce Y and X.
- Y, having a higher reduction potential than X, can also reduce X but cannot reduce Z.

3. **Determining Reactions**:
- Since Y can oxidize X (because Y has a higher reduction potential than X), we can write the half-reaction:
\[ Y^- \rightarrow Y + e^- \]
- This indicates that Y will lose electrons (be oxidized) and can reduce X:
\[ X + e^- \rightarrow X^- \]

4. **Conclusion**:
- Based on the reduction potentials, Y will oxidize X, and since Z has a higher reduction potential than Y, Y cannot oxidize Z. Therefore, the correct conclusion is:
\[ \text{Y will oxidize X and not Z.} \]

### Final Answer:
Y will oxidize X and not Z.