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 but not Z
B. Y will oxidise both X and Z
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 given information about the gas X and the ions Y and Z in the solution, along with their reduction potentials.

### Step-by-Step Solution:

1. **Understanding Reduction Potentials**:
We are given the order of reduction potentials: \( Z > Y > X \). This means that Z has the highest tendency to be reduced, followed by Y, and X has the lowest tendency to be reduced.

**Hint**: Remember that a higher reduction potential means a greater tendency to gain electrons.

2. **Analyzing the Reactions**:
- Since \( Z > Y \), Z will not be oxidized by Y. This means Y cannot reduce Z.
- Since \( Y > X \), Y can reduce X, which means Y will oxidize X.

**Hint**: Compare the reduction potentials to determine which species can oxidize which.

3. **Identifying the Oxidation and Reduction**:
- Y will be reduced (gaining electrons) while X will be oxidized (losing electrons).
- Z cannot be oxidized by Y because Y has a lower reduction potential than Z.

**Hint**: Oxidation involves losing electrons, while reduction involves gaining electrons.

4. **Conclusion**:
Based on the above analysis, we can conclude that:
- Y will reduce X.
- Z will remain unaffected as Y cannot oxidize Z.

5. **Final Answer**:
The correct option is that Y will reduce X, and Z will not be reduced by Y.

### Summary of the Solution:
- The order of reduction potentials is \( Z > Y > X \).
- Y can reduce X but cannot oxidize Z.
- Therefore, the correct conclusion is that Y will oxidize X.