Standard reduction potentials at `25^(@) C" of " Li ^(+)abs(Li,Ba^(2+))Ba,Na^(+)abs(Na and Mg^(2+))Mg" are " -3.05, -2.90, -2.71` and `- 2.37`
volt respectively. Which of the following is the strongest oxidising agent?

Options

A. `Na^(+)`
B. `Li^(+)`
C. `Ba^(2+)`
D. `Mg^(2+)`

Answer

To determine which of the given ions is the strongest oxidizing agent, we need to analyze the standard reduction potentials provided for each ion. The stronger the oxidizing agent, the higher (less negative) its standard reduction potential will be.

### Step-by-Step Solution:

1. **List the Standard Reduction Potentials**:
- \( \text{Li}^+ + e^- \rightarrow \text{Li} \) : \( E^\circ = -3.05 \, \text{V} \)
- \( \text{Ba}^{2+} + 2e^- \rightarrow \text{Ba} \) : \( E^\circ = -2.90 \, \text{V} \)
- \( \text{Na}^+ + e^- \rightarrow \text{Na} \) : \( E^\circ = -2.71 \, \text{V} \)
- \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \) : \( E^\circ = -2.37 \, \text{V} \)

2. **Identify the Highest Standard Reduction Potential**:
- The values of the standard reduction potentials are:
- \( -3.05 \, \text{V} \) (Li)
- \( -2.90 \, \text{V} \) (Ba)
- \( -2.71 \, \text{V} \) (Na)
- \( -2.37 \, \text{V} \) (Mg)

3. **Compare the Values**:
- The least negative value among these potentials is \( -2.37 \, \text{V} \) (for Mg).
- The more positive the standard reduction potential, the stronger the oxidizing agent.

4. **Conclusion**:
- Since \( \text{Mg}^{2+} \) has the highest standard reduction potential of \( -2.37 \, \text{V} \), it is the strongest oxidizing agent among the given options.

### Final Answer:
The strongest oxidizing agent is \( \text{Mg}^{2+} \).