Question

The correct order of oxidizing power of the given ions is Given: \( E^{\circ}_{\text{Mg}^{2+}/\text{Mg}} = -2.36 \, \text{V} \) and \( E^{\circ}_{\text{Cl}_2/\text{Cl}^-} = +1.36 \, \text{V}\)

• A. VO\(_2^+\) < Cr\(_2\)O\(_7^{2-}\) < MnO\(_4^-\)
• B. VO\(_2^+\) < MnO\(_4^-\) < Cr\(_2\)O\(_7^{2-}\)
• C. MnO\(_4^-\) < Cr\(_2\)O\(_7^{2-}\) < VO\(_2^+\)
• D. Cr\(_2\)O\(_7^{2-}\) < VO\(_2^+\) < MnO\(_4^-\)

Hint:

To determine the order of oxidizing power, compare the standard electrode potentials. The more positive the reduction potential, the stronger the oxidizing agent.

Answer 1

The Correct Option is A

The oxidizing power of an ion is directly related to its standard electrode potential. The more positive the reduction potential, the stronger the oxidizing agent. The given standard electrode potentials of the ions are as follows: - \( \text{VO}_2^+ \) has a potential of \( +1.0 \, \text{V} \)
- \( \text{Cr}_2\text{O}_7^{2-} \) has a potential of \( +1.33 \, \text{V} \)
- \( \text{MnO}_4^- \) has a potential of \( +1.51 \, \text{V} \) Since the higher the potential, the stronger the oxidizing agent, we can arrange them in order of increasing oxidizing power: \( \text{VO}_2^+<\text{Cr}_2\text{O}_7^{2-}<\text{MnO}_4^-. \)