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Base on the values given, select correct statements
$\begin{array}{ll} \frac{1}{2} F_{2} + e^{-} ⟶ F^{-} E^{\circ} = 2 .87 V \\ \frac{1}{2} {Cl}_{2} + e^{-} ⟶ {Cl}^{-} E^{\circ} = 1 .40 V \\ \frac{1}{2} {Br}_{2} + e^{-} ⟶ {Br}^{-} E^{\circ} = 1 .09 V \\ \frac{1}{2} I_{2} + e^{-} ⟶ I^{-} E^{\circ} = 0 .62 V \end{array}$

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Oxidizing power is in order $F_{2} < {Cl}_{2} < {Br}_{2} + I_{2}$

Oxidizing power is in order $F_{2} > {Cl}_{2} > {Br}_{2} > I_{2}$

Oxidizing power of I₂ is maximum

Oxidizing power of F₂ is maximum

answer is B, D.

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Detailed Solution

Higher the value of reduction potential, more stronger is the oxidizing power.

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