In aqueous solution, fluorine is a stronger o | vedclass

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(C) The oxidizing power of a halogen in aqueous solution depends on the overall energy change involved in the process: $\frac{1}{2}X_2(g) + e^- \right

Vedclass · Accepted Answer

Ionization potential

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(C) The oxidizing power of a halogen in aqueous solution depends on the overall energy change involved in the process: $\frac{1}{2}X_2(g) + e^- \rightarrow X^-(aq)$. This process involves three steps: $(1)$ Enthalpy of dissociation $(\frac{1}{2} \Delta H_{diss})$, $(2)$ Electron affinity $(\Delta H_{eg})$, and $(3)$ Enthalpy of hydration $(\Delta H_{hyd})$. Ionization potential is related to the removal of an electron, which is not involved in the reduction process of halogens to halide ions. Therefore, it is not a factor determining the oxidizing strength in aqueous solution.

In aqueous solution, fluorine is a stronger oxidizing agent than chlorine. This is due to all of the following factors $EXCEPT$:

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