Fluorine is a stronger oxidising agent than c | vedclass

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Question

(B) The standard electrode potential $(E^\circ)$ of a halogen is determined by the sum of three energy changes: sublimation energy,dissociation energy

Vedclass · Accepted Answer

Electron affinity

Vedclass · Answer

(B) The standard electrode potential $(E^\circ)$ of a halogen is determined by the sum of three energy changes: sublimation energy,dissociation energy,and hydration energy. Specifically,the process involves: $\frac{1}{2} X_2(g) \rightarrow X(g)$ (dissociation),$X(g) + e^- \rightarrow X^-(g)$ (electron gain enthalpy),and $X^-(g) + aq \rightarrow X^-(aq)$ (hydration enthalpy). Fluorine has a very high hydration enthalpy due to its small size,which compensates for its high dissociation energy and makes it a stronger oxidising agent. Electron affinity is not the sole deciding factor,and in fact,the electron gain enthalpy of chlorine is more negative than that of fluorine. Therefore,the statement is attributed to these factors except for electron affinity.

Column $A$	Column $B$
$(i) \; Na_{2}O$	$(a) \; Neutral$
$(ii) \; Al_{2}O_{3}$	$(b) \; Basic$
$(iii) \; N_{2}O$	$(c) \; Acidic$
$(iv) \; Cl_{2}O_{7}$	$(d) \; Amphoteric$

Fluorine is a stronger oxidising agent than chlorine in aqueous solution. This is attributed to many factors except

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