Question

Which of the following represents correct order of oxidizing power of halogens with water

• A. I$_2$>Br$_2$>Cl$_2$>F$_2$
• B. Cl$_2$>F$_2$>Br$_2$>I$_2$
• C. F$_2$>Cl$_2$>I$_2$>Br$_2$
• D. F$_2$>Cl$_2$>Br$_2$>I$_2$

Hint:

Oxidizing power of halogens is determined by their standard reduction potentials ($E^\ominus$).
Higher $E^\ominus$ value means stronger oxidizing agent.
Trend for $E^\ominus$ (and thus oxidizing power) down Group 17: F$_2$>Cl$_2$>Br$_2$>I$_2$.
Fluorine is the strongest oxidizing agent among all elements.
Reactions with water also reflect this trend: F$_2$ oxidizes water to O$_2$. Cl$_2$ and Br$_2$ disproportionate. I$_2$ reacts very slightly.

Answer 1

The Correct Option is D

Halogens (X$_2$) can act as oxidizing agents. Their oxidizing power is related to their tendency to gain electrons (i.e., their standard reduction potentials). The general reaction of a halogen X$_2$ acting as an oxidizing agent is: X$_2$ + 2e$^-$ $\rightarrow$ 2X$^-$ The standard reduction potentials ($E^\ominus$) for halogens are: F$_2$(g) + 2e$^-$ $\rightarrow$ 2F$^-$(aq) \quad $E^\ominus = +2.87 \text{ V}$ Cl$_2$(g) + 2e$^-$ $\rightarrow$ 2Cl$^-$(aq) \quad $E^\ominus = +1.36 \text{ V}$ Br$_2$(l) + 2e$^-$ $\rightarrow$ 2Br$^-$(aq) \quad $E^\ominus = +1.07 \text{ V}$ (or +1.09 V) I$_2$(s) + 2e$^-$ $\rightarrow$ 2I$^-$(aq) \quad $E^\ominus = +0.54 \text{ V}$ A higher (more positive) standard reduction potential indicates a stronger tendency to be reduced, and thus a stronger oxidizing agent. Based on these $E^\ominus$ values, the decreasing order of oxidizing power is: F$_2$>Cl$_2$>Br$_2$>I$_2$. When halogens react with water, they can oxidize water or disproportionate. \begin{itemize} \item F$_2$: Fluorine is the strongest oxidizing agent. It oxidizes water vigorously: $2\text{F}_2\text{(g)} + 2\text{H}_2\text{O(l)} \rightarrow 4\text{HF(aq)} + \text{O}_2\text{(g)}$ (Fluorine oxidizes O from -2 in H$_2$O to 0 in O$_2$). \item Cl$_2$: Chlorine reacts with water to form HCl and HOCl (hypochlorous acid). This is a disproportionation reaction for Cl$_2$ (0 to -1 in HCl and +1 in HOCl), but in this context, Cl$_2$ can also oxidize other substances in water. Its oxidizing power is less than F$_2$. $\text{Cl}_2\text{(g)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCl(aq)} + \text{HOCl(aq)}$ \item Br$_2$: Bromine reacts with water similarly to chlorine, but to a lesser extent. $\text{Br}_2\text{(l)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HBr(aq)} + \text{HOBr(aq)}$ \item I$_2$: Iodine has limited solubility and reactivity with water. The reaction is very slight. Iodine is the weakest oxidizing agent among the common halogens. In fact, I$^-$ can be oxidized to I$_2$ by stronger oxidizing agents like Cl$_2$ or Br$_2$. $\text{I}_2\text{(s)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HI(aq)} + \text{HOI(aq)}$ (equilibrium lies far to the left). \end{itemize} The question asks for "oxidizing power of halogens with water". This refers to their general strength as oxidizing agents, which is best compared using their standard reduction potentials. The specific reactions with water illustrate this trend. Fluorine's ability to oxidize water to O$_2$ highlights its supreme oxidizing power. The other halogens' reactions with water are more complex (disproportionation), but their overall oxidizing strength follows the $E^\ominus$ trend. So, the correct decreasing order of oxidizing power is F$_2$>Cl$_2$>Br$_2$>I$_2$. This matches option (d). \[ \boxed{\text{F}_2 \text{>Cl}_2 \text{>Br}_2 \text{>I}_2} \]