Question

Which reaction is not feasible?

• A. $2\mathrm{KI}+{\mathrm{Br}}_2\to 2\mathrm{KBr}+{\mathrm{I}}_2$
• B. $2\mathrm{KBr}+{\mathrm{I}}_2\to 2\mathrm{KI}+{\mathrm{Br}}_2$
• C. $2\mathrm{KBr}+{\mathrm{Cl}}_2\to 2\mathrm{KCl}+{\mathrm{Br}}_2$
• D. $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{F}}_2\to 4\mathrm{HF}+{\mathrm{O}}_2$

Answer 1

The correct option is B

$2\mathrm{KBr}+{\mathrm{I}}_2\to 2\mathrm{KI}+{\mathrm{Br}}_2$

Explanation for correct option :

(B) $2\mathrm{KBr}+{\mathrm{I}}_2\to 2\mathrm{KI}+{\mathrm{Br}}_2$

• For a reaction to be feasible, it should be spontaneous which implies that there should be no need for extra energy to be put into the reaction.
• In the above reaction, it is shown that the bromine ion is getting oxidized by iodine.
• This is chemically not feasible as bromine is a better and stronger oxidizing agent than iodine.
• The electrochemical series shows that bromine has a higher electrode potential (oxidizing potential) than iodine.
• Therefore, ${\mathrm{Br}}^{-}$ ion cannot be oxidized to ${\mathrm{Br}}_2$ by ${\mathrm{I}}_2$ as bromine is a better oxidizing agent.
• In all other cases, it is feasible because oxidation is done by the reactant having a higher electrode potential.

Explanation for incorrect options :

(A) $2\mathrm{KI}+{\mathrm{Br}}_2\to 2\mathrm{KBr}+{\mathrm{I}}_2$

• Bromine is a stronger oxidizing agent as compared to Iodine.
• Therefore it can easily convert ${\mathrm{I}}^{-}$ion into ${\mathrm{I}}_2$.

(C) $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{F}}_2\to 4\mathrm{HF}+{\mathrm{O}}_2$

• Fluorine is the best oxidizing agent having the highest electronegativity.
• It can easily replace the ions due to its highly reactive nature.

(D) $2\mathrm{KBr}+{\mathrm{Cl}}_2\to 2\mathrm{KCl}+{\mathrm{Br}}_2$

• Chlorine is more reactive than Bromine but lesser than Fluorine.
• In the above reaction, it can easily replace ${\mathrm{Br}}^{-}$ ion. Therefore this reaction is also feasible.

Hence, option (B) is correct.