Question

Which will liberate bromine from a solution of potassium bromide?
A. ${{\text{I}}_{\text{2}}}$
B. ${\text{C}}{{\text{l}}_{\text{2}}}$
C. ${\text{S}}{{\text{O}}_{\text{2}}}$
D. HI

Answer 1

First we will check the oxidation state of bromine in the reactant and in the product to determine whether it is getting oxidized to reduce. After determining this we can find the reducing agent if bromine is getting oxidized or the oxidized agent if bromine is getting reduced.

Complete step-by-step answer:
We have a solution of potassium bromide and we have to form bromine.
${\text{KBr}}\, \to \,{\text{B}}{{\text{r}}_{\text{2}}}$
The oxidation state of bromine in potassium bromide is $- 1$ and in bromine is zero.
So, during the reaction of potassium bromide the oxidation state of bromine will change from $- 1 \to 0$ it means bromine will lose electrons so oxidation of bromine will take place. It also means that we have to find out the oxidising agent.

According to the electrochemical series the element having high reduction potential gets reduced and works as an oxidising agent.
The reduction potential of ${\text{C}}{{\text{l}}_{\text{2}}}$is $+ 1.36$ V.
The reduction potential of ${\text{B}}{{\text{r}}_{\text{2}}}$is $+ 1.08$ V.
The reduction potential of ${{\text{I}}_{\text{2}}}$is $+ 0.535$ V.
So, the reduction potential of ${\text{C}}{{\text{l}}_{\text{2}}}$ is high than the ${\text{B}}{{\text{r}}_{\text{2}}}$ so, ${\text{C}}{{\text{l}}_{\text{2}}}$ can act as oxidizing agent for the formation of ${\text{B}}{{\text{r}}_{\text{2}}}$.
${\text{2}}\,{\text{KBr}}\,{\text{ + }}\,{\text{C}}{{\text{l}}_{\text{2}}}\, \to 2{\text{KCl}}\,{\text{ + }}\,{\text{B}}{{\text{r}}_{\text{2}}}\,$
The reaction of potassium bromide with ${\text{S}}{{\text{O}}_{\text{2}}}$ in water gives the salt of sulphuric acid and hydrogen bromide.
${\text{KBr}}\,{\text{ + }}\,{\text{S}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,{\text{HBr}}\,{\text{ + }}\,{\text{KHS}}{{\text{O}}_{\text{3}}}$
The reaction of potassium bromide with HI gives potassium iodide and hydrogen bromide.
${\text{KBr}}\,{\text{ + }}\,{\text{HI}}\, \to \,{\text{HBr}}\,{\text{ + }}\,{\text{KI}}$
So, ${\text{C}}{{\text{l}}_{\text{2}}}$ will liberate bromine from a solution of potassium bromide?

Therefore, option (B) ${\text{C}}{{\text{l}}_{\text{2}}}$is correct.

Note: The species undergoes oxidation work as a reducing agent and the species undergoes reduction works as an oxidizing agent. The gain of an electron is known as reduction and the loss of electron is known as oxidation. The electrochemical series is a series of elements, in which elements are arranged according to their reduction potential. The reduction potential of an element tells the ease of reduction of an element. Due to high reduction potential chlorine can displace the bromine, as well as iodine, from their solution. Similarly, bromine can displace the iodine from its solution.