Question

Consider the following list of reagents :
Acidified ${K_2}C{r_2}{O_7}$, alkaline $KMn{O_4}$, $CuS{O_4}$, ${H_2}{O_2}$, $C{l_2}$, ${O_3}$, $FeC{l_3}$ and $N{a_2}{S_2}{O_3}$. The total number of reagents that can oxidise aqueous iodide to iodine is :
a.) 7
b.) 6
c.) 5
d.) 4

Answer 1

Hint: The reagents that can donate electrons are the oxidised ones and the reagents that can accept electrons are the reduced ones. These reagents that can accept electrons and thereby help the others to get oxidised are called oxidising reagents. These have a tendency to accept electrons.
So, those reagents that can accept the electrons from above will be known as oxidising ones. These normally have free orbitals to accept electrons.

Complete step by step answer:
First, let's understand what is oxidation and what is required for a species to get oxidised.
The oxidation is the addition of oxygen or removal of hydrogen atoms. In terms of electrons, one can say it is the donation of electrons. When any species donates the electrons, it get oxidised while the other one that accepts these electrons gets reduced.
So, for a species to get oxidised; there must be presence of another species that can easily accept electrons and get reduced.
So, from the above reagents; we will see the reagents that have the ability to accept the electrons. They will be able to oxidise iodide to iodine. Now, let us see the reagents one by one.

The first reagent given is Acidified ${K_2}C{r_2}{O_7}$. In this, the Cr is in +6 oxidation state. Thus, it can easily accept the electrons donated by aqueous iodide and thus will oxidise the iodide to iodine. The reaction is written as -
${K_2}C{r_2}{O_7} + {H_2}S{O_4} + 6KI \to 4{K_2}S{O_4} + Cr{(S{O_4})_3} + 3{I_2} + 7{H_2}O$

The second reagent given is alkaline $KMn{O_4}$. This is a very strong oxidising agent. The Mn is in +7 oxidation state. It will oxidise the aqueous iodide. The reaction is-
$2KMn{O_4} + KI + {H_2}O \to KI{O_3} + 2Mn{O_2} + 2KOH$

The third reagent is $CuS{O_4}$. The copper sulphate is also a good oxidising agent and can easily oxidise the aqueous iodide to iodine. So, this is also correct. The reaction is-
$2KMn{O_4} + KI + {H_2}O \to KI{O_3} + 2Mn{O_2} + 2KOH$

The fourth reagent is ${H_2}{O_2}$. This is also a good oxidising agent and can easily oxidise the aqueous iodide to iodine. So, this is also correct. The reaction is-
$2KI + {H_2}{O_2} \to 2KOH + {I_2}$

The fifth reagent given is $C{l_2}$. This is a very strong oxidising agent. It will oxidise the aqueous iodide. The reaction is-
$2KI + C{l_2} \to 2KCl + {I_2}$

The next reagent is ${O_3}$. The ozone can oxidise aqueous iodide to iodine. The reaction is-
$2KI + {O_3} + {H_2}O \to 2KOH + {I_2} + {O_2}$

The other reagent given is $FeC{l_3}$. The ferric chloride is also a strong oxidising agent. It will oxidise the aqueous iodide. The reaction is-
$2FeC{l_3} + 2KI \to 2FeC{l_2} + {I_2} + 2KCl$

The last reagent given is $N{a_2}{S_2}{O_3}$. This can not oxidise aqueous iodide to iodine.
So, the total reagents that can oxide iodide to iodine are seven.
So, the correct answer is “Option A”.

Note: It must be noted that some reagents have oxidation state in +n state where n is 1, 2, 3 etc. So, these have vacant orbitals where they can accept electrons. These can easily accept electrons to become neutral. So, they are very good reducing agents.