Question

For the reaction between $\text{KMnO}_{\text{4}}^{{}}$ and ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$, the number of electrons transferred per mole of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$is?
A) One
B) Two
C) three
D) Four

Answer 1

Hint: This $\text{KMnO}_{\text{4}}^{{}}$is a powerful oxidizing agent in neutral, acidic, and alkaline medium. The ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$is also an oxidizing agent but in the reaction of $\text{KMnO}_{\text{4}}^{{}}$with the ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.The ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$loses its electron and acts as a reducing agent. In an acidic medium, the $\text{MnO}_{\text{4}}^{-}$ ions involve the addition of 5 electrons and the two peroxide ions give out two electrons.

Complete step by step solution:
The redox reaction is a reaction which is a combination of oxidation and reduction. The one species accepts the electrons and reduces while the other species donates the electron and itself undergoes the oxidation.
During the reaction, we observe the rise in the oxidation state of the reducing agent and fall in the oxidation state of the oxidising agent.
The potassium permanent ions $\text{KMnO}_{\text{4}}^{{}}$ and hydrogen peroxide ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ are readily available in the laboratory. The permanent ion and the hydrogen peroxide react mainly in an acidic medium maintained by sulphuric acid ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ .
The reaction of $\text{KMnO}_{\text{4}}^{{}}$with the ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is also an example of a redox reaction.
The general and unbalanced reaction scheme can be shown as follows:
$\text{KMnO}_{\text{4}}^{{}}\text{ + }{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\to \text{ }{{\text{O}}_{\text{2}}}\text{ }+\text{ }{{\text{H}}_{\text{2}}}\text{O + MnS}{{\text{O}}_{\text{4}}}\text{ + }{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
Here, the $\text{KMnO}_{\text{4}}^{{}}$reacts with the ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ presence of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ to generate the oxygen ${{\text{O}}_{\text{2}}}\text{ }$ , ${{\text{H}}_{\text{2}}}\text{O}$, manganese sulfate $\text{MnS}{{\text{O}}_{\text{4}}}$ , and potassium sulfate ${{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$.
Let's balance the redox reaction.
Here, the oxidizing agent which itself undergoes the reduction is $\text{KMnO}_{\text{4}}^{{}}$ or $\text{MnO}_{\text{4}}^{-}$ and reducing agent which itself undergoes the oxidation is ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ or ${{\text{O}}^{\text{-1}}}$ .
Here, $\text{MnO}_{\text{4}}^{-}$ gains the five electrons and reduced from $\text{M}{{\text{n}}^{\text{+7}}}\to \text{M}{{\text{n}}^{\text{+2}}}$ . The reduction half-reaction is as follows:
$\text{MnO}{}_{\text{4}}^{-}\text{+8}{{\text{H}}^{\text{+}}}\text{+5}{{\text{e}}^{\text{-}}}\to \text{M}{{\text{n}}^{\text{2+}}}\text{+4}{{\text{H}}_{\text{2}}}\text{O}$ (1)
The hydrogen peroxide, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ or ${{\text{O}}^{\text{-1}}}$ releases the two electrons and oxidized from ${{\text{O}}^{\text{-1}}}\to {{\text{O}}_{\text{2}}}$. The oxidation half-reaction is as follows:
$\text{2}{{\text{O}}^{\text{-1}}}\to {{\text{O}}_{\text{2}}}\text{ + 2}{{\text{e}}^{-}}$ (2)
Each ${{\text{O}}^{\text{-1}}}$ion donates a single electron and produces ${{\text{O}}_{\text{2}}}$ gas.
Now, multiple equations (1) by 2 and equation (2) by 5 and add the reduction and oxidation half-reaction. We have,
\[\begin{matrix}
\text{2MnO}{}_{\text{4}}^{-}\text{+16}{{\text{H}}^{\text{+}}}\text{+10}{{\text{e}}^{\text{-}}}\to 2\text{M}{{\text{n}}^{\text{2+}}}\text{+8}{{\text{H}}_{\text{2}}}\text{O} \\
\text{ + 10 }{{\text{O}}^{\text{-1}}}\to 5{{\text{O}}_{\text{2}}}\text{ + 10}{{\text{e}}^{-}} \\
\overline{\text{ 2MnO}{}_{\text{4}}^{-}\text{ + 10 }{{\text{O}}^{\text{-1}}}\text{ + 16}{{\text{H}}^{\text{+}}}\to 2\text{M}{{\text{n}}^{\text{2+}}}\text{ + 8}{{\text{H}}_{\text{2}}}\text{O + }5{{\text{O}}_{\text{2}}}\text{ }} \\
 {} \\
\end{matrix}\]
Or by adding the necessary ions and the radicals we get,
$\text{2KMnO}_{\text{4}}^{{}}\text{ + 5}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{ + 3}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\to \text{ 5}{{\text{O}}_{\text{2}}}\text{ }+\text{ 8}{{\text{H}}_{\text{2}}}\text{O + 2MnS}{{\text{O}}_{\text{4}}}\text{ + }{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
From the oxidation half-reaction, the number of electrons transferred by the one-mole hydrogen peroxide ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$is equal to two.

Hence, (B) is the correct option.

Note: It may be noted that $\text{KMnO}_{\text{4}}^{{}}$ titrations are carried out only in the presence of dil. ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$. This is because the oxygen produced from the reaction of $\text{KMnO}_{\text{4}}^{{}}$ with dil. ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is used for oxidizing the reducing agent. $\text{KMnO}_{\text{4}}^{{}}$ do not produce any oxygen of its own to act as an oxidizing agent. However, it may be noted that the hydrochloric acid $\text{HCl}$ or nitric acid $\text{HN}{{\text{O}}_{\text{3}}}$ cannot be used in place of dil. ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$. In case $\text{HCl}$ used, the oxygen produced by the reaction of $\text{KMnO}_{\text{4}}^{{}}$and dil. $\text{HCl}$ will partially use up to oxidize $\text{HCl}$ to chlorine. Hence, $\text{KMnO}_{\text{4}}^{{}}$ titrations are carried out only in the presence of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$.