Question

Identify the oxidising and reducing agents in the following reactions:

$3{\mathrm{MnO}}_2+4\mathrm{Al}\to 3\mathrm{Mn}+2{\mathrm{Al}}_2{\mathrm{O}}_3$

Answer 1

• Oxidation means loss of electron in this oxidation state increases.
• Reduction means gain of electron in this oxidation state decreases.
• The one who gets oxidised act as a reducing agent and the one who get reduced act as an oxidising agent.
• Reaction:

$3{\mathrm{MnO}}_2\left(\mathrm{s}\right)+4\mathrm{Al}\left(\mathrm{s}\right)\to 3\mathrm{Mn}\left(\mathrm{s}\right)+2{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$
(Manganese oxide) (Aluminium) (Manganese) (Aluminium oxide)

• Calculation of oxidation state:

${\mathrm{MnO}}_2$ $\mathrm{Al}{}_2\mathrm{O}{}_3$

Oxidation state of $\mathrm{Mn}$ will be: Oxidation state of $\mathrm{Al}$ will be:

Let $\mathrm{x}$ be the oxidation number of $\mathrm{Mn}$ Let $\mathrm{x}$ be the oxidation number of $\mathrm{Al}$

Oxidation state of $\mathrm{O}$ is $-2$ Oxidation state of $\mathrm{O}$ is $-2$

so for ${\mathrm{MnO}}_2$ $\mathrm{x}+2(-2)=0$ so for $\mathrm{Al}{}_2\mathrm{O}{}_3$ $2\mathrm{x}+3(-2)=0$

$\mathrm{x}-4=0$ $2\mathrm{x}=6$

$\mathrm{x}=+4$ $\mathrm{x}=+3$

Oxidation state of $\mathrm{Mn}$ in ${\mathrm{MnO}}_2$ is $+4$. Oxidation state of $\mathrm{Al}$ in $\mathrm{Al}{}_2\mathrm{O}{}_3$ is $+3$.

⦁ Here oxidation state of $\mathrm{Al}$ in $\mathrm{Al}$ is $0$ and that of $\mathrm{Mn}$ in ${\mathrm{MnO}}_2$ is$+4$ at reactant side but oxidation state of $\mathrm{Al}$ in $\mathrm{Al}{}_2\mathrm{O}{}_3$is $+3$ and that of $\mathrm{Mn}$ in $\mathrm{Mn}$ is $0$ at product side.
⦁ So, we can say that $\mathrm{Al}$ is getting oxidised since its oxidation number is increasing from $0$ to $+3$ and ${\mathrm{MnO}}_2$ is getting reduced since its oxidation number is decreasing from $+4$ to $0$.
⦁ The one who gets oxidised act as a reducing agent and the one who get reduced act as an oxidising agent.
⦁ So here $\mathrm{Al}$ is a reducing agent and ${\mathrm{MnO}}_2$is an oxidising agent.