Question

Which of the following reactions is an example of auto-reduction?
(A) $F{e_3}{O_4} + 4CO \to 3Fe + 4C{O_2}$
(B)$CuO + C \to 2Cu + CO$
(C) $Cu_{(aq)}^{ + 2} + F{e_{(s)}} \to C{u_{(s)}} + Fe_{(aq)}^{( + 2)}$
(D)$C{u_2}O + \dfrac{1}{2}C{u_2}S \to 2Cu + \dfrac{1}{2}S{O_2}$

Answer 1

The auto reduction process is widely in metallurgical industries to extract metals such as Hg, Pb, Cu, etc. from its ore. In an auto reduction reaction the reduction process takes place without any help of reducing agent.

Complete answer:
When a molecule or an atom loses an electron that molecule /atom is said to undergo oxidation. When a molecule/atom gains an electron in course of reaction that molecule/atom is said to undergo reduction. Oxidizing agent is the substance or chemical species that oxidizes another substance with which it reacts and itself gets reduced whereas, reducing agent is the substance or chemical species that reduces another substance with which it reacts and itself gets oxidized.
Auto-reduction can be defined as the process where the substance is reduced without influence of any reducing agent in course of reaction is known as auto-reduction. The word ‘auto’ means self and reduction means gaining an electron.
Let’s look into options
Option A is:
$F{e_3}{O_4} + 4CO \to 3Fe + 4C{O_2}$
$F{e^{ + 3}}$ is oxidized to Fe and C is reduced from +2 oxidation to +4 oxidation state. The chemical reaction where both oxidation and reduction takes place simultaneously is said to be a redox reaction. Therefore this reaction is a redox reaction. This option is wrong.
Option B:
$CuO + C \to 2Cu + CO$
$C{u^{ + 2}}$ is oxidized to Cu by the influence of oxidizing agent C and itself getting reduced. Therefore, $C{u^{ + 2}}$ is oxidized. This exactly meets the definition of oxidation process. Thus, this is the wrong option.
Option C is :
$C{u^{ + 2}} + Fe \to Cu + F{e^{ + 2}}$
$C{u^{ + 2}}$ is oxidized to C by the influence of oxidizing agent Fe and Fe itself getting reduced. Therefore, Cu is oxidized. This exactly meets the definition of oxidation. Thus, this is the wrong option.
Option D is:
$C{u_2}O + \dfrac{1}{2}CuS \to 2Cu + \dfrac{1}{2}S{O_2}$
Metals with less the electropositive character like Hg, Pb, Cu etc. when the sulphides of these metals are heated in presence of air, it forms oxides.
$2C{u_2}S + 3{O_2} \to 2C{u_2}O + 3S{O_2}$
The oxide of copper reacts with remaining sulphide to give the metal and $S{O_2}$.
$2C{u_2}O + C{u_2}s \to 6Cu + S{O_2}$
Here there is no reducing agent but the copper is reduced.

Thus, option D is the correct answer.

Note:
If you want to know whether the element is oxidized or reduced. Firstly, write the oxidation number of all the reactants and products. Then compare the oxidation number of an element on both reactant and product. Then predict the reaction. Metals with a less electropositive character like Hg, Pb, Cu etc. show when the sulphides of these metals are mixed with the oxides of the same metals, they undergo auto reduction.