Question

What is a redox reaction? Identify the substances oxidized and the substances reduced in the following reactions.
$1.$ $2PbO + C \to 2Pb + C{O_2}$
$2.$ $Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}$

Answer 1

Hint: Redox reactions are very common and vital to some of the functions of life such as; photosynthesis, respiration, rusting etc. Redox reactions involve transfer of electrons between two species.

Complete answer:
A redox (reduction-oxidation) reaction is any chemical reaction where the oxidation state of a molecule, atom or ion changes by gaining or losing an electron. We can also define redox reactions as the reactions in which one species is oxidized and another is reduced.
Now, we’ll learn about oxidation and reduction;
Oxidation: It involves loss of electrons as a result the oxidation state will increase. The oxidized half in the redox reaction loses its electrons.
Reduction: It involves gain of electrons as a result of which the oxidation state decreases. The reduced half in the redox reaction gains electrons.
The electrons given by the oxidized half are gained by the reduced half in the redox reaction. Hence, there is transfer of electrons.
In order to find out which substance is oxidized and which one is reduced, we’ll check the oxidation state. If the oxidation state increases, the substance is oxidized and if it decreases, the substance is surely reduced.
Let’s see the first reaction;
$2PbO + C \to 2Pb + C{O_2}$
The oxidation state of $Pb$ in $PbO$ is ; $x + ( - 2) = 0$ , $x = 2$
While the oxidation state of $Pb$(in product side) is; $x = 0$
The oxidation state decreases ( $2$ to $0$ ), thus $PbO$ is reduced to $Pb$
The oxidation state of $C$ (reactant side) is; $x = 0$
And the oxidation state of $C$ in $C{O_2}$ is; $x + 2( - 2) = 0$ , $x = 4$
The oxidation state increases ( $0$ to $4$ ), thus $C$ is oxidized to $C{O_2}$
Second reaction is;
$Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}$
The oxidation state of $Mn$ in $Mn{O_2}$ (reactant side) is; $x + 2( - 2) = 0$ , $x = 4$
And the oxidation state of$Mn$ in $MnC{l_2}$ (product side) is; $x + 2( - 1) = 0$ , $x = 2$
The oxidation state decreases ( $4$ to $2$ ), thus $Mn{O_2}$ is reduced to $MnC{l_2}$
The oxidation state of $Cl$ in $HCl$ (reactant side) is ; $x + 1 = 0$ , $x = - 1$
And the oxidation state of $Cl$ in product side is; $x = 0$
The oxidation state increases ( $ - 1$ to $0$) , thus $HCl$ is oxidized to $C{l_2}$

Note:
We have seen that the substance oxidized loses its electron that results in increase of its oxidation state, while the substance reduced gains the electrons that results in decrease in its oxidation state.