Question

Which of the following is a redox reaction?
A. $\text{NaOH}+\text{HCl}\to \text{NaCl}+{{\text{H}}_{2}}\text{O}$
B. $\text{AgN}{{\text{O}}_{3}}+\text{KI}\to \text{AgI}+\text{KN}{{\text{O}}_{3}}$
C. $\text{Ba}{{\text{O}}_{2}}+{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to \text{BaS}{{\text{O}}_{4}}+{{\text{H}}_{2}}{{\text{O}}_{2}}$
D. $\text{SnC}{{\text{l}}_{2}}+\text{HgC}{{\text{l}}_{2}}\to \text{SnC}{{\text{l}}_{4}}+\text{Hg}$

Answer 1

Hint: Check the oxidation states of elements in reactants side and products side. If the oxidation state of a particular element in reactants side and products side is same, then, it will not be a redox reaction. If the oxidation state of a particular element in reactants side and products side is different, then, it will be a redox reaction.

Complete step by step answer:
Let us discuss this question option wise, because by that only we can identify the redox reaction.
A. $\text{NaOH}+\text{HCl}\to \text{NaCl}+{{\text{H}}_{2}}\text{O}$: This is a neutralisation reaction.
The oxidation states of elements in reactants side:
- The oxidation state of alkali metals like sodium is fixed which is +1.
So, the oxidation state of hydroxide ions will be -1.
- The oxidation state of ${{\text{H}}^{+}}$ ion is +1.
So, the oxidation state of chlorine will be -1.
The oxidation state of elements in products side:
- The oxidation state of alkali metals like sodium is fixed which is +1.
So, the oxidation state of chlorine will be -1. Sodium chloride is formed as a salt.
- The oxidation state of ${{\text{H}}^{+}}$ ion is +1.
So, the oxidation state of hydroxide ions will be -1.
As, there is no change in oxidation states of elements while going through the reaction. So, it is not a redox reaction.
B. $\text{AgN}{{\text{O}}_{3}}+\text{KI}\to \text{AgI}+\text{KN}{{\text{O}}_{3}}$: This is a precipitation reaction.
The oxidation states of elements in reactants side:
- The oxidation state of silver metal is fixed which is +1.
So, the oxidation state of nitrate ion will be -1.
- The oxidation state of alkali metals like potassium ion is fixed which is +1.
So, the oxidation state of iodine will be -1.
The oxidation state of elements in products side:
- The oxidation state of silver metal is +1.
So, the oxidation state of iodine will be -1. Silver iodide forms the precipitates of yellow colour.
- The oxidation state of potassium is +1.
So, the oxidation state of nitrate ion will be -1.
As, there is no change in oxidation states of elements while going through the reaction. So, it is not a redox reaction.
C. $\text{Ba}{{\text{O}}_{2}}+{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to \text{BaS}{{\text{O}}_{4}}+{{\text{H}}_{2}}{{\text{O}}_{2}}$: This is a displacement reaction.
The oxidation states of elements in reactants side:
- The oxidation state of alkali earth metals like barium is fixed which is +2.
So, the oxidation state of $\text{O}_{2}^{2-}$ ion will be $\left[ 2\times \text{x} \right]=-2$.
Then, the oxidation state of oxygen will be -1. This ion is peroxide.
- The oxidation state of ${{\text{H}}^{+}}$ ion is +1. The oxidation number of ${{\text{H}}^{+}}$ is $2\times \left( +1 \right)$ or +2.
So, the oxidation state of sulphate ion will be -2.
The oxidation state of elements in products side:
- The oxidation state of barium is fixed which is +1. So, the oxidation state of sulphate will be -2. The compound is barium sulphate.
- The oxidation state of ${{\text{H}}^{+}}$ ion is +1. The oxidation number of ${{\text{H}}^{+}}$ is $2\times \left( +1 \right)$ or +2. So, the oxidation state of oxygen in peroxide ion will be -1. The compound is hydrogen peroxide.
As, there is no change in oxidation state of elements while going through the reaction. So, it is not a redox reaction.
D. $\text{SnC}{{\text{l}}_{2}}+\text{HgC}{{\text{l}}_{2}}\to \text{SnC}{{\text{l}}_{4}}+\text{Hg}$:
The oxidation states of elements in reactants side:
- The oxidation state of chlorine is -1.
So, the oxidation state of tin $\left( \text{Sn} \right)$ will be $\left[ \text{x}+\left( 2\times -1 \right) \right]=0$ or +2.
- The oxidation state of chlorine is -1.
So, the oxidation state of mercury $\left( \text{Hg} \right)$ will be $\left[ \text{x}+\left( 2\times -1 \right) \right]=0$ or +2.
The oxidation state of elements in products side:
- The oxidation state of chlorine is -1.
So, the oxidation state of tin $\left( \text{Sn} \right)$ will be $\left[ \text{x}+\left( 4\times -1 \right) \right]=0$ or +4.
- The oxidation state of elemental mercury will be 0.
As, there is a change in oxidation states of elements like tin from +2 to +4 and mercury from +2 to 0 while undergoing the reaction. So, it is a redox reaction.

$\text{SnC}{{\text{l}}_{2}}+\text{HgC}{{\text{l}}_{2}}\to \text{SnC}{{\text{l}}_{4}}+\text{Hg}$ is a redox reaction, the correct option is option ‘d’.

Note: Here, in the reaction, tin undergoes an oxidation reaction. The oxidation state is increased from +2 to +4. On the other hand, mercury had undergone a reduction reaction. As, the oxidation state is decreased from +2 to 0. In a redox reaction, one will undergo oxidation and another will undergo reduction. This is a rule.