Justify the following reaction as redox reaction $2Na + S\left( s \right) \to N{a_2}S\left( s \right)$
Find out the oxidising and reducing agents.
Answer
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Hint: To answer this question, relate the oxidation states of sodium atom (Na) and sulphur atom (S) on the reactant side and product side. The element which itself gets reduced is called oxidising agent and the element which itself gets oxidised is called reducing agent.
Complete answer:
Let us first look at some important definitions.
Oxidation: An increase in the oxidation number of an element in the given substance is called oxidation.
Reduction: It is defined as the reaction in which an element gains electrons and there is a decrease in the oxidation number of an element in the given substance.
Oxidising agent: The element which itself gets reduced and oxidised the other element in a redox reaction is called oxidising agent. Reducing agent: The element which itself gets oxidised and reduces the other element in a redox reaction is called reducing agent.
Redox reaction: Reaction which involves change in the oxidation state of the interacting species or when oxidation and reduction occur simultaneously in a reaction, the reaction is known as redox reaction.
Now the given reaction is:
$2Na + S\left( s \right) \to N{a_2}S\left( s \right)$
On reactant side:
Oxidation state of sodium (Na) = 0
Oxidation state of sulphur (S) = 0
On product side:
Oxidation state of sodium in $N{a_2}S$ = +1
Oxidation state of sulphur in $N{a_2}S$= -2
Hence, the oxidation state of sodium (Na) is increasing from 0 to +1, and thus it is getting oxidised and will act as a reducing agent. Oxidation state of sulphur (S) is reduced from 0 to -2. Thus, it is getting reduced and will act as an oxidising agent. Also, there is simultaneously both oxidation and reduction, so a given reaction is justified as a redox reaction.
Note: In oxidation reaction, an element loses its electrons whereas in reduction reaction, an element gains electrons. In the given reaction, sodium atom loses its one electron while sulphur atom gains that one electron.
Complete answer:
Let us first look at some important definitions.
Oxidation: An increase in the oxidation number of an element in the given substance is called oxidation.
Reduction: It is defined as the reaction in which an element gains electrons and there is a decrease in the oxidation number of an element in the given substance.
Oxidising agent: The element which itself gets reduced and oxidised the other element in a redox reaction is called oxidising agent. Reducing agent: The element which itself gets oxidised and reduces the other element in a redox reaction is called reducing agent.
Redox reaction: Reaction which involves change in the oxidation state of the interacting species or when oxidation and reduction occur simultaneously in a reaction, the reaction is known as redox reaction.
Now the given reaction is:
$2Na + S\left( s \right) \to N{a_2}S\left( s \right)$
On reactant side:
Oxidation state of sodium (Na) = 0
Oxidation state of sulphur (S) = 0
On product side:
Oxidation state of sodium in $N{a_2}S$ = +1
Oxidation state of sulphur in $N{a_2}S$= -2
Hence, the oxidation state of sodium (Na) is increasing from 0 to +1, and thus it is getting oxidised and will act as a reducing agent. Oxidation state of sulphur (S) is reduced from 0 to -2. Thus, it is getting reduced and will act as an oxidising agent. Also, there is simultaneously both oxidation and reduction, so a given reaction is justified as a redox reaction.
Note: In oxidation reaction, an element loses its electrons whereas in reduction reaction, an element gains electrons. In the given reaction, sodium atom loses its one electron while sulphur atom gains that one electron.
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