
Assertion:In the given chemical equation, sodium is getting oxidised and H, is getting reduced.
\[2Na + {H_2} \to 2NaH\]
Reason: The chemical reactions in which oxidation and reduction take place simultaneously, are called redox reactions.
(a) Both assertion and reason are correct and the reason is the correct explanation for assertion.
(b) Both assertion and reason are correct but the reason is not the correct explanation of assertion.
(c) Assertion is correct but the reason is incorrect
(d) Both assertion and reason are incorrect
Answer
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Hint: We know that, in a chemical reaction, an oxidising agent, also known as an oxidant, gains electrons and is reduced. Also, in a chemical reaction, a reducing agent, also known as a reductant, loses electrons and is oxidised. When a reducing agent is in one of its lower oxidation states, it is referred to as an electron donor.
Complete answer: Here, NaH can be written as \[N{a^ + }{H^ - }\] since it is an ionic compound
\[2Na(s) + {H_2}(g) \to 2N{a^ + }{H^ - }\]
This reaction can be divided into two halves as follows:
\[Na(s) \to N{a^ + } + {e^ - }\]; This is the oxidation half reaction. Here we can see the loss of electrons.
\[{H_2}(s) + 2{e^ - } \to 2{H^ - }\] ; This is the reduction half reaction. Here we can see the gain of electrons.
Hence, we can write the overall reaction as,
\[2Na(s) + {H_2}(g) \to 2N{a^ + }{H^ - }(s)\]
That is, in this reaction, Na is oxidised to\[N{a^ + }\], and hydrogen from \[{H_2}\]and \[{H^ - }\]ion is reduced. As a result, it's a redox reaction.
Note:
The loss of electrons or an increase in the oxidation state of an atom, an ion, or any atoms in a molecule is referred to as oxidation. The gain of electrons or a decrease in the oxidation state of an atom, an ion, or any atoms in a molecule is referred to as reduction (a reduction in oxidation state).
Complete answer: Here, NaH can be written as \[N{a^ + }{H^ - }\] since it is an ionic compound
\[2Na(s) + {H_2}(g) \to 2N{a^ + }{H^ - }\]
This reaction can be divided into two halves as follows:
\[Na(s) \to N{a^ + } + {e^ - }\]; This is the oxidation half reaction. Here we can see the loss of electrons.
\[{H_2}(s) + 2{e^ - } \to 2{H^ - }\] ; This is the reduction half reaction. Here we can see the gain of electrons.
Hence, we can write the overall reaction as,
\[2Na(s) + {H_2}(g) \to 2N{a^ + }{H^ - }(s)\]
That is, in this reaction, Na is oxidised to\[N{a^ + }\], and hydrogen from \[{H_2}\]and \[{H^ - }\]ion is reduced. As a result, it's a redox reaction.
Note:
The loss of electrons or an increase in the oxidation state of an atom, an ion, or any atoms in a molecule is referred to as oxidation. The gain of electrons or a decrease in the oxidation state of an atom, an ion, or any atoms in a molecule is referred to as reduction (a reduction in oxidation state).
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