Justify that the reaction: $2 Na ( s ) + H_2 ( g ) \rightarrow 2 NaH ( s )$ is a redox change.
(N/A) The reaction $2 Na ( s ) + H_2 ( g ) \rightarrow 2 NaH ( s )$ involves the formation of an ionic compound,which can be represented as $Na^{+} H^{-} ( s )$.
This process can be split into two half-reactions:
$1. \text{Oxidation half-reaction: } 2 Na ( s ) \rightarrow 2 Na^{+} ( s ) + 2 e^-$
$2. \text{Reduction half-reaction: } H_2 ( g ) + 2 e^- \rightarrow 2 H^{-} ( s )$
In this reaction,$Na$ loses electrons (is oxidized) and $H_2$ gains electrons (is reduced).
Since both oxidation and reduction occur simultaneously,the reaction is a redox change.
This process can be split into two half-reactions:
$1. \text{Oxidation half-reaction: } 2 Na ( s ) \rightarrow 2 Na^{+} ( s ) + 2 e^-$
$2. \text{Reduction half-reaction: } H_2 ( g ) + 2 e^- \rightarrow 2 H^{-} ( s )$
In this reaction,$Na$ loses electrons (is oxidized) and $H_2$ gains electrons (is reduced).
Since both oxidation and reduction occur simultaneously,the reaction is a redox change.
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