A) Anode
B) Cathode
C) Cathode and anode both
D) None of these
Correct Answer: A
Solution :
NaH ⇌ \[N{{a}^{+}}+{{H}^{-}}\] At anode : \[{{H}^{-}}\to H+{{e}^{-}}\] \[H+H\to {{H}_{2}}\]
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