A) Hydrogen is liberated at the anode
B) Hydrogen is liberated at the cathode
C) No reaction takes place
D) Hydride ion migrates towards cathode
Correct Answer: A
Solution :
\[{{M}^{+}}{{H}^{-}}\to {{M}^{+}}+\underset{\text{Hydride ion}}{\mathop{{{H}^{-}}\ \ }}\,\] \[{{H}^{-}}\to \frac{1}{2}{{H}_{2}}+{{e}^{-}}\](At anode)
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