
Electrolysis of fused sodium hydride liberate hydrogen at the
A. Anode
B. Cathode
C. Cathode and anode both
D. None of these
Answer
162.3k+ views
Hint: Electrolysis is a process through which chemical decomposition of an electrolyte in a solution is done by electric current. An electrolytic cell is a vessel in which the process of electrolysis takes place. Two metallic rods called electrodes are placed in the electrolytic solution and are connected to the two terminals of a battery with the help of a wire. The two electrodes are called anode ( which connects to the positive terminal) and cathode (which connects to the negative terminal).
Complete Step by Step Answer:
Hydrogen loses one electron to form H- ion whereas sodium gains one electron to form $Na^{+}$ ion. When $H^{-}$ and $Na^{+}$ react, they form $NaH$ (sodium hydride). Hydrogen readily reacts with metals to form its hydride and shows its electronegative nature.
During electrolysis of fused sodium hydride, sodium is liberated in its liquid form whereas hydrogen in its gaseous form. Now in $NaH$, hydrogen is showing its electronegative character, it will be liberated around the electrode which is connected to the positive terminal of the battery which is anode. On the other hand, sodium due its electropositive nature is liberated at cathode.
At cathode:
$2Na^{+}{+} 2e^{-}\longrightarrow 2Na(l)$
At anode:
$2H^{-}\longrightarrow H_{2}(g){+}2e^{-}$
Hence, the correct answer is Option (A).
Note: Hydrogen shows electropositive nature as well while reacting with a non-metal. So, during electrolysis of water where hydrogen shows electropositive character, it is liberated at cathode whereas oxygen is liberated at anode. During electrolysis of water, both oxygen and hydrogen are liberated in gaseous form. Hydrogen liberated through electrolysis is very clean and a good renewable fuel resource.
Complete Step by Step Answer:
Hydrogen loses one electron to form H- ion whereas sodium gains one electron to form $Na^{+}$ ion. When $H^{-}$ and $Na^{+}$ react, they form $NaH$ (sodium hydride). Hydrogen readily reacts with metals to form its hydride and shows its electronegative nature.
During electrolysis of fused sodium hydride, sodium is liberated in its liquid form whereas hydrogen in its gaseous form. Now in $NaH$, hydrogen is showing its electronegative character, it will be liberated around the electrode which is connected to the positive terminal of the battery which is anode. On the other hand, sodium due its electropositive nature is liberated at cathode.
At cathode:
$2Na^{+}{+} 2e^{-}\longrightarrow 2Na(l)$
At anode:
$2H^{-}\longrightarrow H_{2}(g){+}2e^{-}$
Hence, the correct answer is Option (A).
Note: Hydrogen shows electropositive nature as well while reacting with a non-metal. So, during electrolysis of water where hydrogen shows electropositive character, it is liberated at cathode whereas oxygen is liberated at anode. During electrolysis of water, both oxygen and hydrogen are liberated in gaseous form. Hydrogen liberated through electrolysis is very clean and a good renewable fuel resource.
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