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Chemistry
During electrolysis of NaOH
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Q. During electrolysis of $NaOH$
Electrochemistry
A
$H_{2}$ is liberated at cathode
39%
B
$O_{2}$ is liberated at cathode
29%
C
$H_{2}$ is liberated at anode
10%
D
$O_{2}$ is liberated at anode
23%
Solution:
During electrolysis of $NaOH$
At anode; $4OH^{-} \to 2H_{2}O + O_{2} + 4e^{-}$