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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
58%
B
$O_2$ is liberated at cathode
11%
C
$H_2$ is liberated at anode
5%
D
$O_2$ is liberated at anode.
26%
Solution:
During electrolysis of $NaOH$ (molten), $O_2$ if liberated at anode.
$4OH^- \rightarrow 2H_2O + O_2 + 4e^-$