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During electrolysis of NaOHA. `H_(2)` is liberarted at cathodeB. `O_(2)` is liberated at cathodeC. `H_(2)` is liberated at anodeD. `O_(2)` is liberated at anode. |
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Answer» Correct Answer - D During electrolysis of NaOH (molten), `O_(2)` is liberated at anode. `4OH^(-)rarr2H_(2)O+O_(2)+4e^(-)` |
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