Redox Reactions And Electrochemistry Ques 624

Question: In the electrolysis of water, one faraday of electrical energy would liberate

Options:

A) one mole of oxygen

B) one gram atom of oxygen

C) 8 g oxygen

D) 22.4 lit. of oxygen

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Answer:

Correct Answer: C

Solution:

[c] According to the definition 1 F or 96500 C is the charge carried by 1 mol of electrons when water is electrolysed $ 2H_2O\xrightarrow{{}}4{H^{+}}+O_2+4{e^{-}} $

So, 4 Faraday of electricity liberate = 32 g of $ O_2 $ . Thus 1 Faraday of electricity liberate $ =\frac{32}{4}g,of,O_2=8g,of,O_2 $



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