Redox Reactions And Electrochemistry Question 539
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
Show Answer
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 $
BETA
