Time required to completely decompose `2` moles of water using a current of `2` amperes is

Options

A. `35.61` hours
B. `61.53` hours
C. `16.35` hours
D. `53.61` hours

Answer

D

Explanation

Overall reaction during the electrolysis of water is
`underset(4eq.)underset(2mol)(2H_(2)O)rarrunderset(4eq.)underset(2mol)(2H_(2))+underset(4eq.)underset(2mol)(O_(2))`
Note that `1` mol of `O_(2) (32 g)` contains `4` equivalents of `O_(2)` [8 g is one equivalent of `O_(2)`] and accoding to law of equivalents, the number of equivalents of different substances are always equal.
According the Faradays law
No. of equivalents of substance decomposed by electrolysis
= No. of faradays of electricity passed
Thus, to decompse `2` moles of `H-(2)O` (i.e. `4` equivalents of `H_(2)O`), we need `4` faradays or `4 xx 96500` coulombs.
The relationship between charge, current, and time usually is written as
`I = (Q)/(t)`
where `I` is current, `t` is time and `Q` is charge measured respectively in coulombs, amperes and seconds.
Therefore
`t = (Q)/(I) = (4 xx 96500 C)/(2 amp)`
`= 193000` seconds
`= (193000 s) ((1 hrs)/(3600 s))`
`= 53.61 hr`