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

In the electrolysis of water, one faraday of electrical energy would liberate
  1. one mole of oxygen
  2. one gram atom of oxygen
  3. $8 \mathrm{~g}$ oxygen
  4. $22.4$ lit. of oxygen

Solution

According to the definition $1 \mathrm{~F}$ or $96500 \mathrm{C} 1 \mathrm{~s}$ the charge carried by $1 \mathrm{~mol}$ of electrons when water is electrolysed
$2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_{2}+4 \mathrm{e}^{-}$
So, 4 Faraday of electricity liberate $=32 \mathrm{~g}$ of $\mathrm{O}_{2}$ Thus 1 Faraday of electricity liberate
$=\frac{32}{4} \mathrm{~g}$ of $\mathrm{O}_{2}=8 \mathrm{~g}$ of $\mathrm{O}_{2}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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