A) \[11.2\ d{{m}^{3}}\]
B) \[5.6\,d{{m}^{3}}\]
C) \[22.4\ d{{m}^{3}}\]
D) \[1.0\ d{{m}^{3}}\]
Correct Answer: B
Solution :
Reaction for electrolysis of water is \[2{{H}_{2}}O\] ⇌ \[4{{H}^{+}}+2{{O}^{2-}}\] \[2{{O}^{2-}}\to {{O}_{2}}+4{{e}^{-}}\] \[4{{e}^{-}}+4{{H}^{+}}\to 2{{H}_{2}}\] \[\because \] \[n=4\] so 4 Faraday charge will liberate 1 mole \[=22.4\ d{{m}^{3}}\]oxygen \[\therefore \] 1 Faraday charge will liberate \[\frac{22.4}{4}=5.6\ d{{m}^{3}}\ {{O}_{2}}\].
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