In order to separate oxygen from one mole of | vedclass

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(A) The chemical reaction for the electrolysis of water is: $H_2O(l) \rightarrow H_2(g) + \frac{1}{2} O_2(g)$.In this reaction,oxygen is in the $-2$ o

Vedclass · Accepted Answer

$1.93 	imes 10^5$

Vedclass · Answer

(A) The chemical reaction for the electrolysis of water is: $H_2O(l) ightarrow H_2(g) + \frac{1}{2} O_2(g)$.In this reaction,oxygen is in the $-2$ oxidation state in $H_2O$ and becomes $0$ in $O_2$.For $\frac{1}{2} 	ext{ mole of } O_2$,the change in oxidation state is $2$ electrons per mole of $O$ atom.Since we have $1 	ext{ mole of } H_2O$,it contains $1 	ext{ mole of } O$ atoms.The reaction for the oxidation of oxygen is: $O^{2-} ightarrow \frac{1}{2} O_2 + 2e^-$.Thus,$n = 2$ moles of electrons are required for $1 	ext{ mole of } H_2O$.Using Faraday's law,$Q = n 	imes F = 2 	imes 96487 \ C \approx 192974 \ C$.Rounding to significant figures,$Q = 1.93 	imes 10^5 \ C$.

In order to separate oxygen from one mole of $H_2O$,the required quantity of electricity in coulombs is:

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