Calculate the volume of hydrogen at STP obtained by passing a current of 0.4 ampere through acidified water for 30 minutes.

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Solution

$H_{2} O ⟶ {2 H}^{+} + \frac{1}{2} O_{2} + {2 e}^{-} {2 H}^{+} + {2 e}^{-} ⟶ H_{2} 2 F g i v e n 1 m o l e H_{2} q = i \times t = 0.4 \times 30 \times 60 = 720 720 C g i v e n = \frac{720}{193000} m o l e H_{2} = 0.00373 m o l e H_{2} = 0.00373 \times 22400 = 83.6 m l H_{2} o f S T P$

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Calculate the volume of hydrogen at STP obtained by passing a current of 0.4 ampere through acidified water for 30 minutes.

H2O⟶2H++12O2+2e−2H++2e−⟶H22Fgiven1moleH2q=i×t=0.4×30×60=720720Cgiven=720193000moleH2=0.00373moleH2=0.00373×22400=83.6mlH2ofSTP

$H_{2} O ⟶ {2 H}^{+} + \frac{1}{2} O_{2} + {2 e}^{-} {2 H}^{+} + {2 e}^{-} ⟶ H_{2} 2 F g i v e n 1 m o l e H_{2} q = i \times t = 0.4 \times 30 \times 60 = 720 720 C g i v e n = \frac{720}{193000} m o l e H_{2} = 0.00373 m o l e H_{2} = 0.00373 \times 22400 = 83.6 m l H_{2} o f S T P$