The volume of H2 gas at NTP obtained by passing 4 amperes through acidified H2O for 30 minutes is

Question

The volume of H2 gas at NTP obtained by passing 4 amperes through acidified H2O for 30 minutes is (A) 0.836 L (B) 0.0432 L (C) 0 .1672 L (D) 5.6 L

Tardigrade Admin · Accepted Answer

Quantity of electricity passed = 4 A × (30 × 60s) = 7200 C 2H2O + 2e-arrow H2 + 2OH- 2 × 96500 C liberate H2 = 22.4L at STP 7200 C will liberate = (22.4/2× 96500) × 7200 C = 0.836 L

Q. The volume of $H_{2}$ gas at NTP obtained by passing $4$ amperes through acidified $H_{2} O$ for $30$ minutes is

0.836 L

0.0432 L

0 .1672 L

5.6 L

Quantity of electricity passed
$= 4 A \times (30 \times 60 s) = 7200 C$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$
$2 \times 96500 C$ liberate $H_{2} = 22.4 L$ at $STP$
$7200 C$ will liberate $= \frac{22.4}{2 \times 96500} \times 7200 C$
$= 0.836 L$

Q. The volume of H2​ gas at NTP obtained by passing 4 amperes through acidified H2​O for 30 minutes is