Question

When an electric current is passed through acidulated, water, 112 mL of hydrogen gas at NTP collects at the cathode in 965 seconds. The current passed, in amperes, is

Answer 1

22,400 mL of hydrogen at STP (or NTP) = 2g
So 112 mL of hydrogen at STP $=\frac{2 g \times 112 m L}{22,400 m L}=10^{- 2}g$
$2H^{+}+\underset{2 F}{2 e^{-}} \rightarrow \underset{1 \text{ mol}}{H_{2}}$
$=2\times 96500 \, C=2g$
2 g hydrogen is deposited by $=2\times 96500C$
So $10^{- 2}$ g hydrogen will be deposited by
$=\frac{2 \times 96500 \times 1 0^{- 2} g}{2 g}=965 \, C$
Charge $\left(Q\right)=it$
So $i=\frac{Q}{t}=\frac{965 \text{ C}}{965 \text{ s}}=1\text{ C}\text{s}^{- 1}=1 \, A$