When electric current is passed through acidified water for 1930 s, 1120 mL of `H_(2)` gas is collected (at STP) at the cathode. What is the current passed in amperes?

Options

A. `0.05`
B. `0.50`
C. `5.0`
D. 50

Answer

C

Explanation

Moles of `H_(2)` collected = `(1120)/(22400) = (1)/(20)mol`
`1mol H_(2) H_(2) equiv 2F ("coulomb")`
`(1)/(20)molH_(2)equiv 2Fxx(1)/(20) = 0.1F` Coulomb
Thus `0.1F = 1930 xxi rightarrow 9650 = 2xx965xxi Rightarrow i= 5A`