Question

A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 2.50 min , 14.4 mL of water-saturated H2(g) is collected. The temperature and pressure of the system are 25 ∘C and 715 torr .

Calculate the magnitude of the current in amperes if the number of moles of hydrogen gas produced is 5.36×10−4 mol .

2H2O----> 2H2(g) + O2(g)

Answer 1

If there is nothing but hydrogen (H2) and water vapor in the gas collected,first we
must find out how much water vapor is in the gas. At 25°C (298°K) the vapor
pressure of water is 23.75 torr,

so the partial pressure of H2 in the gas = 715 torr - 23.75 torr = 691.25 torr
At STP the volume of H2 = 14.4 ml * (691.25 / 760) * (273 / 298) = 12 ml
Since 1 mole of gas occupies 22.4 liters at STP,

moles of H2 = (1 / 22.4 *1000) *12 ml = 5.36 * 10^-4 moles

One mole of H2 contains 6.023 * 10^23 molecules, and we have 5.36 * 10^-4 mole

molecules of H2 = 3.228 * 10^20 molecules,

and each molecule requires two electrons to form H2. So we have a total of 6.456 * 10^20 electrons (H) that have flowed .during these 2.5 minutes (2 minutes and 30 seconds, or 150 seconds).

One ampere is defined as a flow rate of one coulomb of electrons per second,
1C = 6.24 * 10^18 electrons per second.

A total of 6.456 * 10^20 electrons = 103.4 C
103.4 C flowed over a time of 150 seconds, so the flow rate in amperes
A = 103.4 C / 150 s, = 0.689 A = (689 mA).