Question #a4ee5

The first thing to do here is to write the reduction half-reaction that describes the reduction of hydrogen ions, #"H"^(+)#, which you'll sometimes see written as #"H"_3"O"^(+)#, to hydrogen gas, #"H"_2#.

#2"H"_ ((aq))^(+) + color(blue)(2)"e"^(-) -> "H"_ (2(g))#

This half-reaction takes place at the cathode. You don't really need to write the oxidation half-reaction to answer this question, so I'll skip that part.

The important part to notice here is that you need #color(blue)(2)# moles of electrons to produce #1# mole of hydrogen gas. Keep this in mind.

Now, you know that you must produce #"1.12 cm"^3# of hydrogen gas per second under STP conditions. I'll use the old definition of STP conditions since it's very likely that this was the definition given to you.

So, STP conditions were defined as a pressure of #"1 atm"# and a temperature of #0^@"C"#. Under these conditions for pressure and temperature, one mole of any ideal gas occupies #"22.4 L"#.

This means that your electrolysis must produce

#1.12 color(red)(cancel(color(black)("cm"^3))) * (1 color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * (1 color(red)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("dm"^3)))) * "1 mole H"_2/(22.4color(red)(cancel(color(black)("L")))) = 5.0 * 10^(-5)"moles H"_2#

Use the aforementioned mole ratio to calculate the number of moles of electrons needed

#5.0 * 10^(-5) color(red)(cancel(color(black)("moles H"_2))) * (color(blue)(2)color(white)(a)"moles e"^(-))/(1color(red)(cancel(color(black)("mole H"_2)))) = 1.0 * 10^(-4)"moles e"^(-)#

Now, this many moles of electrons must be delivered by the cathode per second. As you know, an ampere is defined as one coulomb per second.

#"1 A" = "1 C"/"1 s"#

One coulomb is simply the magnitude of the charge carried by #6.242 * 10^(18)# electrons. Use Avogadro's number to calculate the number of electrons present in that many moles

#1.0 * 10^(-4) color(red)(cancel(color(black)("moles e"^(-)))) * (6.022 * 10^(23)"e"^(-))/(1color(red)(cancel(color(black)("mole e"^(-))))) = 6.022 * 10^(19)"e"^(-)#

This means that you have

#6.022 * 10^(19)color(red)(cancel(color(black)("e"^(-)))) * "1 C"/(6.242 * 10^(18)color(red)(cancel(color(black)("e"^(-))))) = "9.65 C"#

Therefore, you can say that if you need #"9.65 C"# per second, you need to have a current of

#"9.65 C"/"1 s" = color(green)(|bar(ul(color(white)(a/a)color(black)("9.65 A")color(white)(a/a)|)))#

ALTERNATIVELY

You can get the answer a little quicker by using Faraday's constant, #F#, which gives you the magnitude of the charge carried by one mole of electrons.

#F = "96,485.33 C mol"^(-1)#

You have

#1.0 * 10^(-4)color(red)(cancel(color(black)("moles e"^(-)))) * "96,485.33 C"/(1color(red)(cancel(color(black)("mole e"^(-))))) = "9.65 C"#

In one second, you will once again get

#"9.65 C"/"1 s" = color(green)(|bar(ul(color(white)(a/a)color(black)("9.65 A")color(white)(a/a)|)))#