The half reactions for this reaction are
#2"H"_"2""O" + "2e"^-) rarr "O"_2(g) + "4H"^+"(aq)" + "4e"^-#
#"2H"^+ + "2e"^-) rarr "H"_2#
#2"H"_"2""O" + "2e"^-) rarr "O"_2(g) + "(""4H"^+"(aq)" + "4e"^-)")"#
#"4H"^+ + "4e"^-) rarr 2"H"_2#
Overall reaction#" "2H_2O rarr 2"H"_2 + O_2#
This means that for every #2# mol of electrons 1 mole of #"O"_2# and #2# mol of #"H"_2# are formed
So if think clearly this is #"stoichiometry"# actually
And we are assuming that the electrolysis is done under standard conditions
#"moles of "O_2 = "weight"/ "molar mass"#
#"2 mol" = "32g"/16#
If 2 mol of #"O"_2# are produced then if it's not a #100%# yield and a #50%#(which is in your case) yield for hydrogen .
This is because actually 4g #H_2# should be produced. Maybe it got burnt up.Or the electrode working to make hydrogen was electric resistive.
But if #100%# of #O_2# is formed then the electricity used is the same.
#2"H"_"2""O" + "2e"^-) rarr "O"_2(g) + "(""4H"^+"(aq)" + "4e"^-)")"#
Multiply both sides with #2#
#4"H"_"2""O" + "4e"^-) rarr 2"O"_2(g) + "8H"^+"(aq)" + "8e"^-)#
#4# electrons are required.
Now you calculate the Coulombs using the electrons required
#x" coulombs" xx "1 mol of electrons"/ "96500C" = 2C#
#x" coulombs" = "96500C" * "2C"#
#= 193000C#
Now amps could be different
