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
