Question 6e25b

May 23, 2017

$193000 C$

Explanation:

The half reactions for this reaction are

$2 {\text{H"_"2""O" + "2e"^-) rarr "O"_2(g) + "4H"^+"(aq)" + "4e}}^{-}$

${\text{2H"^+ + "2e"^-) rarr "H}}_{2}$

$2 \text{H"_"2""O" + "2e"^-) rarr "O"_2(g) + "(""4H"^+"(aq)" + "4e"^-)")}$

${\text{4H"^+ + "4e"^-) rarr 2"H}}_{2}$

Overall reaction${\text{ "2H_2O rarr 2"H}}_{2} + {O}_{2}$

This means that for every $2$ mol of electrons 1 mole of ${\text{O}}_{2}$ and $2$ mol of ${\text{H}}_{2}$ are formed

So if think clearly this is $\text{stoichiometry}$ actually

And we are assuming that the electrolysis is done under standard conditions

$\text{moles of "O_2 = "weight"/ "molar mass}$

$\frac{\text{2 mol" = "32g}}{16}$

If 2 mol of ${\text{O}}_{2}$ 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.

$2 \text{H"_"2""O" + "2e"^-) rarr "O"_2(g) + "(""4H"^+"(aq)" + "4e"^-)")}$

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

$x \text{ coulombs" xx "1 mol of electrons"/ "96500C} = 2 C$

$x \text{ coulombs" = "96500C" * "2C}$

$= 193000 C$

Now amps could be different