of those reactions as well. This means that two electrons would be transferred in
each half reaction for one mole of water. The experimental number of electrons
transferred is found by multiplying the average current by the time the current ran in
seconds, which is equal to the coulombs of charge that passed through the solution.
Then you divide this amount by the moles of hydrogen gas (which is equal to the
moles of water) multiplied by 96,485. 96,485 is Faraday’s constant, which is equal to
the coulombs per mole of an electron. On the bottom of the equation you have units
of coulombs/ e

* mol of H
2
O. This is equal to mol of H
2
O * coulombs over e

. When
you bring the denominator of the equation to the numerator you get coulombs * e

over mol of H
2
O * coulombs. The coulombs cancel out and you get e

over mol of H
2
O
or electrons per mole of water.
Reference equation:
e
−
¿
mol H
2
O
e
−
¿
coulombs
¿
¿
96,485
¿
¿
=
Q
(
coulombs
)
¿
z
¿
E.
Reflection
How did the number of electrons transferred during the reaction that you measured
compare to that which should have theoretically been transferred? What may account
for any differences?
The experimentally determined number of electrons transferred during my electrolysis
of water reaction was 2.19, while the theoretical value of transferred electrons for this
reaction should have been 2. The experimental value that I obtained has a percent error
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 Spring '14
 Chemistry, Atom, Electron, Mole, Reaction, Electrons, half reactions