20.10 – 20.11 Student
Notes
B. Aqueous Electrolysis
1. Water can be
oxidized or reduced in a half reaction.
a. Reduce
water to
b.
Oxidize water to
2.
Electrolysis of
Sulfuric Acid Solutions
a.
Possible
half-reactions at the cathode
b. Possible
half-reaction at the anode
c. Overvoltage
–
d. Net cell
reaction is the _______________. H2SO4 was used
3. Electrolysis
of Sodium Chloride Solutions
a. Depends on
the concentration of the Cl- (according to
the Nernst equation): high concentration of Cl-
then ____is the product; low concentration of Cl-
then _____is the product
b. Chlor-alkali membrane cell –
c. Chlor-alkali mercury cell –
4. Electroplating
of Metals
a. Electrogalvanizing – (zinc electroplating) –
b. Purify
some metals
c. Predicting
the Half-Reactions in an Aqueous Electrolysis
What
do you expect to be the half reactions in the electrolysis of aqueous copper
(II) sulfate?
C. Stoichiometry
of Electrolysis
1. Faraday’s
Laws
a.
b.
2. One Faraday
(9.65 x 104 C) is equivalent
3. Electric
charge =
Coulombs =
Amperes – A –
4. Calculating
the Amount of Charge from the Amount of Product in an Electrolysis
When
an aqueous solution of copper (II) sulfate, CuSO4, is electrolyzed,
copper metal is deposited.
Cu2+ (aq) + 2 e-
ŕ Cu (s)
(The
other electrode reaction gives oxygen: 2 H2O ŕ O2 + 4 H+ + 4 e-.)
If a constant current was passed for 5.00 h and 404 mg of copper metal was
deposited, what was the current?
5. Calculating
the Amount of Product from the Amount of Charge in an Electrolysis
When
an aqueous solution of potassium iodide is electrolyzed using platinum
electrodes, the half-reactions are
2 I- (aq) ŕ I2 + 2 e-
2 H2O (l) + 2 e- ŕ H2 (g) + 2
How
many grams of iodine are produced when a current of 8.52 mA
flows through the cell for 10.0 min?