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PRACTICE QUIZ (ELECTROCHEMISTRY)
1. (2 pts.) Combine the following half-reactions into a
balanced overall equation:
H2S(aq) ----> 2H+(aq) + S(s) + 2e-
6e- + 14H+(aq) + Cr2O72-(aq) ----> 2Cr3+(aq) + 7H2O(l)
2. (4 pts.) How many grams of Al will be produced from Al2O3
in 15.0 hrs. using a current of 5.46 amps?
3. (1 pt. each) Circle T (true) or F (false).
T F a. In the reaction 2OH- + 4ClO- + S2O32- ----> 4Cl- + 2SO42- + H2O,
the oxidizing agent is ClO-.
T F b. In the electrolysis of molten NaCl, the anode half-reaction is
Na+ + e- ----> Na.
T F c. The cathode in the lead storage battery when it is discharging
is Pb. The equation is:
Pb(s) + PbO2(s) + 2H+ + 2HSO4- ----> 2PbSO4(s) + 2H2O
T F d. Electrons are always lost at the anode.
4. Given the following reduction potentials,
(1) Cl2 + 2e- ----> 2Cl- E = 1.36 v
(2) Fe3+ + e- ----> Fe2+ E = 0.77 v
(3) Fe2+ + 2e- ----> Fe E = -0.44 v
(4) Mg2+ + 2e- ----> Mg E = -2.37 v
a. (1 pt.) Which is the strongest reducing agent?__________
b. (5 pts.) Draw a diagram for a working voltaic cell using
half-reactions (1) and (3). Label all parts, including
direction of electron flow, direction of anion flow, the
anode, the cathode, the ions in solution, and the salt bridge.
Calculate E for the cell.
5. (4 pts.) When 20.0 g of pure Cr is dissolved in an acidic
KMnO4 solution, the Cr is converted to Cr3+ and the MnO4-
to Mn2+. How many mL of 2.75 M KMnO4 solution are required
to react completely with the Cr? (HINT: balance the reaction.)
Cr ----> Cr3+ + 3e-
5e- + 8H+ + MnO4- ----> Mn2+ + 4H2O
KEY
1. (2 pts.) Combine the following half-reactions into a balanced
overall equation:
6e- + 14H+ + Cr2O72- ----> 2Cr3+(aq) + 7H2O
3 (H2S ----> S + 2H+ + 2e-)
__________________________________________________________
6e- + 8H+ + Cr2O72- + 3H2S ----> 3S + 2Cr3+ + 7H2O + 6e-
or 8H+ + Cr2O72- + 3H2S ----> 3S + 2Cr3+ + 7H2O
2. (4 pts.) How many grams of Al will be produced from Al2O3 in
15.0 hrs. using a current of 5.46 amps?
Al2O3 ----> 2Al3+ + 3O2-
Al3+ + 3e- ----> Al
g Al = (5.46 amps)(15.0 hrs.)(3600 sec/hr)(1 coul/amp sec)(1F/96500 coul)
(1 mole Al)/3F x (27.0 g Al)/1 mole Al = 27.5 g Al
3. (1 pt. each) Circle T (true) or F (false).
(T) F a. In the reaction 2OH- + 4ClO- + S2O32- ----> 4Cl- + 2SO42- + H2O,
the oxidizing agent is ClO-.
NOTE HALF-REACTION: 2e- + H2O + ClO- ----> Cl- + 2OH-
T (F) b. In the electrolysis of molten NaCl, the anode half-reaction is
Na+ + e- ----> Na.
T (F) c. The cathode in the lead storage battery when it is discharging
is Pb. The equation is:
Pb(s) + PbO2(s) + 2H+ + 2HSO4- ----> 2PbSO4(s) + 2H2O
NOTE HALF-REACTION: Pb(s) + HSO4- ---> PbSO4(s) + H+ + 2e- anode
(T) F d. Electrons are always lost at the anode.
4. Given the following reduction potentials,
(1) Cl2 + 2e- ----> 2Cl- E = 1.36 v Reducing agents are oxidized;
(2) Fe3+ + e- ----> Fe2+ E = 0.77 v Mg ----> Mg2+ + 2e- (E = 2.37 v)
(3) Fe2+ + 2e- ----> Fe E = -0.44 v Most positive E for oxidation.
(4) Mg2+ + 2e- ----> Mg E = -2.37 v
a. (1 pt.) Which is the strongest reducing agent? Mg
b. (5 pts.) Draw a diagram for a working voltaic cell using
half-reactions (1) and (3). Label all parts, including
direction of electron flow, direction of anion flow, the anode,
the cathode, the ions in solution, and the salt bridge.
Calculate E for the cell.
anode: Fe ----> Fe2+ + 2e- 0.44 v
cathode: Cl2 + 2e- ----> 2Cl- 1.36 v
net: Cl2 + Fe ----> Fe2+ + 2Cl- (1.80) v
Direction of electron flow is "anode to cathode"
Direction of anion flow is "cathode to anode"
Fe is the anode: Fe2+ ions are in the anode solution
Cl2 is the cathode: Cl- ions are in the cathode solution
Salt bridge connnects the anode solution and the cathode solution
Electrons flow through a wire that connects the anode and the cathode
5. (4 pts.) When 20.0 g of pure Cr is dissolved in an acidic
KMnO4 solution, the Cr is converted to Cr3+ and the
MnO4- to Mn2+. How many mL of 2.75 M KMnO4 solution
are required to react completely with the Cr?
(HINT: balance the reaction.)
5(Cr(s) ---> Cr3+ + 3e-)
3(5e- + 8H+ + MnO4- ----> Mn2+ + 4H2O)
15e- + 24H+ + 3MnO4- + 5Cr ---> 5Cr3+ + 3Mn2+ + 12H2O + 15e-
Vol. KMnO4 = 20.0 g Cr x 1 mole Cr x 3 moles MnO4- x 1 mole KMnO4
52.0 g Cr 5 moles Cr 1 mole MnO4-
1.00 L KMnO4
x _______________ = 0.0839 L = 83.9 mL
2.75 mole KMnO4
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