DAT GChem Section 7 Study Exam
Questions Correctly Answered.
What is the oxidizing agent in the following reaction?(7.1)
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) - CORRECT ANSWER MnO4-
The oxidizing agent (or oxidant) is the species that gets reduced (gains electrons)
First of all the oxidant or reductant is always a reactant so Mn2+ can be eliminated.
H+ is not oxidized or reduced.
MnO4- -->Mn2+ : +7 -->+2 MnO4- is reduced and is therefore the oxidizing agent and the correct answer
choice.
Fe2+--> Fe3+ : +2-->+3 Fe2+ is oxidized and is therefore the reducing agent rather than the oxidizing
agent.
What is the oxidation state of the Fe atom in Fe2O3?(7.1) - CORRECT ANSWER +3
What is the oxidation state of sulfur in Al2(SO3)3?(7.1) - CORRECT ANSWER +4
What is the reductant in the following reaction? (7.1)
2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) - CORRECT ANSWER Cu
The reductant (or reducing agent) is the species that gets oxidized (loses electrons). First of all the
oxidant or reductant is always a reactant.
Cu is oxidized and is therefore the reductant (therefore choice B is correct)
What is the cathode in the following reaction? (7.2)
, PbO2(s) + Pb(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) - CORRECT ANSWER PbO2
Reduction occurs at the cathode
PbO2 → PbSO4 : lead goes from +4 → +2 therefore PbO2 is reduced and is the cathode (PbO2 is correct)
Which of the following will result in an increase in the cell potential for the following reaction? (7.3)
IO3-(aq) + 6H+(aq) 5Li(s) → 1/2I2(s) + 3H2O(l) + 5Li+(aq)
Increase in [Li+]
Removing all of the I2 from the reaction mixture
Increase in [IO3-]
Increasing the size of the anode - CORRECT ANSWER Increase in [IO3-]
Shifting the equilibrium to the right (toward the products) according to Le Chateliers Principle results in
an increase in the cell potential.
Increasing [Li+] results in a shift to the left and decreases the cell potential eliminating choice A.
Removing I2 doesnt change the cell potential as solids dont shift equilibria eliminating choice B.
Increasing [IO3-] results in a shift to the right and increases the cell potential and therefore C is the
correct answer choice.
Increasing the size of the anode (Li is the anode) is increasing the size of a solid which dont shift
equilibria eliminating choice D.
Which of the following will result in an increase in the emf in the following reaction? (7.3)
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Questions Correctly Answered.
What is the oxidizing agent in the following reaction?(7.1)
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) - CORRECT ANSWER MnO4-
The oxidizing agent (or oxidant) is the species that gets reduced (gains electrons)
First of all the oxidant or reductant is always a reactant so Mn2+ can be eliminated.
H+ is not oxidized or reduced.
MnO4- -->Mn2+ : +7 -->+2 MnO4- is reduced and is therefore the oxidizing agent and the correct answer
choice.
Fe2+--> Fe3+ : +2-->+3 Fe2+ is oxidized and is therefore the reducing agent rather than the oxidizing
agent.
What is the oxidation state of the Fe atom in Fe2O3?(7.1) - CORRECT ANSWER +3
What is the oxidation state of sulfur in Al2(SO3)3?(7.1) - CORRECT ANSWER +4
What is the reductant in the following reaction? (7.1)
2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) - CORRECT ANSWER Cu
The reductant (or reducing agent) is the species that gets oxidized (loses electrons). First of all the
oxidant or reductant is always a reactant.
Cu is oxidized and is therefore the reductant (therefore choice B is correct)
What is the cathode in the following reaction? (7.2)
, PbO2(s) + Pb(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) - CORRECT ANSWER PbO2
Reduction occurs at the cathode
PbO2 → PbSO4 : lead goes from +4 → +2 therefore PbO2 is reduced and is the cathode (PbO2 is correct)
Which of the following will result in an increase in the cell potential for the following reaction? (7.3)
IO3-(aq) + 6H+(aq) 5Li(s) → 1/2I2(s) + 3H2O(l) + 5Li+(aq)
Increase in [Li+]
Removing all of the I2 from the reaction mixture
Increase in [IO3-]
Increasing the size of the anode - CORRECT ANSWER Increase in [IO3-]
Shifting the equilibrium to the right (toward the products) according to Le Chateliers Principle results in
an increase in the cell potential.
Increasing [Li+] results in a shift to the left and decreases the cell potential eliminating choice A.
Removing I2 doesnt change the cell potential as solids dont shift equilibria eliminating choice B.
Increasing [IO3-] results in a shift to the right and increases the cell potential and therefore C is the
correct answer choice.
Increasing the size of the anode (Li is the anode) is increasing the size of a solid which dont shift
equilibria eliminating choice D.
Which of the following will result in an increase in the emf in the following reaction? (7.3)
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
