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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 el...

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  • August 15, 2024
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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)

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