Download Exam Review 3 Key - Iowa State University

Exam 3 Review
Supplemental Instruction
Iowa State University
Leader: Olivia
Course: CHEM 178
Instructor: Bonaccorsi
Date:
T/F A reductant, or reducing agent, is reduced in a chemical reaction.
F
1. Indicate whether the following balanced equations involve oxidation-reduction. If they do,
identify the elements that undergo changes in oxidation number.
a. PBr3(l) + 3H2O(l)  H3PO3(aq) + 3HBr (aq)
b. NaI (aq) + 3HOCl (aq)  NaIO (aq) + 3HCl (aq)
c. 3SO2 (g) + 2HNO3 (aq) + 2H2O (l)  3H2SO4(aq) + 2NO (g)
d. 2H2SO4 (aq) + 2NaBr (s)  Br2 (l) + SO2 (g) + Na2SO4 (aq) + 2H2O (l)
2. Complete and balance the following equations in acidic solution, and identify the oxidizing
and reducing agents.
a. NO2- (aq) + Cr2O72- (aq)  Cr3+ (aq) NO3- (aq)
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b. Cr2O72- (aq) + CH3OH (aq)  HCO2H (aq) + Cr3+ (aq)
3. Given the following data:
Fe2+ + 2e-  Fe
Eored = -0.44 V
+
Ag + e  Ag
Eored = 0.80 V
Answer the following questions with respect to the reaction
Fe2+ (aq) + 2Ag (s)  Fe (s) + 2Ag+ (aq)
a. What is E0 for the reaction?
b. Is the reaction spontaneous at standard state conditions?
c. What is the value of E at equilibrium?
4. Metallic Mg can be made by the electrolysis of molten MgCl2
a. What is the mass of Mg formed by passing a current of 4.55 A through Molten MgCl2
for 4.5 days?
b. How many minutes are needed to plate out 25.00g of Mg from molten MgCl2 using
3.5 A of current?
5. A voltaic cell is constructed of one half cell consisting of an aluminum strip placed in a
solution of Al(NO3)3, and the other has a nickel strip placed in a solution of NiSO4. The overall
cell reaction is 2Al (s) + 3Ni2+ (aq)  2Al3+ (aq) + 3Ni (s)
a. What is being oxidized and reduced?
b. Write the half reactions that occur for the two half cells.
c. Which electrode is the anode? Which is the cathode?
d. Which way do the electrons flow?
Find the cell potential of a galvanic cell based on the following reduction half-reactions at 25 °C
Cd2+ + 2 e- → Cd E0 = -0.403 V
Pb2+ + 2 e- → Pb E0 = -0.126 V
where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M.
Ecell = E0cell - (RT/nF) x lnQ
Ecell = 0.277 V - 0.013 V x ln(0.100)
Ecell = 0.277 V - 0.013 V x -2.303
Ecell = 0.277 V + 0.023 V
Ecell = 0.300 V
Answer:
The cell potential for the two reactions at 25 °C and [Cd2+] = 0.020 M and [Pb2+] = 0.200 M
is 0.300 volts
Consider a galvanic cell that uses the reaction:
Cu(s) + 2Fe3+(aq) → Cu2+(aq) + 2Fe2+(aq)
What is the potential of a cell at 25 °C that has the following ion concentrations?
[Fe3+] = 1.0 x 10-4 M [Cu2+] = 0.25 M [Fe2+] = 0.20 M
What is Q, reaction quotient?
Notice that solid copper is omitted. Cell is not under standard conditions, so the Nernst
Equation must be used:
E 0 cell = Eo cathode - E o anode
(What should it be under standard conditions) E o cell = 0.77 V – 0.34 V = 0.43 V Total of 2
electrons transferred, n = 2
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Label the voltaic cell: Zn/Cu
Cathode, anode, salt bridge, electron flow, reduction, oxidation, flow of electrons, flow of
anions, flow of cations, which side’s metal strip gains/loses mass, half reactions
Label the electrolytic cell: Zn/Cu
Cathode, anode, salt bridge, electron flow, reduction, oxidation, flow of electrons, flow of
anions, flow of cations, which side’s metal strip gains/loses mass, half reactions
Calculate E°cell, Ecell, and G for the following cell reaction.
3Zn(s) + 2Cr3+(0.0010 M)
E°cell =
Ecell =
ΔG =
3Zn2+(0.010 M) + 2Cr(s)
Calculate the standard reaction entropies for the following reactions using standard molar
entropies.
2 SO2 (g) + O2 (g)  2 SO3 (g)
-187.8 J/mol K
SO3(g) + H2O (l)  H2SO4 (aq) -306.7 J/mok K
S (g) + O2 (g)  SO2 (g)
-124.6 J/mol K
An endothermic reaction has ΔH = 60. kJ and ΔS = −120. J/K. Assuming that ΔH and ΔS do not
vary with temperature, at what temperature will the system be at equilibrium?
ΔG = ΔH – T ΔS
0 = 60 – (T * -120)
T = -500
It will not reach equilibrium
Because the enthalpy change is positive, the entropy change would need to be positive to achieve
equilibrium at some temperature.