QUIZ CHAPTER #18
ANSWER KEY
Student’s Name
1. What is the cell reaction for the following voltaic cell?
Ni(s) | Ni2+(aq) || Y3+(aq) | Y(s)
a) 3 Ni(s) + 2 Y3+(aq) → 2 Y(s) + 3 Ni2+(aq)
b) 2 Y(s) + 3 Ni2+(aq) → 3 Ni(s) + 2 Y3+(aq)
c) Ni(s) + Y3+(aq) → Y(s) + Ni2+(aq)
d) Y(s) + Ni2+(aq) → Ni(s) + Y3+(aq)
2. Consider the following reduction potentials:
Cd2+(aq) + 2e– → Cd(s); E° = –0.40 V
Pb2+(aq) + 2e– → Pb(s); E° = –0.13 V
Cu2+(aq) + 2e– → Cu(s); E° = 0.34 V
Br2(l) + 2e– → 2 Br–(aq); E° = 1.07 V
Under standard-state conditions, which of the following reactions is spontaneous as written?
a) 2 Br–(aq) + Cu2+(aq) → Br2(l) + Cu(s)
b) Cd2+(aq) + Pb(s) → Cd(s) + Pb2+(aq)
c) 2 Br–(aq) + Pb2+(aq) → Br2(l) + Pb(s)
d) Cu2+(aq) + Cd(s) → Cu(s) + Cd2+(aq)
3. Protection of iron (E°(Fe2+/Fe) = − 0.23 V; E°(Fe3+/Fe) = −0.04 V) from corrosion can be
accomplished by making an electrical contact between iron and certain other metals. Metal(s)
that would provide protection is(are):
1. Mg; Mg2+(aq) + 2e– → Mg(s); E° = –2.38 V
a) 1 only
2. Ni; Ni2+(aq) + 2e– → Ni(s); E° = –0.23 V
b) 4 only
3. Zn; Zn2+(aq) + 2e– → Zn(s); E° = –0.76 V
c) 1 and 3
4. Pb; Pb2+(aq) + 2e– → Pb(s); E° = –0.13 V
d) 2 and 4
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4. If an electrolysis plant operates its electrolytic cells at a total current of 1.0×106 A, how long
will it take to produce one metric ton (one million grams) of Mg(s) from seawater containing
Mg2+? (F = 96,485 C (coulombs))
a) 66 min
Mg2+(aq) + 2e– → Mg(s)
b) 0.55 year
2 moles of electrons transferred per every mole of Mg(s)
c) 2.2 h
Molar mass of Mg is 24.31 g/mol
d) 2.2 days
1 mol Mg
2 mol e! 96485 C
𝑄 𝑡𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 = 1.0×10 g Mg×
×
×
= 7. 𝟗3×10! C
24.31 g Mg 1 mol Mg 1 mol e!
!
𝑡=
𝑄 (𝑡𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒) 7. 𝟗3×10! C
=
= 7. 𝟗3×10! s = 1𝟑2 min = 2.2 h
𝐼 (𝑐𝑢𝑟𝑟𝑒𝑛𝑡)
1.0×10! A
5. When an aqueous solution of lithium sulfate, Li2SO4, is electrolyzed, what are the expected
products?
Reduction Half-Reaction
1) Li (aq) + 1e– → Li(s);
+
E° (V)
E1° = –3.04 V
cathode half-reaction
2) 2 H2O(l) + 2e– → H2(g) + 2 OH–(aq);
E2° = –0.83 V
cathode half-reaction
3) 2 H+(aq) + 2e– → H2(g);
E3° = 0.00 V
4) O2(g) + 4 H+(aq) + 4e– → 2 H2O(l); E4° = 1.23 V reverse of the anode half-reaction
5) S2O82–(aq) + 2e– → 2 SO42–(aq);
E5° = 2.01 V reverse of the anode half-reaction
a) Li(s) and H2(g)
b) H2(g), OH–(aq), O2(g), and H+(aq)
c) O2(g), H+(aq), and Li(s)
d) H2(g), OH–(aq), and Li(s)
Cathode: E1° < E2°, therefore half-reaction 2 will occur, producing H2(g) and OH–(aq)
Anode: –E5° < –E4°, therefore half-reaction 4 will occur, producing O2(g) and H+(aq)
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Bonus 1. The following oxidation–reduction reaction occurs in acidic solution:
+7 -2
+2 -2
+6 -2
+5 -2
S2O82–(aq) + NO(g) → SO42–(aq) + NO3–(aq)
oxidizing
agent
reducing
agent
Assign the oxidation numbers for every element, label the oxidizing agent and the reducing
agent, identify the oxidation half-reaction and the reduction half-reaction, and balance the
equation using the half-reaction method. Show all of your work to get full credit.
S2O82–(aq) + 2e– → 2 SO42–(aq)
reduction half-reaction (balanced)
NO(g) → NO3–(aq) + 3e–
oxidation half-reaction (unbalanced)
Balancing the oxidation half-reaction:
a) add 2 H2O to the left side to balance oxygen:
2 H2O(l) + NO(g) → NO3–(aq) + 3e–
b) add 4 H+ to the right side to balance hydrogen:
2 H2O(l) + NO(g) → 4 H+(aq) + NO3–(aq) + 3e–
Balancing the number of electrons in the half-reactions:
reduction:
S2O82–(aq) + 2e– → 2 SO42–(aq)
multiply by 3
oxidation:
2 H2O(l) + NO(g) → 4 H+(aq) + NO3–(aq) + 3e–
multiply by 2
reduction:
3 S2O82–(aq) + 6e– → 6 SO42–(aq)
oxidation:
4 H2O(l) + 2 NO(g) → 8 H+(aq) + 2 NO3–(aq) + 6e–
Add the half-reactions together and cancel the electrons:
3 S2O82–(aq) + 4 H2O(l) + 2 NO(g) → 6 SO42–(aq) + 8 H+(aq) + 2 NO3–(aq)
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Bonus 2. The following oxidation–reduction reaction occurs in basic solution:
+3 -2
+2 -2 +1
+6 -2
Cr2O3(s) → Cr(OH)2(aq) + CrO42–(aq)
Cr in Cr2O3 is both oxidizing and reducing agent
Assign the oxidation numbers for every element, label the oxidizing agent and the reducing
agent, identify the oxidation half-reaction and the reduction half-reaction, and balance the
equation using the half-reaction method. Show all of your work to get full credit.
Hint. Sometimes in chemical reactions the same element acts as both oxidizing and reducing
agents.
Reduction half-reaction:
Cr2O3(s) + 2e– → 2 Cr(OH)2(aq) reduction half-reaction (unbalanced)
a) add 1 H2O to the left side to balance oxygen:
H2O(l) + Cr2O3(s) + 2e– → 2 Cr(OH)2(aq)
b) add 2 H+ to the left side to balance hydrogen:
2 H+(aq) + H2O(l) + Cr2O3(s) + 2e– → 2 Cr(OH)2(aq)
Oxidation half-reaction:
Cr2O3(s) → 2 CrO42–(aq) ) + 6e–
oxidation half-reaction (unbalanced)
a) add 5 H2O to the left side to balance oxygen:
5 H2O(l) + Cr2O3(s) → 2 CrO42–(aq) ) + 6e–
b) add 10 H+ to the right side to balance hydrogen:
5 H2O(l) + Cr2O3(s) → 10 H+(aq) + 2 CrO42–(aq) ) + 6e–
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Balancing the number of electrons in the half-reactions:
reduction:
2 H+(aq) + H2O(l) + Cr2O3(s) + 2e– → 2 Cr(OH)2(aq)
multiply by 3
oxidation:
5 H2O(l) + Cr2O3(s) → 10 H+(aq) + 2 CrO42–(aq) ) + 6e–
no change
reduction:
6 H+(aq) + 3 H2O(l) + 3 Cr2O3(s) + 6e– → 6 Cr(OH)2(aq)
oxidation:
5 H2O(l) + Cr2O3(s) → 10 H+(aq) + 2 CrO42–(aq) ) + 6e–
Add the half-reactions together and cancel the electrons:
6 H+(aq) + 3 H2O(l) + 3 Cr2O3(s) + 5 H2O(l) + Cr2O3(s) →
→ 6 Cr(OH)2(aq) + 10 H+(aq) + 2 CrO42–(aq)
Group all Cr2O3(s) and all H2O(l) on the left side together and subtract 6 H+(aq) from both sides:
4 Cr2O3(s) + 8 H2O(l) → 6 Cr(OH)2(aq) + 4 H+(aq) + 2 CrO42–(aq)
Add 4 OH-(aq) to both sides:
4 Cr2O3(s) + 8 H2O(l) + 4 OH-(aq) → 6 Cr(OH)2(aq) + 4 H+(aq) + 4 OH-(aq) + 2 CrO42–(aq)
Combine 4 H+(aq) and 4 OH-(aq) on the right side into 4 H2O(l) and subtract 4 H2O(l) from both
sides:
4 Cr2O3(s) + 4 H2O(l) + 4 OH-(aq) → 6 Cr(OH)2(aq) + 2 CrO42–(aq)
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