Chem1000A
Spring 2005
Assignment 7 - Solutions
Two or three questions will be marked from each assignment.
DUE ON March 9, 2005 (Wednesday) 1:00 PM
To be dropped off at my office (E860)
1. Determine the oxidation states in the following sulfur compounds (only one resonance structure is
given). Use the original definition of the oxidation state (by cleaving bonds). For (b) and (c),
distinguish between the oxidation states for individual sulfur atoms and an average oxidation state for
sulfur (i.e., the average of the two – or more – values). You would obtain these average oxidation
states if you use the ‘simple rules’, that are usually used for oxidation states.
(a) peroxodisulfate: S2O82-II
-II
:O :
:O :
-II
-II
-I -I..
..
..
..
O
S
O
O
S
..
..
.. ..-VI O
-VI
:O
:O:
.. : - ..
-II
-II
(b) S2F4
-I..
: :
+I.. +III
.. F ..-I
: S S ..F:
:..
F:
:..
F:
-I
-I
(c) disulfite S2O52-II -II
:
:
:
-II.. O :O
O
.. S S : +III
+V
:O: :O
.. : - ..
-II -II
(d) dithionate S2O62-II -II
:O : :O :
..
+V +V..
-II O
.. -II
.. S S O
:O: :O
.. : - ..
-II -II
2. Determine oxidation states for all elements in the following compounds:
(a) H-O-O-F
H+I-O-I-O+I-F-I
(b) H3C-O-O-F
H+I3C-II-O-I-O+I-F-I
(c) glycol, H2C(OH)CH(OH)CH2(OH):
H
H
H
H
H
..
..
O
..
.. H
O
.... H
O
.. H
(d) ClO3
-
H+I2C-I(O-IIH+I)C0H+I(O-IIH+I)C-IH+I2(O-IIH+I)
Cl+VO-II3-
1-4
3. Balance the following redox reactions and show that the final reaction equation have the correct
electron, material, and charge balance.
(a) In acidic aqueous solution: HMnO4- → MnO2 + MnO4H+IMn+VIO-II4- → Mn+IVO-II2 + Mn+VIIO-II4oxidation half-reaction (unbalanced): HMnO4- → MnO2
reduction half-reaction (unbalanced): HMnO4- → MnO4principle element balance:
oxidation half-reaction (unbalanced): HMnO4- → MnO2
reduction half-reaction (unbalanced): HMnO4- → MnO4oxygen balance:
oxidation half-reaction (unbalanced): HMnO4- → MnO2 + 2 H2O
reduction half-reaction (unbalanced): HMnO4- → MnO4hydrogen balance (acidic solution):
oxidation half-reaction (unbalanced): 3H+ + HMnO4- → MnO2 + 2 H2O
reduction half-reaction (unbalanced): HMnO4- → MnO4- + H+
Inserting electrons:
oxidation half-reaction (balanced): 3H+ + HMn+VIO4- → Mn+IVO2 + 2 H2O + 2ereduction half-reaction (balanced): e- + HMn+VIO4- → Mn+VIIO4- + H+ | ×2
electron balance:
oxidation half-reaction (balanced): 3H+ + HMn+VIO4- → Mn+IVO2 + 2 H2O + 2ereduction half-reaction (balanced): 2e- + 2 HMn+VIO4- → 2Mn+VIIO4- + 2H+
combination: 3HMnO4- + 2e- + 3H+ → 2 MnO4- + 2H+ +MnO2 + 2 H2O + 2esimplification: 3HMnO4- + H+ → 2MnO4- + MnO2 + 2 H2O
material balance:4H, 3 Mn, 12O|
3Mn, 12O, 4H
charge balance:(3-) + (+) = 2- |
(2-)
correct electron, material, and charge balance!
(b) In acidic aqueous solution: Zn + NO3- → Zn2+ + NH4+
Zn0 + N+VO-II3- → Zn+II 2- + N-IIIH+I4+
oxidation half-reaction (unbalanced): Zn → Zn2+
reduction half-reaction (unbalanced): NO3- → NH4+
principle element balance:
oxidation half-reaction (unbalanced): Zn → Zn2+
reduction half-reaction (unbalanced): NO3- → NH4+
oxygen balance:
oxidation half-reaction (unbalanced): Zn → Zn2+
reduction half-reaction (unbalanced): NO3- → NH4+ + 3H2O
hydrogen balance (acidic solution):
oxidation half-reaction (unbalanced): Zn → Zn2+
reduction half-reaction (unbalanced): 10H+ + NO3- → NH4+ + 3H2O
Inserting electrons:
| ×4
oxidation half-reaction (balanced): Zn0 → Zn+II 2+ + 2e+
+V
-III
reduction half-reaction (balanced): 8e + 10H + N O3 → N H4+ + 3H2O
electron balance:
oxidation half-reaction (balanced): 4Zn → 4Zn 2+ + 8ereduction half-reaction (balanced): 8e- + 10H+ + NO3- → NH4+ + 3H2O
combination: 4Zn + 8e- + 10H+ + NO3- → NH4+ + 4Zn2+ + 3 H2O + 8esimplification: 4Zn + 10H+ + NO3- → NH4+ + 4Zn2+ + 3 H2O
2-4
material balance:4Zn, 10 H, 1N, 3 O| 1N, 10 H, 4 Zn, 3 O
charge balance:(10+) + (1-) = 9+ | (1+) + (8+) = 9+
correct electron, material, and charge balance!
(c) In basic aqueous solution: ZrO(OH)2 + SO32- → Zr + SO42Zr+IVO-II(O-IIH+I)2 + S+IVO-II32- → Zr0 + S+VIO-II42oxidation half-reaction (unbalanced): SO32- → SO42reduction half-reaction (unbalanced): ZrO(OH)2 → Zr
principle element balance:
oxidation half-reaction (unbalanced): SO32- → SO42reduction half-reaction (unbalanced): ZrO(OH)2 → Zr
oxygen balance:
oxidation half-reaction (unbalanced): H2O + SO32- → SO42reduction half-reaction (unbalanced): ZrO(OH)2 → Zr + 3H2O
hydrogen balance (basic solution – 1st step):
oxidation half-reaction (unbalanced): H2O + SO32- → SO42- + 2H+
reduction half-reaction (unbalanced): 4H+ + ZrO(OH)2 → Zr + 3H2O
hydrogen balance (basic solution – 2nd step):
oxidation half-reaction (unbalanced): H2O + SO32- + 2OH- → SO42- + 2H2O
reduction half-reaction (unbalanced): 4H2O + ZrO(OH)2 → Zr + 3H2O + 4 OHInserting electrons:
oxidation half-reaction (balanced): H2O + S+IVO32- + 2OH- → S+VIO42- + 2H2O +2e- | ×2
reduction half-reaction (balanced): 4e- + 4H2O + Zr+IVO(OH)2 → Zr0 + 3H2O + 4 OHelectron balance:
oxidation half-reaction (balanced): 2H2O + 2SO32- + 4OH- → 2SO42- + 4H2O +4ereduction half-reaction (balanced): 4e- + 4H2O + ZrO(OH)2 → Zr + 3H2O + 4 OHcombination: 6H2O +4e- +2SO32- + 4OH- + ZrO(OH)2 → 2SO42- + 4H2O +4e- + Zr + 3H2O + 4OHsimplification: 2SO32- + ZrO(OH)2 → 2SO42- + H2O + Zr
material balance: 1 Zr, 2 S, 2 H, 9 O| 1S, 9 O, 1 Zr, 2 H
charge balance:
4- | 4correct electron, material, and charge balance!
(d) In basic aqueous solution: P4 → HPO32- + PH3
P04 → H+IP+IIIO-II32- + P-IIIH+I3
oxidation half-reaction (unbalanced): P4 → HPO32reduction half-reaction (unbalanced): P4 → PH3
principle element balance:
oxidation half-reaction (unbalanced): P4 → 4 HPO32reduction half-reaction (unbalanced): P4 → 4PH3
oxygen balance:
oxidation half-reaction (unbalanced): 12H2O + P4 → 4HPO32reduction half-reaction (unbalanced): P4 → 4PH3
hydrogen balance (basic solution – 1st step):
oxidation half-reaction (unbalanced): 12H2O + P4 → 4HPO32- + 20 H+
reduction half-reaction (unbalanced): 12H+ + P4 → 4PH3
hydrogen balance (basic solution – 2nd step):
oxidation half-reaction (unbalanced): 20OH- +12H2O + P4 → 4 HPO32- + 20 H2O
reduction half-reaction (unbalanced): 12H2O + P4 → 4PH3 + 12 OH3-4
Inserting electrons:
oxidation half-reaction (balanced): 20OH- +12H2O + P04 → 4HP+IIIO32- + 20 H2O +12ereduction half-reaction (balanced): 12e- + 12H2O + P04 → 4P-IIIH3 + 12 OHelectron balance:
oxidation half-reaction (balanced): 20OH- +12H2O + P04 → 4HP+IIIO32- + 20 H2O +12ereduction half-reaction (balanced): 12e- + 12H2O + P04 → 4P-IIIH3 + 12 OHcombination: 12e- + 12H2O + 20OH- +12H2O +2P4 → 4HPO32- + 20H2O +12e- + 4PH3 + 12OHsimplification: 4H2O + 8OH- +2P4 → 4HPO32- + 4PH3 | /2
simplification: 2H2O + 4OH- + P4 → 2HPO32- + 2PH3
material balance: 8 H, 6 O, 4 P | 8 H, 6 O, 4 P
charge balance:
4- | 4correct electron, material, and charge balance!
4-4