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Homework 3
SHOW all work, please BOX your final balanced equation for each set below (numbers 1 and 2) Insert
spaces or perform balancing on a separate sheet of paper. Don’t forget to clearly label the ox and red
species!
1.) Balance the following chemical reactions as indicated AND identify the reducing and oxidizing
species: (insert spaces as needed or perform on a separate sheet of paper)
Remember, a redox reaction (basic) is balanced like an acidic redox reaction. The half reactions
are pulled out. The atoms are balanced as is. Water is used to balance the number of oxygens,
hydrogen ions are used to balance the number of H’s. Electrons are used to balance charge. The
reactions are multiplied by a factor to make sure that the numbers of electrons lost = numbers
of electrons gained. The reactions are added together. THEN OH-1 ions are added to BOTH
sides. The number of OH-1 ions added needs to equal the number of H+1 ions present. This
will create additional waters (which need to be reduced!) and you will now have an excess of
OH-1 ions on one side indicating the basic solution 
a. MnO4-1(aq) + SO3-2 (aq)  MnO2 (s) + SO4-2 (aq) [basic]
Mn+7 O-2 S+4 O-2
Mn+4 O-2 S+6 O-2
Mn+7 or MnO4-1 is reduced therefore it is the oxidizing agent
S+4 or SO3-2 is oxidized and therefore is the reducing agent
MnO4-1 → MnO2
MnO4-1 → MnO2 + 2H2O
4H+1 + MnO4-1 → MnO2 + 2H2O
2(3e-1 + 4H+1 + MnO4-1 → MnO2 + 2H2O)
6e-1 + 8H+1 + 2MnO4-1 → 2MnO2 + 4H2O
SO3-2 → SO4-2
H2O + SO3-2 → SO4-2
H2O + SO3-2 → SO4-2 + 2H+1
(H2O + SO3-2 → SO4-2 + 2H+1 + 2e-1)3
3H2O + 3SO3-2 → 3 SO4-2 + 6H+1 + 6e-1
6e-1 + 8H+1 + 2MnO4-1 → 2MnO2 + 4H2O
3H2O + 3SO3-2 → 3 SO4-2 + 6H+1 + 6e-1
2H+1 + 2MnO4-1 + 3SO3-2 → 2MnO2 + H2O + 3 SO4-2
Now OH-1 needs to be added to BOTH sides. The same numbers of OH-1 will be added as there
are H+1 ions.
2OH-1 + 2H+1 + 2MnO4-1 + 3SO3-2 → 2MnO2 + H2O + 3 SO4-2 + 2OH-1
2H2O + 2MnO4-1 + 3SO3-2 → 2MnO2 + H2O + 3 SO4-2 + 2OH-1
Now cancel any duplicate waters (since you made water by adding H+1 and OH-1 you may now
have waters on both sides of the → again – reduce!!)
H2O + 2MnO4-1 + 3SO3-2 → 2MnO2 + 3 SO4-2 + 2OH-1
b. Zn(s) + NO3-1 (aq)  Zn(OH)4-2 (aq) + NH3 (g) [basic]
Zn is oxidized and is therefore the reducing agent
NO3-1 is reduced and is therefore the oxidizing agent
Zn → Zn(OH)4-2
4H2O + Zn → Zn(OH)4-2
4H2O + Zn → Zn(OH)4-2 + 4H+1
4H2O + Zn → Zn(OH)4-2 + 4H+1 + 2e-1
NO3-1 → NH3
NO3-1 → NH3 + 3H2O
9H+1 + NO3-1 → NH3 + 3H2O
8e-1 + 9H+1 + NO3-1 → NH3 + 3H2O
4(4H2O + Zn → Zn(OH)4-2 + 4H+1 + 2e-1)
8e-1 + 9H+1 + NO3-1 → NH3 + 3H2O
16H2O + 4Zn → 4Zn(OH)4-2 + 16H+1 + 8e-1
8e-1 + 9H+1 + NO3-1 → NH3 + 3H2O
13H2O + 4Zn + NO3-1 → 4Zn(OH)4-2 + 7H+1 + NH3
7OH-1 + 13H2O + 4Zn + NO3-1 → 4Zn(OH)4-2 + NH3 + 7H+1 + 7OH-1
7OH-1 + 13H2O + 4Zn + NO3-1 → 4Zn(OH)4-2 + NH3 + 7H2O
7OH-1 + 6H2O + 4Zn + NO3-1 → 4Zn(OH)4-2 + NH3
2.) Balance the following chemical reactions as indicated AND identify the reducing and oxidizing
species: (insert spaces as needed or perform on a separate sheet of paper)
a. ClO3-1(aq) + I-1 (aq)  I2 (s) + Cl-1 (aq) [acidic]
Cl+1 O-2 I-1
I0
Cl-1
Cl+1 or ClO3-1 is being reduced therefore it is the oxidizing agent
I-1 is being oxidized therefore it is the reducing agent
ClO3-1 → Cl-1
ClO3-1 → Cl-1 + 3H2O
6H+1 + ClO3-1 → Cl-1 + 3H2O
6e-1 + 6H+1 + ClO3-1 → Cl-1 + 3H2O
I-1 → I2
2I-1 → I2
2I-1 → I2 + 2e-1
3(2I-1 → I2 + 2e-1)
6e-1 + 6H+1 + ClO3-1 → Cl-1 + 3H2O
6I-1 → 3I2 + 6e-1
6H+1 + ClO3-1 + 6I-1 → Cl-1 + 3H2O + 3I2
6H
1Cl
3O 6 I (+6 + -1 + -6) = -1
6H 1Cl 3O 6 I (-1 + 0 + 0) = -1
b. Cr2O7-2(aq) + Zn (s)  Zn+2 (aq) + Cr+3(aq) [acidic]
Cr+6 O-2
Zn0
Zn+2
Cr+3
Cr+6 or Cr2O7-2 is reduced, therefore it is the oxidizing agent
Zn0 is oxidized, therefore it is the reducing agent
Cr2O7-2 → Cr+3
Cr2O7-2 → 2Cr+3 + 7H2O
14H+1 + Cr2O7-2 → 2Cr+3 + 7H2O
6e-1 + 14H+1 + Cr2O7-2 → 2Cr+3 + 7H2O
Zn → Zn+2
Zn → Zn+2 + 2e-1
6e-1 + 14H+1 + Cr2O7-2 → 2Cr+3 + 7H2O
3(Zn → Zn+2 + 2e-1)
6e-1 + 14H+1 + Cr2O7-2 → 2Cr+3 + 7H2O
3Zn → 3Zn+2 + 6e-1
14H+1 + 3Zn + Cr2O7-2 → 3Zn+2 + 2Cr+3 + 7H2O
3.) Write the electron configurations – DO NOT USE THE NOBLE GAS CONFIGURATION!! – for the
following:
a. Cr 1s2 2s2 2p6 3s2 3p6 4s2 3d4 →
1s2 2s2 2p6 3s2 3p6 4s1 3d5
b. Cu 1s2 2s2 2p6 3s2 3p6 4s2 3d9 →
1s2 2s2 2p6 3s2 3p6 4s1
c. F
1s2 2s2 2p5
3d10
4.) Write a set of quantum numbers (n, l, ml, and ms) for the 1st electron, the 2nd electron, the 5th electron,
and the 6th electron for fluorine. (HINT!! Orbital box diagrams might help!)
5
1
6
2
1s
2s
2p
Note: The 1st and 2nd electron MUST have the same ml value because they are in the same “box” (orbital)
while the 5th and the 6th electron MUST have different ml values since they are not in the same “box”
(orbital).
1st e: n = 1
l=0
ml = 0
ms = + ½
2nd e: n = 1
l=0
ml = 0
ms = - ½
5th e:
6th e:
n=2
n=2
l=1
l=1
ml = -1
ml = 0
ms = + ½
ms = + ½
5.) Write the electron configurations – DO NOT USE THE NOBLE GAS CONFIGURATION!! – for
the following
0
a. Cu+2 1s2 2s2 2p6 3s2 3p6 4s2 3d9 → 1s2 2s2 2p6 3s2 3p6 4s
3d9
6
b. Cl-1 1s2 2s2 2p6 3s2 3p5 → 1s2 2s2 2p6 3s2 3p
6.) Which of the following quantum number sets are allowed? If there is an error identify AND correct
it by changing the l number only
a. n= 1, l = 1, ml= 0 NOT VALID:
b. n= 6,
l
l = 4,
l cannot be the same numerically as n! Therefore l = 0
ml= -5 NOT VALID: the ml value is outside the –
the value must = 5
l to + l range. Therefore,