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- Balance the following redox reactions using the oxidation number method.
a. SnCl2 + HgCl2 → SnCl4 + HgCl
initial
final
change
Coefficient
Total e-
Sn
+2
→
+4
2
×
1
=
2
Hg
+2
→
+1
1
×
2
=
2
Place a “1” in front of compounds containing Sn, and a “2” in front of compounds with Hg: Answer: 1 SnCl2 + 2 HgCl2 → 1 SnCl4 + 2 HgCl b. HNO3 + H2S → NO + S + H2O
initial
final
change
Coefficient
Total e-
N
+5
→
+2
3
×
2
=
6
S
-2
→
0
2
×
3
=
6
Place a “2” in front of compounds containing N, and a “3” in front of compounds with S. Then balance for hydrogen and oxygen. Answer: 2 HNO3 + 3 H2S → 2 NO + 3 S + 4 H2O c. NaClO + H2S → NaCl + H2SO4
initial
final
change
Coefficient
Total
e-Cl
+1
→
-1
2
×
4
=
8
S
-2
→
+6
8
×
1
=
8
Answer: 4 NaClO + 1 H2S → 4 NaCl + 1 H2SO4 d. CdS + I2 + HCl → CdCl2 + HI + S
Because one of the atoms undergoing oxidation or reduction has a subscript (I2) we will account for the number of atoms of each element when preparing our summary chart:
initial
final
change
no. atoms
No.
e -
Coefficient
Total e-
S
-2
→
0
2
2
×
1
=
2
I
0
→
-1
1
×
2
(in I2)=
2
×
1
=
2
Place the balancing coefficients into the equation in front of the elements undergoing oxidation and reduction. For iodine, the 1 will go in front of the diatomic I 2 because these were the atoms being counted.
1 CdS + 1 I2 + HCl → CdCl2 + HI + 1 S
Then balance the rest of the equation. First balance for iodine atoms, then for Cd and H:
Answer: 1 CdS + 1 I2 + 2 HCl → 1 CdCl2 + 2 HI + 1 S
e. I2 + HNO3 → HIO3 + NO2 + H2O
Once again, because one of the atoms undergoing oxidation or reduction has a subscript (I2) we will account for the number of atoms of each element when preparing our summary chart:
initial
final
change
no. atoms
No.
e -
Coefficient
Total
e -I
0
→
+5
5
×
2
(in I2)
10
×
1
=
10
N
+5
→
+4
1
=
1
×
10
=
10
The “1” for iodine is placed in front of the diatomic iodine; the “10” goes in front of both nitrogens. Then balance for iodine on both sides of the equation, then for all other atoms.
Answer: 1 I2 + 10 HNO3 → 2 HIO3 + 10 NO2 + 4 H2O
f. MnO4- + H+ + Cl- → Mn2+ + Cl2 + H2O
initial
final
change
no. atoms
No.
e -
Coefficient
Total
e -Mn
+7
→
+2
5
5
×
2
=
10
Cl
-1
→
0
1
×
2
(Cl2)=
2
×
5
=
10
Because of the diatomic chlorine (Cl2) we multiply the change in oxidation number for chlorine by 2. We then determine what coefficients are needed to balance for electrons. The “5” for chlorine will be placed in front of the diatomic chlorine. Then balance both sides of the equation for chlorine, then for all other atoms. Answer: 2 MnO4- + 16 H+ + 10 Cl- → 2 Mn2+ + 5 Cl2 + 8 H2O
- Balance the following half-reactions for both atoms and electrons by adding the appropriate number of electrons to the correct side of the equation. Also identify each as either an oxidation or reduction.
a. Pb2+ → Pb Pb2+ + 2e- → Pb reduction b. Cl2 → Cl- Cl2 + 2 e- → 2 Cl- reduction c. Fe3+ → Fe2+ Fe3+ + e- → Fe2+ reduction d. N2O + H2O → NO + H+ N2O + H2O → 2NO + 2 H+ + 2e-
oxidation
- Break each equation into two half-reactions. Identify each half-reaction as oxidation or reduction.
a. Cu + 2 H+ → Cu2+ + H2 Cu → Cu2+ + 2 e- oxidation 2 H+ + 2 e- → H2 reduction b. 2 Al + 3 S → Al2S3 2 Al → 2Al3+ + 6 e- oxidation 3S + 6e- → 3 S2- reduction
- Balance the following equations using the half-reaction method.
a. Na + Br2 → NaBr
Step 1
Step 2
Step 3
Write the two balanced half-reactions, removing any spectator ions:
Balance for electrons
Add the half-reactions, replacing any spectator ions that were removed and/or recombining compounds
Na → Na+ + e-
× 2
2 Na → 2 Na+ + 2e-
Br2 + 2 e- → 2 Br-
Br2 + 2 e- → 2 Br-
add together:
2 Na + Br2 → 2 Na+ + 2 Br-
reform compound:
2 Na + Br2 → 2 NaBr
b. Zn + S → ZnS
Step 1
Step 2
Step 3
Write the two balanced half-reactions, removing any spectator ions:
Balance for electrons
Add the half-reactions, replacing any spectator ions that were removed and/or recombining compounds
Zn → Zn2+ + 2 e-
Zn → Zn2+ + 2 e-
S + 2 e- → S2-
S + 2 e- → S2-
added together:
Zn + S → Zn2+ + S2-
reform compound:
Zn + S → ZnS
c. Ag + Cr2O72- + H+ → Ag+ + Cr3+ + H2O
For each half-reaction, remember to balance for atoms first, then add electrons to balance for charge.
Step 1
Step 2
Step 3
Write the two balanced half-reactions, removing any spectator ions:
Balance electrons
Add the half-reactions, replacing any spectator ions that were removed and/or recombining compounds
Ag → Ag+ + e-
× 6
6 Ag → 6 Ag+ + 6 e-
Cr2O72- + 14 H+ + 6 e- → 2 Cr3+ + 7 H2O
added together: 6Ag + Cr2O72- + 14 H+ → 6Ag+ 2Cr3+ + 7H2O