South Pasadena • AP Chemistry
Name
1 ▪ Describing Matter
Period
1.5
PROBLEMS
–
Date
REDOX REACTIONS
Balance the following reaction using the half-reaction method.
1. HCl (aq) + K2Cr2O7 (aq)  KCl (aq) + CrCl3 (aq) + H2O (ℓ) + Cl2 (g)
(+1)(–1)
(+1)(+6)(–2)
(+1)(–1)
(+3)(–1)
(+1)(–2)
(0)
Ox:
Red:
2 HCl → Cl2 + 2 H+ + 2 e–
6 e– + 6 H+ + K2Cr2O7 + 8 HCl→ 2 KCl + 2 CrCl3 + 7 H2O
Ox:
Red:
6 HCl → 3 Cl2 + 6 H+ + 6 e–
6 e + 6 H + K2Cr2O7 + 8 HCl→ 2 KCl + 2 CrCl3 + 7 H2O
–
+
14 HCl + K2Cr2O7 → 2 KCl + 2 CrCl3 + 7 H2O + 3 Cl2
2. FeCl2 (aq) + KMnO4 (aq) + HCl (aq)  FeCl3 (aq) + KCl (aq) + MnCl2 (aq) + H2O (ℓ)
(+2)(–1)
(+1)(+7)(–2)
(+1)(–1)
(+3)(–1)
(+1)(–1)
(+2)(–1)
Ox:
Red:
FeCl2 + HCl → FeCl3 + H+ + e–
5 e– + 5 H+ + KMnO4 + 3 HCl → MnCl2 + KCl + 4 H2O
Ox:
Red:
5 FeCl2 + 5 HCl → 5 FeCl3 + 5 H+ + 5 e–
5 e– + 5 H+ + KMnO4 + 3 HCl → MnCl2 + KCl + 4 H2O
5 FeCl2 + KMnO4 + 8 HCl → 5 FeCl3 + KCl + MnCl2 + 4 H2O
3. S2 (aq) + MnO4 (aq)  S (s) + MnO2 (s)
(–2)
(+7)(–2)
(0)
(basic solution)
(+4)(–2)
Ox:
Red:
S2– → S + 2 e–
3 e– + 4 H+ + MnO4– → MnO2 + 2 H2O
Ox:
Red:
3 S2– → 3 S + 6 e–
6 e + 8 H + 8 OH + 2 MnO4– → 2 MnO2 + 4 H2O + 8 OH–
–
–
+
3 S2– + 2 MnO4– + 4 H2O → 3 S + 2 MnO2 + 8 OH–
4. CuS (s) + NO3 (aq)  Cu2+ (aq) + S (s) + NO (g)
(+2)(–2)
(+5)(–2)
(+2)
(0)
(acidic)
(+2)(–2)
Ox:
Red:
CuS → S + Cu2+ + 2 e–
3 e– + 4 H+ + NO3– → NO + 2 H2O
Ox:
Red:
3 CuS → 3 S + 3 Cu2+ + 6 e–
6 e– + 8 H+ + 2 NO3– → 2 NO + 4 H2O
3 CuS + 8 H+ + 2 NO3– → 3 Cu2+ + 3 S + 2 NO + 4 H2O
5. HNO3 (aq) + S (s)  NO2 (g) + H2SO4 (aq) + H2O (ℓ)
(+1)(+5)(–2)
(0)
(+4)(–2)
(+1)(+6)(–2)
(+1)(–2)
Ox:
Red:
S + 4 H2O → H2SO4 + 6 H+ + 6 e–
e– + H+ + HNO3 → NO2 + H2O
Ox:
Red:
S + 4 H2O → H2SO4 + 6 H+ + 6 e–
6 e + 6 H + 6 HNO3 → 6 NO2 + 6 H2O
–
+
6 HNO3 + S → 6 NO2 + H2SO4 + 2 H2O
(+1)(–2)
6. KMnO4 (aq) + HCl (aq) + H2S (g)  KCl (aq) + MnCl2 (aq) + S (s) + H2O (ℓ)
(+1)(+7)(–2)
(+1)(–1)
(+1)(–2)
(+1)(–1)
(+2)(–1)
(0)
(+1)(–2)
Ox:
Red:
H2S → S + 2 H+ + 2 e–
5 e– + 5 H+ + KMnO4 + 3 HCl → KCl + MnCl2 + 4 H2O
Ox:
Red:
5 H2S → 5 S + 10 H+ + 10 e–
10 e– + 10 H+ + 2 KMnO4 + 6 HCl → 2 KCl + 2 MnCl2 + 8 H2O
2 KMnO4 + 6 HCl + 5 H2S → 2 KCl + 2 MnCl2 + 5 S + 8 H2O
7. FeCl3 (aq) + H2S (g)  FeCl2 (aq) + HCl (aq) + S (s)
(+3)(–1)
(+1)(–2)
(+2)(–1)
(+1)(–1)
(0)
Ox:
Red:
H2S → S + 2 H+ + 2 e–
e– + H+ + FeCl3 → FeCl2 + HCl
Ox:
Red:
H 2S → S + 2 H + + 2 e –
2 e + 2 H + 2 FeCl3 → 2 FeCl2 + 2 HCl
–
+
2 FeCl3 + H2S → 2 FeCl2 + 2 HCl + S
8. NaCl (aq) + H2SO4 (aq) + MnO2 (s)  Na2SO4 (aq) + MnSO4 (aq) + H2O (ℓ) + Cl2 (g)
(+1)(–1)
Ox:
Red:
(+1)(+6)(–2)
(+4)(–2)
(+1)(+6)(–2)
(+2)(+6)(–2)
(+1)(–2)
2 NaCl + H2SO4 → Cl2 + Na2SO4 + 2 H+ + 2 e–
2 e– 2 H+ + MnO2 + H2SO4 → MnSO4 + 2 H2O
2 NaCl + 2 H2SO4 + MnO2 → Na2SO4 + MnSO4 + 2 H2O + Cl2
9. HMnO4 (aq) + HCl (aq)  MnCl2 (aq) + H2O (ℓ) + Cl2 (g)
(+1)(+7)(–2)
(+1)(–1)
(+2)(–1)
(+1)(–2)
(0)
Ox:
Red:
2 HCl → Cl2 + 2 H+ + 2 e–
5 e– + 5 H+ + HMnO4 + 2 HCl → MnCl2 + 4 H2O
Ox:
Red:
10 HCl → 5 Cl2 + 10 H+ + 10 e–
10 e– + 10 H+ + 2 HMnO4 + 4 HCl → 2 MnCl2 + 8 H2O
2 HMnO4 + 14 HCl → 2 MnCl2 + 8 H2O + 5 Cl2
10. MnO4 (aq) + SO32 (aq)  Mn2+ (aq) + SO42 (aq) (acidic)
(+7)(–2)
(+4)(–2)
(+2)
(+6)(–2)
Ox:
Red:
H2O + SO32– → SO42– + 2 H+ + 2 e–
5 e– + 8 H+ + MnO4– → Mn2+ + 4 H2O
Ox:
Red:
5 H2O + 5 SO32– → 5 SO42– + 10 H+ + 10 e–
10 e + 16 H+ + 2 MnO4– → 2 Mn2+ + 8 H2O
–
6 H+ + 2 MnO4– + 5 SO32– → 2 Mn2+ + 5 SO42– + 3 H2O
11. H2O2 (aq) + I (aq)  H2O (ℓ) + I2 (aq) (acidic)
(+1)(–1)
Ox:
Red:
(–1)
(+1)(–2)
(0)
2 I– → I 2 + 2 e–
2 e– + 2 H+ + H2O2 → 2 H2O
H2O2 + 2 H+ + 2 I– → 2 H2O + I2
(0)
12. AsO33 (aq) + I2 (aq)  AsO43 (aq) + I– (aq)
(+3)(–2)
(0)
Ox:
Red:
(+5)(–2)
(basic)
(–1)
H2O + AsO33– + 2 OH– → AsO43– + 2 H+ + 2 OH– + 2 e–
2 e– + I 2 → 2 I –
AsO33– + I2 + 2 OH– → AsO43– + 2 I– + H2O
13. Cr (s) + ClO4 (aq)  CrO2 (aq) + ClO3 (aq)
(0)
(+7)(–2)
(+3)(–2)
(basic)
(+5)(–2)
Ox:
Red:
Cr + 2 H2O → CrO2– + 4 H+ + 3 e–
2 e– + 2 H+ + ClO4– → ClO3– + H2O
Ox:
Red:
2 Cr + 4 H2O → 2 CrO2– + 8 H+ + 6 e–
6 e– + 6 H+ + 3 ClO4– → 3 ClO3– + 3 H2O
2 Cr + 3 ClO4– + H2O + 2 OH– → 2 CrO2– + 3 ClO3– + 2 H+ + 2 OH–
2 Cr + 3 ClO4– + 2 OH– → 2 CrO2– + 3 ClO3– + H2O
AP Chemistry 2010B #3
A sample of ore containing the mineral tellurite, TeO2 , was dissolved in acid. The resulting solution was then
reacted with a solution of K2Cr2O7 to form telluric acid, H2TeO4 . The unbalanced chemical equation for the
reaction is given below.
…TeO2(s) + …Cr2O72–(aq) + …H+(aq) → …H2TeO4(aq) + …Cr3+(aq) + …H2O(ℓ)
(a) Identify the molecule or ion that is being oxidized in the reaction.
Te in TeO2 is oxidizied (from +4 to +6)
(b) Give the oxidation number of Cr in the Cr2O72–(aq) ion.
+6
(c) Balance the chemical equation given above by writing the correct lowest whole-number coefficients on the
dotted lines.
Ox:
Red:
6 H2O + 3 TeO2 → 3 H2TeO4 + 6 H+ + 6 e–
6 e– + 14 H+ + Cr2O72– → 2 Cr3+ + 7 H2O
3 TeO2 + Cr2O72– + 8 H+ → 3 H2TeO4 + 2 Cr3+ + H2O
In the procedure described above, 46.00 mL of 0.03109 M K2Cr2O7 was added to the ore sample after it was
dissolved in acid. When the chemical reaction had progressed as completely as possible, the amount of unreacted
(excess) Cr2O72–(aq) was determined by titrating the solution with 0.110 M Fe(NO3)2 . The reaction that occurred
during the titration is represented by the following balanced equation.
6 Fe2+(aq) + Cr2O72–(aq) + 14 H+(aq) → 2 Cr3+(aq) + 6 Fe3+(aq) + 7 H2O(ℓ)
A volume of 9.85 mL of 0.110 M Fe(NO3)2 was required to reach the equivalence point.
(d) Calculate the number of moles of excess Cr2O72–(aq) that was titrated.
9.85 mL 
0.110 mol Fe2+ 1 mol Cr2O72–
2–
 1000 mL   6 mol Fe2+  = 0.000181 mol Cr2O7 in excess
(e) Calculate the number of moles of Cr2O72–(aq) that reacted with the tellurite.
46.00 mL 
0.03109 mol Cr2O72–
2–
1000 mL

 = 0.001430 mol Cr2O7 added
0.00143 mol Cr2O72– added – 0.000181 mol Cr2O72– excess = 0.001249 mol Cr2O72– reacted
(f) Calculate the mass, in grams, of tellurite that was in the ore sample.
Molar Mass = 1(127.60) + 2(16.00) = 159.60 g/mol
3 mol TeO2  159.60 g TeO2
0.001249 mol Cr2O72– 
1 mol Cr2O72–  1 mol TeO2  = 0.5980 g TeO2