CHEM 120: REDOX & BALANCING EQUATIONS-QUESTIONS AND ANSwERS
wITH RATIONALES/GRADED A+/2026 UpDATE/100% CORRECT
/INSTANT DOwNLOAD
QUESTION 1
Balance the following equation in acidic solution:
MnO4⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Answer:
MnO4⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
RATIONALE:
Mn in MnO4⁻ is +7, reduced to +2 in Mn²⁺ (gain 5 e⁻). Fe²⁺ is oxidized to Fe³⁺ (loss 1 e⁻).
Need 5 Fe²⁺ to supply 5 e⁻. In acidic medium, balance O with H₂O and H with H⁺. Left:
4 O → right: 4 H₂O, so add 8 H⁺ to left. Check charge: left: -1 + 5(2+) + 8 = +17; right:
2+ + 5(3+) = +17.
---
QUESTION 2
Balance in basic solution: Cl₂ → Cl⁻ + ClO⁻
Answer:
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
RATIONALE:
Disproportionation: one Cl (0) reduced to Cl⁻ (-1), other Cl (0) oxidized to ClO⁻ (Cl is
+1). Write half-reactions: Reduction: Cl₂ + 2e⁻ → 2Cl⁻. Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ +
, 2H₂O + 2e⁻ (in basic). Multiply reduction by 1, oxidation by 1, add: 2Cl₂ + 4OH⁻ → 2Cl⁻
+ 2ClO⁻ + 2H₂O. Divide by 2: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.
---
QUESTION 3
Identify oxidizing agent: Zn + CuSO₄ → ZnSO₄ + Cu
Answer:
CuSO₄ (Cu²⁺ is the oxidizing agent)
RATIONALE:
Zn (0) → Zn²⁺ (oxidation, loses 2 e⁻). Cu²⁺ (+2) → Cu (0) (reduction, gains 2 e⁻).
Oxidizing agent is reduced: Cu²⁺.
---
QUESTION 4
Balance: C₂H₅OH + O₂ → CO₂ + H₂O
Answer:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
RATIONALE:
Carbon: 2 C on left → 2 CO₂. Hydrogen: 6 H on left (C₂H₅OH has 5+1=6) → 3 H₂O.
Oxygen: left from C₂H₅OH: 1 O, right from CO₂: 4 O, from H₂O: 3 O → total 7 O right, so
need 6 more O from O₂ → 3 O₂ molecules. Check: left O = 1+6=7, right O=4+3=7.
---
wITH RATIONALES/GRADED A+/2026 UpDATE/100% CORRECT
/INSTANT DOwNLOAD
QUESTION 1
Balance the following equation in acidic solution:
MnO4⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Answer:
MnO4⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
RATIONALE:
Mn in MnO4⁻ is +7, reduced to +2 in Mn²⁺ (gain 5 e⁻). Fe²⁺ is oxidized to Fe³⁺ (loss 1 e⁻).
Need 5 Fe²⁺ to supply 5 e⁻. In acidic medium, balance O with H₂O and H with H⁺. Left:
4 O → right: 4 H₂O, so add 8 H⁺ to left. Check charge: left: -1 + 5(2+) + 8 = +17; right:
2+ + 5(3+) = +17.
---
QUESTION 2
Balance in basic solution: Cl₂ → Cl⁻ + ClO⁻
Answer:
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
RATIONALE:
Disproportionation: one Cl (0) reduced to Cl⁻ (-1), other Cl (0) oxidized to ClO⁻ (Cl is
+1). Write half-reactions: Reduction: Cl₂ + 2e⁻ → 2Cl⁻. Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ +
, 2H₂O + 2e⁻ (in basic). Multiply reduction by 1, oxidation by 1, add: 2Cl₂ + 4OH⁻ → 2Cl⁻
+ 2ClO⁻ + 2H₂O. Divide by 2: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.
---
QUESTION 3
Identify oxidizing agent: Zn + CuSO₄ → ZnSO₄ + Cu
Answer:
CuSO₄ (Cu²⁺ is the oxidizing agent)
RATIONALE:
Zn (0) → Zn²⁺ (oxidation, loses 2 e⁻). Cu²⁺ (+2) → Cu (0) (reduction, gains 2 e⁻).
Oxidizing agent is reduced: Cu²⁺.
---
QUESTION 4
Balance: C₂H₅OH + O₂ → CO₂ + H₂O
Answer:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
RATIONALE:
Carbon: 2 C on left → 2 CO₂. Hydrogen: 6 H on left (C₂H₅OH has 5+1=6) → 3 H₂O.
Oxygen: left from C₂H₅OH: 1 O, right from CO₂: 4 O, from H₂O: 3 O → total 7 O right, so
need 6 more O from O₂ → 3 O₂ molecules. Check: left O = 1+6=7, right O=4+3=7.
---
