Explanation
To balance redox reactions using the oxidation number method:
- Assign oxidation numbers to all elements.
- Identify changes in oxidation numbers to find which species are oxidized and reduced.
- Express these changes as electron transfers.
- Balance electrons lost and gained, then balance the remainder of the equation by inspection.
Question a)
Concepts
Oxidation number assignment, redox balancing, oxidation-reduction, electron transfer, balancing in acidic medium.
Step-By-Step Solution
Assign oxidation numbers:
- H2SO4: S is +6, O is -2, H is +1.
- HI: I is -1, H is +1.
- S: S is 0.
- I2: I is 0.
- H2O: H is +1, O is -2.
Redox changes:
- S: +6 (in H2SO4) → 0 (in S)
- Change: +6→0 (gain 6 electrons per S atom)
- I: −1 (in HI) → 0 (in I2)
- Change: −1→0 (lose 1 electron per I atom)
Write skeletal equation: H2SO4+HI→S+I2+H2O
Now balance electrons:
- 1 molecule H2SO4 contains 1 S, which gains 6 electrons.
- Each HI gives 1 I−, which loses 1 electron.
- I2 forms from 2 I−, so let's use 6 HI to supply 6 electrons:
So, H2SO4+6HI→S+3I2+4H2O
Check atoms:
- H: H2SO4 gives 2, 6HI gives 6 = 8. 4H2O gives 8.
- S: 1 on both sides.
- I: 6HI gives 6 I, 3I2 gives 6.
- O: 4H2O gives 4 O, H2SO4 gives 4.
Balanced!
Final Answer
H2SO4+6HI→S+3I2+4H2O
Question b)
Concepts
Oxidation number assignment, redox balancing, acidic redox reaction, electron transfer.
Step-By-Step Solution
Assign oxidation numbers:
- HBr: Br is -1.
- H2SO4: S is +6.
- Br2: Br is 0.
- SO2: S is +4.
- H2O: H is +1, O is -2.
Redox changes:
- Br: −1 (in HBr) → 0 (in Br2) : loses 1 electron
- S: +6 (in H2SO4) → +4 (in SO2) : gains 2 electrons
Match electron exchange:
- For every 2 Br− oxidized to Br2 (lose 2 electrons), one S6+ is reduced to S4+ (gains 2 electrons).
Thus: 2HBr+H2SO4→Br2+SO2+2H2O
Check balance:
- H: 2+2=4 on left; 2×2=4 on right
- Br: 2 on left; 2 in Br2
- S: 1 both sides
- O: 4 in H2SO4, 2 in SO2, 2×1=2 in H2O; 2+2=4
Final Answer
2HBr+H2SO4→Br2+SO2+2H2O
Question c)
Concepts
Oxidation number assignment, balancing redox reactions in aqueous acid, electron transfer, ionic equations.
Step-By-Step Solution
Assign oxidation numbers:
- V3+: V is +3
- I2: I is 0
- H2O: H is +1, O is -2
- VO2+: (VO2+), V is +5 (O2: -2times2=-4, net +1, so V is +5)
- I−: I is -1
- H+: H is +1
Redox changes:
- V: +3 (in V3+) → +5 (in VO2+) (lose 2 electrons per V)
- I: 0 (in I2) → −1 (in I−) (gain 2 electrons per I2 molecule)
Thus:
- 1 V3+ oxidized (loses 2e)
- 1 I2 reduced (gains 2e)
Start with skeleton: V3++I2+H2O→VO2++I−+H+
Balance I: I2 gives 2 I−
Balance O, H (since VO2+ has 2 O): V3++I2+2H2O→VO2++2I−+4H+
Check atoms:
- V: 1 both sides
- I: 2 both sides
- O: 2×1=2 in H2O, 2 in VO2+
- H: 2×2=4 on left, 4 on right
Final Answer
V3++I2+2H2O→VO2++2I−+4H+