Use the oxidation number method to balance these equations:

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:

• H2​SO4​: S is +6, O is -2, H is +1.
• HI: I is -1, H is +1.
• S: S is 0.
• I2​: I is 0.
• H2​O: H is +1, O is -2.

Redox changes:

• S: +6 (in H2​SO4​) → 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: H2​SO4​+HI→S+I2​+H2​O

Now balance electrons:

• 1 molecule H2​SO4​ 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, H2​SO4​+6HI→S+3I2​+4H2​O

Check atoms:

• H: H2​SO4​ gives 2, 6HI gives 6 = 8. 4H2​O gives 8.
• S: 1 on both sides.
• I: 6HI gives 6 I, 3I2​ gives 6.
• O: 4H2​O gives 4 O, H2​SO4​ gives 4.

Balanced!

Final Answer

H2​SO4​+6HI→S+3I2​+4H2​O

Question b)

Concepts

Oxidation number assignment, redox balancing, acidic redox reaction, electron transfer.

Step-By-Step Solution

Assign oxidation numbers:

• HBr: Br is -1.
• H2​SO4​: S is +6.
• Br2​: Br is 0.
• SO2​: S is +4.
• H2​O: H is +1, O is -2.

Redox changes:

• Br: −1 (in HBr) → 0 (in Br2​) : loses 1 electron
• S: +6 (in H2​SO4​) → +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+H2​SO4​→Br2​+SO2​+2H2​O

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 H2​SO4​, 2 in SO2​, 2×1=2 in H2​O; 2+2=4

Final Answer

2HBr+H2​SO4​→Br2​+SO2​+2H2​O

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
• H2​O: 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​+H2​O→VO2+​+I−+H+

Balance I: I2​ gives 2 I−

Balance O, H (since VO2+​ has 2 O): V3++I2​+2H2​O→VO2+​+2I−+4H+

Check atoms:

• V: 1 both sides
• I: 2 both sides
• O: 2×1=2 in H2​O, 2 in VO2+​
• H: 2×2=4 on left, 4 on right

Final Answer

V3++I2​+2H2​O→VO2+​+2I−+4H+