Question

Розставити коефіцієнти методом електронного балансу, укажіть процеси окиснення та відновлення, речовини, які є відновником та окисником: Na 2 SO 3  + KIO 3  + H 2 SO 4  = Na 2 SO 4  +I 2  + K 2 SO 4  + H 2 O

Answer 1

Відповідь:

5Na2SO3  + 2KIO3  + H2SO4  = 5Nа2SO4  +I2  + K2SO4  + H2O

S⁺⁴-2e=S⁺⁶  ║5║відновник

2I⁺⁵+10e=I2⁰║1║ окисник

Пояснення:

Answer 2

Balancing the Equation using the Method of Electron Balance

To balance the equation using the method of electron balance, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. Let's balance the given equation step by step:

1. Identify the oxidation and reduction half-reactions: - Oxidation half-reaction: Na2SO3 → Na2SO4 - Reduction half-reaction: KIO3 → I2

2. Balance the atoms other than hydrogen and oxygen in each half-reaction: - Oxidation half-reaction: 2Na2SO3 → Na2SO4 - Reduction half-reaction: 2KIO3 → I2

3. Balance the oxygen atoms by adding water molecules (H2O) to the side that needs more oxygen: - Oxidation half-reaction: 2Na2SO3 + O2 → 2Na2SO4 - Reduction half-reaction: 2KIO3 → I2 + 3H2O

4. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen: - Oxidation half-reaction: 2Na2SO3 + O2 → 2Na2SO4 + 2H2O - Reduction half-reaction: 2KIO3 + 6H+ → I2 + 3H2O

5. Balance the charges by adding electrons (e-) to the side that needs more negative charge: - Oxidation half-reaction: 2Na2SO3 + O2 → 2Na2SO4 + 2H2O + 4e- - Reduction half-reaction: 2KIO3 + 6H+ + 10e- → I2 + 3H2O

6. Multiply the half-reactions by appropriate coefficients to make the number of electrons equal in both half-reactions: - Oxidation half-reaction: 2Na2SO3 + O2 → 2Na2SO4 + 2H2O + 4e- - Reduction half-reaction: 5KIO3 + 15H+ + 10e- → 5I2 + 15H2O

7. Finally, add the balanced half-reactions together: - 2Na2SO3 + O2 + 5KIO3 + 15H+ → 2Na2SO4 + 2H2O + 4e- + 5I2 + 15H2O - Simplifying the equation: - 2Na2SO3 + O2 + 5KIO3 + 15H+ → 2Na2SO4 + 2H2O + 5I2 + 15H2O

Processes of Oxidation and Reduction

In the balanced equation, the oxidation process occurs in the oxidation half-reaction, where Na2SO3 is oxidized to Na2SO4. The sulfur in Na2SO3 undergoes an increase in oxidation state from +4 to +6.

The reduction process occurs in the reduction half-reaction, where KIO3 is reduced to I2. The iodine in KIO3 undergoes a decrease in oxidation state from +5 to 0.

Substances Acting as Reducing and Oxidizing Agents

In the given equation, the substance acting as the reducing agent is Na2SO3. It donates electrons and gets oxidized from an oxidation state of +4 to +6.

The substance acting as the oxidizing agent is KIO3. It accepts electrons and gets reduced from an oxidation state of +5 to 0.

Please note that the balanced equation and the processes of oxidation and reduction mentioned above are based on the given equation and the method of electron balance.