How to Balance Complex Redox Reactions in Acidic Solution?

[Essan1995]

Hi everyone,

I’ve been practicing redox reaction balancing, but I'm struggling with some of the more complex equations—especially in acidic solutions. I know the basic steps (separate into half-reactions, balance atoms and charges, etc.), but when it comes to reactions involving multiple elements and polyatomic ions, I get stuck.

Do you have any tips, tricks, or step-by-step examples that help make the process clearer?

Thanks in advance for your help!

[dinosaurstun]

To balance a redox reaction in an acidic solution, I usually separate the half-reactions, balance the atoms, then add water and H⁺ ions to balance the oxygen and hydrogen. I also make sure the number of electrons transferred is equal between the half-reactions. This makes it easier for me to handle complex reactions, you can refer to it.

[ChenBeier]

Example
Permanganate reaction with sulfite to manganese 2+ and sulfate in acidic conditions
1. You separate the redox pair

MnO₄- / Mn 2+ and SO₃ 2- / SO₄ 2-

2. So if oxygen is involved like here you add so many H+ on reactant side to creating water on product side

MnO₄- + 8 H+ => Mn 2+ + 4 H₂O

3. Balance the charge with electrons, here 7+ to 2+ means add 5 electrons

MnO₄- + 8 H+ + 5e- => Mn 2+ + 4 H₂O Reduction is ready.

4. For oxidation do it vise versa.
SO₃ 2- + H₂O => SO₄ 2- + 2 H+

Balance with electrons
SO₃ 2- + H₂O => SO₄ 2- + 2 H+ + 2 e- Oxidation ready

5. Now find the lowest common multiple of the electrons,
It's 10 in this case 5×2

6. Multiply both equations accordingly

2 MnO₄- + 16 H+ + 10 e- => 2 Mn 2+ + 8 H₂O

5 SO₃ 2- + 5 H₂O => 5 SO₄ 2- + 10 H+ + 10 e-

7. Add both equations

2 MnO₄- + 16 H+ + 10 e- + 5 SO₃ 2- + 5 H₂O => 2 Mn 2+ + 8 H₂O +5 SO₄ 2- + 10 H+ +10 e-

8. Erase H+ ,H₂O and e-

2 MnO₄- + 6 H+ + 5 SO₃ 2- => 2 Mn 2+ + 3 H₂O +5 SO₄ 2-

Finished