Question

Balance the following redox reactions
1) SO_3-(aq) + MnO_4-(aq) -> SO₄^2-(aq) + Mn²⁺ (aq) in acidic solution

2) H₂O_2(aq) + ClO₂(aq) -> ClO₂-(aq) + O₂(g) in basic solution

Answer 1

(1)

SO₃^2-(aq) + MnO₄^-(aq) -> SO₄^2-(aq) + Mn²⁺ (aq)

(i) Separate the oxidation half and reduction half reactions.

Oxidation half reaction                                                                       Reduction half reaction

SO₃^2-(aq) --------> SO₄^2-(aq)                                           MnO₄^-(aq) ---------> Mn²⁺ (aq)

(ii) Balance atoms ohter than O and H

SO₃^2-(aq) --------> SO₄^2-(aq)                                           MnO₄^-(aq) ---------> Mn²⁺ (aq)

(iii) To balance Oxygens in acidic medium add H2O molecules to the side where O's are required.

SO₃^2-(aq) + H₂O (l) --------> SO₄^2-(aq)                          MnO₄^-(aq) ---------> Mn²⁺ (aq) + 4 H₂O (l)

(iv) To balance hydrogens i acidic medium add H⁺ ions to the side where H's are required.

SO₃^2-(aq) + H₂O (l) --------> SO₄^2-(aq) + 2H⁺ (aq.)     MnO₄^-(aq) + 8H⁺(aq.) --------->Mn²⁺ (aq)+4H₂O (v) To balance charge add electrons to the side positive charge is more.

SO₃^2-(aq)+H₂O(l) --------> SO₄^2-(aq)+2H⁺(aq.)+2e     MnO₄^-(aq)+8H⁺(aq.)+5e ------->Mn²⁺ (aq)+4H₂O

(vi) To balance electrons multiply OHR by 5 and RHR by 2

5SO₃^2-(aq)+5H₂O(l)--->5SO₄^2-(aq)+10H⁺(aq.)+10e 2MnO₄^-(aq)+16H⁺(aq.)+10e --->2Mn²⁺(aq)+8H₂O

(vii) Add OHR and RHR to get balanced equation

5SO₃^2-(aq)+5H₂O(l)--->5SO₄^2-(aq)+10H⁺(aq.)+10e

2MnO₄^-(aq)+16H⁺(aq.)+10e --->2Mn²⁺(aq)+8H₂O

Balanced equation:

5SO₃^2-(aq) + 2MnO₄^-(aq) + 6H⁺(aq.) ------> 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3 H₂O (l)