The following reaction takes place in an acidic solution.
Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g) (unbalanced)
Write the reduction and oxidation half-reactions (without electrons).

2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)

Given

Reaction(unbalanced)

Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)

Required

Half reaction

Solution

1. Add coefficient(equalizing atoms in reaction)

2. Adding H₂O on the O-deficient side.  

3. Adding H⁺ on the H-deficient side.  

4. Adding e⁻

5. Equalizing the number of electrons and sum the all the half-reaction

1.MnO₄⁻(aq) = Mn²⁺(aq) reduction

2.MnO4(aq) = Mn2+(aq) + 4H2O

3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O

4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O

Cl⁻(aq) = Cl₂(g) oxidation

1. 2Cl-(aq) = Cl2 (g)

2-3 none

4. 2Cl-(aq) = Cl2 (g) + 2e-

5.

MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2

2Cl-(aq) = Cl2 (g) + 2e-                                    x5

2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O

10Cl-(aq) = 5Cl2 (g) + 10e-    

2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)