Objective C: Balancing Oxidation Reduction Reactions

Solutions to Objective C HW
Solutions to More Objective C HW

Example: MnO₄^-1 + Cl^-1 --> Mn⁺² + Cl₂

The basic idea is that, in each half reaction, you have to balance

(1) the participating atoms(if necessary),

(2) the oxidation numbers by using electrons,

(3) the charge(if necessary)by using H⁺¹ or OH^-1,

(4) and the number of hydrogens with water(if necessary).

(5) Then you combine the two half reactions algebraically to get the electrons to cancel.

Steps 1 and 2

Figure out what is being oxidized and reduced. Make sure that those participating atoms are balanced and write each half reaction.

MnO₄^-1 + 5e--> Mn⁺² (5 electrons because Mn is being reduced from +7 to 2)

2Cl^-1 --> Cl₂ + 2e (2 electrons because each Cl is being oxidized from –1 to 0)

Step 3

Balance the total charge on each side of equation with H⁺¹ if they specify acidic conditions and with OH^-1 if conditions are alkaline.

Acidic situation:

MnO₄^-1 + 5e + 8H⁺¹ --> Mn⁺²

2Cl^-1 --> Cl₂ + 2e

Basic(alkaline) situation:

MnO₄^-1 + 5e --> Mn⁺² + 8OH^-1

2Cl^-1 --> Cl₂ + 2e

Step 4

Balance the number of hydrogens on each side of the equation with water.

Acidic situation:

MnO₄^-1 + 5e + 8H⁺¹ --> Mn⁺² + 4H₂O

2Cl^-1 --> Cl₂ + 2e

Basic(alkaline) situation:

MnO₄^-1 + 5e + 4H₂O--> Mn⁺² + 8OH^-1

2Cl^-1 --> Cl₂ + 2e

Step 5

Manipulate the equations so that the sum of the two will get the electrons to cancel out.

Acidic situation:

2[MnO₄^-1 + 5e + 8H⁺¹ --> Mn⁺² + 4H₂O] = 2MnO₄^-1 + 10e + 16H⁺¹ --> 2Mn⁺² + 8H₂O

5[2Cl^-1 --> Cl₂ + 2e] = 10Cl^-1 --> 5Cl₂ + 10e

Final answer ( acidic solution): 2MnO₄^-1 + 16H⁺¹ + 10Cl^-1-->5Cl₂ +2Mn⁺² + 8H₂O