BALANCING REDOX REACTIONS

Oxidation-reduction reactions - were originally associated with reactions where oxygen, O₂, was a reactant. It's not surprising that this reactive gas, comprising 21% of the Earth's atmosphere, plays an essential role in biochemical and environmental processes. When you consume "carbs", you are eating carbohydrates (hydrates of carbon) that have the empirical formula of CH₂O. Now, the oxygen that is absorbed through your lungs reacts with these "carbs" and forms CO₂ (which you exhale), water and  E🗲E R G Y . The reaction below shows the oxidation of glucose (C₆H₁₂O₆ or 6CH₂O) by oxygen.

C₆H₁₂O_{6 (aq)} + 6 O_{2 (aq)} → 6 H₂O_(l) + 6 CO_{2 (g)} + Energy

2 Na_(s) + Cl_{2 (g)} → 2 NaCl_(s)

Na_(s) → Na⁺_(s) + e^–        

Cl_{2 (g)} + 2 e^– → 2 Cl^–_(s)      

In the oxidation reaction, electrons are "lost" by the reactant (Na_(s)) while the reduction reaction shows electrons "gained" by Cl_{2 (g)}. Stated another way, sodium functions as a reducing agent since its electrons "reduce" chlorine. Likewise, chlorine functions as an oxidizing agent by removing electrons from sodium. Half-reactions, like other reactions, are balanced by mass (elements) and charge.

Cl₂^o_(g) + 2 e^– → 2 Cl^–_(s)
     (charge = -2)      (charge = -2)  

There are two mnemonics for remembering oxidation and reduction: Leo says Ger and Oil Rig.

In the reactions above, the two half-reactions were written from the balanced molecular equation . . . . it turns out we can do the  reverse ↺ . In fact, seemingly complicated molecular equations are easily balanced from their redox half-reactions. But first, we must be able to determine which element gains electrons and which one loses electrons in the redox reaction. To do this, we must know the "charge" of each reactant element and the "charge" of each product element . . . . only compounds containing elements whose "charge" changes during the reaction are included in the half-reactions. Some "charges" are easily determined - in the example above, both sodium and chlorine are in their elemental (uncharged) state and have a charge of zero. The product of the reaction is NaCl_(s) where Na has a +1 charge and Cl has a -1 charge. The initial half-reactions are . . . .

Na^o_(s) → Na⁺_(s) + 1 e^–   (oxidation)
Cl₂^o_(g) + 2 e^– → 2 Cl^–_(s)   (reduction)

In the oxidation half-reaction, Na^o → Na⁺, a single electron has been lost. In the reduction half-reaction, Cl₂^o → 2 Cl^–, two electrons have been gained – one electron by each Cl atom (in Cl₂^o). While each half-reaction is currently balanced by atom and charge, it's impossible for the Cl_{2 (g)} in the reduction half-reaction to gain 2 electrons if the reducing agent, Na_(s), only loses 1 electron.

Number of electrons lost (oxidation) = Number of electrons gained (reduction)

To account for this inequality, the oxidation half-reaction is multiplied by 2 and added to the reduction half-reaction . . .

2 Na^o_(s) → 2 Na⁺_(s) + 2 e^–   (oxidation)
  Cl₂^o_(g) + 2 e^– → 2 Cl^–_(s)     (reduction)

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                   2 Na^o_(s) + Cl₂^o_(g) → 2 NaCl^o_(s) (molecular equation)

In the reaction between sodium and chlorine, the charges of the reactant elements (Na and Cl₂) and the product ions (Na⁺ and Cl^–) were easily assigned . . . . elements have a charge of 0 and the ion charges are based on their position in the Periodic Table. However, when the reactant or product consists of two or more elements, each element is assigned an . . . . the hypothetical charge of an atom in a compound if all of its bonds to other atoms are assumed to be 100% ionic..

Consider the reaction where MnO₄^– is reduced to MnO₂, the oxidation numbers of the elements are . . .

Mn+7O-2_{4 (aq)}– → Mn+4O-2_{2 (s)}

While the oxidation number of O-2 reflects its normal charge in an ionic compound, it is the assignment of +7 and +4 for manganese that needs further explanation (see Assigning Oxidation Numbers).

Applying the oxidation rules to the reactants and products in this reaction gives +7 and +4 oxidation numbers for manganese.

Mn+7+7O-2-8 = – 1_{4 (aq)}– → Mn+4+4O-2-4 = 0_{2 (s)}

It is only by assigning oxidation numbers to atoms involved in a redox reaction that we can determine which atoms are changing oxidation states. In this reaction, manganese is being reduced (gaining electrons) . . . . Mn+7 + 3 e^– → Mn+4. However, we can't split Mn away from the atoms it is bonded to, so we must write the reduction half-reaction as . . . .

Mn+7O_{4 (aq)}– + 3 e^– → Mn+4O_{2 (s)}

While the added electrons balance the oxidation numbers in the oxidation half-reaction above, the atoms are not balanced. The section below will guide you through the process of balancing both charge and atoms in redox reactions.