Half reaction method of balancing Redox chemical equations

Redox reactions (Oxidation-reduction reactions) are comprised of two half reactions - oxidation reactions and reduction reactions. These reactions involve the transfer of electron from one species to the other species. Loss of electrons is known as oxidation and gain of electrons is known as reduction. A half reaction happens when two such reactions are coupled so that the electron released by one reactant is accepted by the other half reactant and thus completing the reaction. For example:

Zn + Cu²⁺ → Zn²⁺ + Cu; is a redox reaction.

 Zn is oxidized in this reaction .Cu²⁺ is reduced. For balancing redox reactions half reaction method will be suitable.

Balancing of Redox Equations:

During redox reactions, there is a change in oxidation number of the elements due to the transference of electrons. The number of electrons lost during oxidation is equal to the number of electrons gained during reduction. This forms the basic principle of balancing redox equations.

Balancing  Redox Equations by Ion Electron or Half Reaction Method.

The various steps involved in the balancing of redox reactions according to this method are:

• Indicate the oxidation number of each atom involved in the reaction. Identify the elements which undergo a change in oxidation number.
• Divide the skeleton redox equation into half reactions: Oxidation half and Reduction half. In each half reaction, balance the atom which undergoes change in oxidation number.
• Add electrons to whichever side is necessary in order to make up for the difference in oxidation number in each half reaction.
• Balance oxygen atoms by adding required number of water molecules to the side which is falling short of O atoms in each half reaction.
• This step is meant for ionic equations only. It involves the balancing of H atoms in each half reaction.
  • For acidic medium: Add required number of H+ ions to the side falling short of H atoms.
  • For basic medium: Add required number of H₂O molecules to the side falling short of H atoms and equal number of OH- ions to the other side.
• Equalise the number of electrons lost and gained by multiplying the half reactions with suitable integer and add them to get the final equation.

The application of various steps described above has been illustrated as follows by balancing the redox equation representing the reaction between Iodine and Nitric acid.

HNO₃ + I₂ → HIO₃+NO ₂+H₂O

Step 1: Indication of oxidation numbers of each atom.

H      N     O₃   +   I₂   →  H     I     O₃     +    N   O ₂   +H₂   O

(+1) (+5) (-2)     (0)        (+1) (+5) (-2)      (+4) (-2)  (+1) (-2) (Oxidation Numbers of individual elements)

Thus, only nitrogen and iodine undergo change in oxidation number.

Step 2: Division into two half cell reactions and balancing the atoms undergoing change in Oxidation Number.

0            +5

I₂→ 2H I O

  +5         +4     (+5 and +4 are the oxidation numbers of Nitrogen)

HNO₃→ NO₂

Step 3: Addition of electrons to make up the difference in Oxidation Number.

0            +5             (0 and +5 are the oxidation numbers of Iodine)

I₂→ 2H I O₃ +10 e^-

(Each I atom loses 5e^- therefore, two iodine atoms would lose 10e^- )

   +5                     +4   (+5 and +4 are the oxidation numbers of nitrogen)

HNO₃+ e^- → NO₂

(Each N atom gains 1 electron)

Step 4: Balancing of O by adding required number of H molecules to the side falling short of O atoms.

I₂+ 6H₂O→ 2HIO₃+10e^-

HNO3 +e^-→ NO₂+H₂O

Step 5: Not required because the equation is not ionic.

Step 6: To equalise the electrons multiply reduction half reaction by 10 and then add the two.

I₂+ 6H₂O → 2HIO₃ + 10e^-

[HNO₃+e^-→NO₂+H ₂O] × 10

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I₂ + 10HNO₃→2HIO₃ +10NO₂+4H₂O

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Examples:

1.When sodium reacts with chlorine, sodium loses an electron forming ion (Na+) and that electron is accepted by chlorine atom to form a chloride ion (Cl-).The resulting molecule is Na+Cl- i.e.  NaCl. In this process, each atom completes its set of eight electrons in the valence shell.

2Na → 2 Na⁺ + 2e- (Oxidation half)

Cl₂ + 2e- → 2Cl^-      (Reduction half)

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Net Reaction: 2Na + Cl₂ → 2(Na⁺ + Cl^-) i.e. 2NaCl (Redox Reaction)

2 In the reduction of ferric chloride to ferrous chloride by nascent hydrogen, the three positive charges carried by the ferric ion are reduced to two positive charges by a gain of one electron. This electron is supplied by the nascent hydrogen atom which is automatically changed to a hydrogen ion by the loss of one electron.

H → H⁺ + e^-   (Oxidation half)

Fe³⁺ + 3Cl^- + e^-     → Fe²⁺ + 3Cl^-   (Reduction half)

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Net Reaction: Fe³⁺ + 3Cl^- + H→ Fe²⁺ + 2Cl^-+ H⁺+ Cl^- (Redox Reaction)

Ions which do not take part in reaction are known as spectator ions and in the net reaction may be disregarded. Here Cl^- ions are disregarded.

Net reaction: Fe³⁺ + H → Fe²⁺ + H⁺

3. Similarly when a metal displaces another metal from its salt solution, metal which displaces another metal is said to be oxidised and the one which is displaced is said to be reduced. Thus, when iron displaces copper from the solution of copper sulphate, Fe is oxidised and Cu⁺⁺ ion is reduced.

Fe → Fe²⁺+2e- (Oxidation half)

Cu ²⁺+ SO₄^2- +2e- → Cu + SO₄^2-  (Reduction half)

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Net Reaction: Fe + Cu ²⁺→ Fe²⁺+ Cu (Redox Reaction)

SO₄^-- ion is a spectator ion, as it does not take part in chemical reaction.

All the above illustrations clearly show that when one substance is oxidised, the other substance is reduced at the same time. Thus oxidation and reduction go hand in hand. As Oxidation means loss of electrons and Reduction means gain of electrons, any oxidation process when written in a reverse becomes a reduction process.