Forming Ionic Equation from the Half Equation

  1. Ionic equation of a redox reaction can be formed from the half equations of the reaction.
  2. When writing the ionic equation, make sure that the number of electrons in both the oxidation reaction and reduction reaction are balance.


Example:
Redox reaction between potassium iodide and potassium manganate (VII)

Half equations
2I- → I2 + 2e -------(1)

MnO4- +  8H+ + 5e → Mn2+ +  4H2O -------(2)

To make the number of electrons in both chemical equation equal
(1) x 5
10I- → 5I2 + 10e
(2) x 2
2MnO4- +  16H+ + 10e → 2Mn2+ +  8H2O
Ionic Equations
Add the 2 equations together. Exclude the electrons.


10I- + 2MnO4- +  16H+ → 5I2 2Mn2+ +  8H2O



Example:
Redox reaction between iron(II) sulphate and potassium dicromate(VI)

Half equations

Fe2+ → Fe3+ + e -------(1)

Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O -------(2)

To make the number of electrons in both chemical equation equal
(1) x 5
6Fe2+ → 6Fe3+ + 6e
(2)
Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
Ionic Equations
Add the 2 equations together. Exclude the electrons.

6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O



Example:
Redox reaction between iron(III) nitrate and sulphur dioxide gas
Half equations

Fe3+ → Fe2+ + e -------(1)

SO2 + 2H2O → SO42- + 4H+ + 2e -------(2)

To make the number of electrons in both chemical equation equal
(1) x 2
2Fe3+ → 2Fe2+ + 2e

(2)
SO2 + 2H2O → SO42- + 4H+ + 2e

Ionic Equations
Add the 2 equations together. Exclude the electrons.

2Fe3+ + SO2 + 2H2→ 2Fe2+ SO42- + 4H+



Example:
Redox reaction between iron(III) chloride and hydrogen sulphide gas
Half equations

Fe3+ → Fe2+ + e -------(1)

H2S → 2H+ + S + 2e -------(2)

To make the number of electrons in both chemical equation equal
(1) x 2
2Fe3+ → 2Fe2+ + 2e

(2)
H2S → 2H+ + S + 2e

Ionic Equations

Add the 2 equations together. Exclude the electrons.

2Fe3+ + H2S → 2Fe2+ + 2H+ + S



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