Ionic Equation

Ionic Equation

Molecular and Ionic equations

On adding silver nitrate solution to a sodium chloride solution we get a white precipitate of silver chloride. Sodium chloride (a salt) is produced on mixing sodium hydroxide and hydrochloric acid in equivalent proportion. These reactions are usually represented by the following chemical equations:

NaCl +AgNO3 = AgCl +NaNO3
NaOH + HCL = NaCl + H2O
These equations are termed molecular equation. If all strong bases and acids and most salts are completely ionized, it follows that molecular equations given above are not strictly appropriate when no molecules are actually present. Chemists differ somewhat in their views on writing such equations. Some argue that since it is simpler to write molecular equations, there is no harm in writing such equations in the molecular form as long as the chemist is fully aware of the ionic character of the reacting substances. Others insist on expressing highly ionized compounds only in the form of ionic symbols which is obviously more exact. It is of course essential to know the ionic or non-ionic nature of the compounds in advance.

 Some rules that should follow while writing ionic equations are given below:

a) All strong electrolytes are expressed in ionic symbols if they are soluble in water and all weak electrolytes and covalent substances are expressed in the molecular form.
b)     Any electrolyte which is highly insoluble in water is generally written in the molecular form to indicate its insolubility. Actually, this is incorrect since many such solids are ionic in crystal form as well.
c)  In addition to the atoms which must balance on both sides of the equation, the ionic charges must also balance.
Example: Potassium chloride and silver nitrate react to produce potassium nitrate (soluble) and silver chloride (insoluble).
Molecular: KCl + AgNO3 = AgCl + KNO3
Ionic:  K+ + Cl + NO3 = AgCl + K+ + NO3

The Ion-Electron method of balancing Equation for Oxidation-Reduction Method (by use of Half reactions)

 In this method the reaction is split up into two half-reactions. In one half-reaction the oxidizing agent picks up electrons and gets reduced, and in the other reducing agent is oxidized by supplying electrons. The two half-reaction are balanced separately and added in such a way that the electrons on the left of one and on the right of the other cancel out.
This produced seems to imply that electrons are produced in redox reaction and travel the solution from the reducing agent to the other species which is reduced. This is not true.

Following steps should be followed while balancing the equation by this method:

a)  Separate the oxidizing and reducing agents.
b)  Write down one half-equation for the oxidizing agent changing into its reduced form.
c)  Write down the other half-equation showing the reducing agent changing into its oxidized form.
d)  Balance the atoms other than H and O for each half-reaction by adjusting coefficients, if necessary.
e)  Balance the oxygen atoms on the two sides by adding H2O to the side deficient in oxygen.
f)  Balance the hydrogen atoms on the two sides by adding H+ to the side deficient in hydrogen.
g) Equalize the charge on both sides by adding electrons (e) to the side deficient in negative charge.
h)  If the reaction proceeds in basic solution, add enough OH on both side of the half-reaction to get rid of H+ appearing there. Combine H+ and OH to give H2O and remove H2O duplication.

i)  Add these two balanced half-reactions in such a way that the electrons appearing on the right of one half-reaction and on the left of the other cancel. For this each half-reaction will be multiply by some number before addition.


Example: write balanced equation for the oxidation of ferrous to ferric ion by dichromate ion in acid solution. The dichromate ion by dichromate yields cr3+.

For one half-Reaction

a) Writing down the reactant and product of half-reaction for the oxidizing agent changing into its reduced form:
       Cr2O72- = Cr3+
b) Balancing the atoms other than oxygen:
       Cr2O72- = 2Cr3+
c) Balancing oxygen atoms by adding 7H2O on the right:
       Cr2O72- = 2Cr3+ + 7H2O
d) Balancing hydrogen atoms by adding 14H+ on the left:
       Cr2O72- + 14H+ = 2Cr3+ + 7H2O
e) Equalizing the charge by adding 6 electrons on the left:
             Cr2O72- + 14H+ + 6e= 2Cr3+ + 7H2O……. (1)

For the other Half-Reaction

a) Writing down the reactant and product of the half-reaction for the reducing agent changing into its oxidized form.
b) Equalizing the charge by adding one electron on the right:
          Fe2+ = Fe3+ + e…………… (2)

Adding two half-reactions

    Multiplying the half-reaction (2) by 6 and adding to the half-reaction (1)
                     Cr2O72- + 14H+ + 6e= 2Cr3+ + 7H2O
                           6Fe2+ = 6Fe3+ +6 e
               Cr2O72- + 6Fe2++ 14H+ = 2Cr3++6Fe3+ + 7H2O

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