Q. 25

# Balance the following ionic equations

(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}→ Cr^{3+} + I_{2} + H_{2}O

(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}→ Cr^{3+} + Fe^{3+} + H_{2}O

(iii) MnO^{-}_{4} + SO^{2-}_{3} + H^{+}→ Mn^{2+} + S O^{2-}_{4} + H_{2}O

(iv) MnO_{4}^{-} + H^{+} + Br^{-}→ Mn^{2+} + Br_{2} + H_{2}O

Answer :

Option (i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}→ Cr^{3+} + I_{2} + H_{2}O

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H^{+} on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. Multiply oxidation reaction by 3.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charge on the LHS is +6 which matches with the RHS of the equation thus the reaction is balanced.

Option (ii)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H^{+} on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Verify whether all charges are balanced.

LHS = 6×+2 + 1× -2 + 14 = 24

RHS = 6×+3 + 2×+3 + 7×0 = 24

Thus the sum total is same the solve reaction is correct.

Option (iii)

Step (i) we generate the unbalanced skeleton:

Here Mn undergoes reduction and S oxidation.

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since the atoms are balanced the reaction is same as previous step.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H^{+} on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen and 1 H_{2}O molecule on LHS of oxidation reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

LHS = 2×-1+6+5×-2= -6

RHS = 2×+2+3× 0+5×-2 = -6

Equal charges on both sides imply balanced equation.

Option (iv)

Step (i) we generate the unbalanced skeleton:

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since the atoms are balanced the reaction is same as previous step.

Oxidation:

Reduction:

^{+} on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charge on LHS and RHS is equal.

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