Q. 23

# Balance the following equations by the oxidation number method.

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

(ii)

(iii)

(iv) MnO_{2} + C_{2} O^{2-}_{4}→ Mn^{2+} + CO_{2}

Answer :

(i)

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 carefullymake 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 (ii) given is wrong in sense of molecules given there, the correct equation is:

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. On the RHS of oxidation reaction the charge is -12 so we balance it by adding 12 H^{+} ions. And for reduction reaction we add 2 H^{+} ions on LHS.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 6 oxygen atoms in oxidation reaction, so add 3 water molecule to RHS of the equation to balance oxygen and 1 oxygen atom in reduction 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 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.

Finally always check the balance of charge for equation

LHS = 2×0 + 10×-1 + 8 = -2

RHS = 2×-1 + 10×0 + 4×0 = -2

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

Given Option (iii) is wrong equation, corrected equation is stated as

I_{2} + 2S_{2} O_{3}^{2-}→2I^{-} + S_{4} O^{2-}_{6}

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. Since the charges are balanced we can proceed to next step.

Step (v) Balance the oxygen atoms. The oxygen atoms are balanced in both the reactions so we don’t need to add water molecule to adjust oxygen.

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. Since the electrons gained and lost is 2 in both reactions we don’t need to multiply by any coefficients.

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.

The charges on both sides of equation are -4 so the reaction is balanced.

Option (iv) MnO_{2} + C_{2} O^{2-}_{4}→ Mn^{2+} + CO_{2}

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. Here Mn undergoes reduction, and C undergoes oxidation

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. On the RHS of oxidation reaction the charge is balanced, for reduction reaction we add 4 H^{+} ions on LHS.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in oxidation reaction, which are balanced and we add 2 water molecules in RHS of reduction reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. They are same so we can directly add the equations.

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

The charge on LHS is +2 which matches with the charge on the RHS, thus the equation is balanced.

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