Question

Consider the following reactions. reaction

MnO4−(aq) + Cl−(aq) → MnO2(s) + ClO3−(aq) in basic solution

Balance each equation under the specified conditions

Answer 1

MnO₄^- + Cl^- ------> MnO₂ + ClO^3-

The reaction can bbe balcaned by half reaction method. The half reactions are

(i) MnO₄^-     ------> MnO₂   

(ii) Cl^-   ------> ClO₃^-

Balancig the equations (i),

MnO₄^-     ------> MnO₂ + 2 H₂O (O atom is balanced by adding 2 H₂O molecule)

MnO₄^- + 4 H⁺ + ------> MnO₂ + 2 H₂O ( H is balanced by adding 4H+ on left side of the equation)

MnO₄^- + 4 H⁺   + 4 OH^- ------> MnO₂ + 2 H₂O + 4OH^- ( the medium is basic so adding 4 OH- on both sides of the equaiton )

MnO₄^- + 4 H₂O ------> MnO₂ + 2 H₂O + 4OH^- ( 4H⁺ + 4 OH^- = 4 H2O)

MnO₄^- + 2 H₂O ------> MnO₂ + 4OH^-   

MnO₄^- + 2 H₂O + 3e ------> MnO₂   + 4OH^- (iii) ( balancing the charge by adding 3e on left side)

Blancing the equaiton (ii),

Cl^- + 3 H₂O ------> ClO₃^- + 6 H+ ( balancing the O atom by adding H2O molecule and then balancing the number of H atom by adding H+ ion)

Cl^- + 3 H₂O + 6 OH- ------> ClO₃^- + 6 H+ + 6OH- ( the medium is basic so adding both sides by OH- ion )

Cl^- + 3 H₂O + 6 OH- ------> ClO₃^- + 6 H₂O

Cl^- + 6 OH- ------> ClO₃^- + 3 H₂O + 5e (iv) ( balancing the charge on right sides of equaiton by adding 5e)

(iii) =>   MnO₄^- + 2 H₂O + 3e ------> MnO₂   + 4OH^-

(iv) => Cl^- +   6 OH- ------> ClO₃^- + 3 H₂O + 5e

multiplying eqn (iii) by 5 and (iv) by 3 we get;

(iii) x 5 => 5 MnO₄^- + 10 H₂O + 15e ------> 5 MnO₂   + 20 OH^-

(iv) x 3 => 3 Cl^- +   18 OH- ------> 3 ClO₃^- + 9 H₂O + 15e

adding both the equation we get,

5 MnO₄^- + H₂O + 3 Cl^-   ------> 5 MnO₂   + 3 ClO₃^- + 2 OH^- ( this is the balanced equation)