Question

Balance the following redox equation by half reaction method :

H₂C₂O_4(aq) + MnO₄^Θ_(aq) → CO_2(g) + Mn^2⊕_(aq)(acidic)

Answer 1

H₂C₂O_4(aq) + MnO₄^Θ_(aq) → CO_2(g) + Mn^2⊕_(aq)(acidic)

H₂C₂O_4(aq) + MnO₄^-_(aq) → CO_2(g) + Mn²⁺_(aq)

Step 1 :

Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products.

Divide the equation into two half equations.

Step 2 : 

Balance the atoms except O and H in each half equation. 

Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3 : 

Balance H atoms by adding H⁺ ions to the side with less H. 

Hence,

Add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ions to the left side of reduction half equation.

Step 4 : 

Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5 : 

Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. 

Then add two half equation.

Add two half equations :

5H₂C₂O_4(aq) + 2MnO₄^- _(aq) + 6H⁺_(aq)→ 10CO₂ + 2Mn²⁺_(aq) + 8H₂O_(l)

The equation is balanced is terms of number of atoms and the charges.

Hence, balanced equation :

5H₂C₂O_4(aq) + 2MnO₄^- _(aq) + 6H⁺_(aq)→ 10CO₂ + 2Mn²⁺_(aq) + 8H₂O_(l)