Question

Balance the following redox equation by half reaction method :

Bi(OH)_3(s) + SnO^2-_2(aq) ⟶ SnO^2-_3(aq) + Bi_(s)

Answer 1

Bi(OH)_3(s) + SnO^2-_2(aq) ⟶ SnO^2-_3(aq) + Bi_(s)

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 half equations for O atoms by adding H₂O to the side with less O atoms. 

Add 1H₂O to left side of oxidation half equation and 3H₂O 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 3H⁺ ions to the left side of reduction half equation.

Step 4 :

Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5 : 

Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. 

Then add two half equation.

Reaction occurs in basic medium. 

However, 

H⁺ ions cancel out and the reaction is balanced. 

Hence, 

No need to add OH^- ions. 

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

Hence, balanced equation :