Question

Balance the following redox equation in a basic solution BH, - + CIO, - - H,BO, +CI-

Answer 1

Explanation by half reaction method

first we have to write the given unbalanced reaction

BH₄^- + ClO₃^- → Cl^- + H₂BO₃^-

now in Step 2. firstly we have to Separate the redox reaction into half-reactions and mention the oxidation numbers for each atom in the equations.

B(-5)H(+1)₄^- + Cl(+5)O(-2)₃^- → Cl(-1)^- + H(+1)₂B(+3)O(-2)₃^-

Oxidation reaction

B(-5)H(+1)₄^- → H(+1)₂B(+3)O(-2)₃^-

Reduction reaction

Cl(+5)O(-2)₃^- →Cl(-1)^-

Step 3. now we have to Balance the atoms in both half reaction.

a) Balance all other atoms except hydrogen and oxygen.

oxidation

BH₄^- → H₂BO₃^-

Reduction

ClO₃^- → Cl^-

b) Balance the oxygen atoms.and Check the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

Oxidation

BH₄^- + 3H₂O → H₂BO₃^-

Reduction

ClO₃^- → Cl^- + 3H₂O

c) Balance the hydrogen atoms and Check if there are the same numbers of hydrogen atoms on the left and right side, if they aren't equilibrate these atoms by adding protons (H⁺).

Oxidation

BH₄^- + 3H₂O → H₂BO₃^- + 8H⁺

Reduction

ClO₃^- + 6H⁺ → Cl^- + 3H₂O

d) For reactions in a basic medium, add one OH^- ion to each side for every H⁺ ion present in the equation. The OH^- ions must be added to both sides of the equation to keep the charge and atoms balanced. Combine OH^- ions and H⁺ ions that are present on the same side to form water.

Oxidation

BH₄^- + 3H₂O + 8OH^- → H₂BO₃^- + 8H₂O

Reduction

ClO₃^- + 6H₂O → Cl^- + 3H₂O + 6OH^-

Step 4. Balance the charge. To balance the charge, add electrons (e^-) to the more positive side to equal the less positive side of the half-reaction. It doesn't matter what the charge is as long as it is the same on both sides.

O:

BH₄^- + 3H₂O + 8OH^- → H₂BO₃^- + 8H₂O + 8e^-

R:

ClO₃^- + 6H₂O + 6e^- → Cl^- + 3H₂O + 6OH^-

Step 5. now balanced the electron gain equivalent to electron lost. The electrons lost in the oxidation reaction will must be equal the electrons gained in the reduction reaction.

and to make the balanced reaction multiply the coefficients of all species by integers that producing the lowest common multiple between the half-reactions

oxidation

BH₄^- + 3H₂O + 8OH^- → H₂BO₃^- + 8H₂O + 8e^-

| *3

Reduction

ClO₃^- + 6H₂O + 6e^- → Cl^- + 3H₂O + 6OH^-

| *4

Oxidation

3BH₄^- + 9H₂O + 24OH^- → 3H₂BO₃^- + 24H₂O + 24e^-

Reduction

4ClO₃^- + 24H₂O + 24e^- → 4Cl^- + 12H₂O + 24OH^-

Step 6. Add the half-reactions together.now two half reaction add both of side and write by combining them.

3BH₄^- + 4ClO₃^- + 33H₂O + 24OH^- + 24e^- → 3H₂BO₃^- + 4Cl^- + 36H₂O + 24e^- + 24OH^-

Step 7. Simplify the equation. The same species on opposite sides of the arrow can be canceled. so final balanced equation will be

3BH₄^- + 4ClO₃^- → 3H₂BO₃^- + 4Cl^- + 3H₂O