Balance the following redox reaction occurring in basic solution: ClO−(aq)+Cr(OH)4−(aq)→CrO42−(aq)+Cl−(aq)

Answer 1

To balance the redox reaction occurring in a basic solution, follow these steps:

1. Determine the oxidation states of each element in the reactants and products:

ClO⁻: Cl (+1), O (-2)

Cr(OH)₄⁻: Cr (+3), O (-2), H (+1)

CrO₄²⁻: Cr (+6), O (-2)

Cl⁻: Cl (-1)

2. Identify the elements that undergo changes in their oxidation states:

Cl: +1 to -1 (reduction)

Cr: +3 to +6 (oxidation)

3. Balance the atoms undergoing redox changes separately:

Reduction: ClO⁻ → Cl⁻

Oxidation: Cr(OH)₄⁻ → CrO₄²⁻

4. Balance the electron transfer:

For the reduction half-reaction, 2 electrons are needed:

ClO⁻ + 2e⁻ → Cl⁻

For the oxidation half-reaction, 3 electrons are released:

Cr(OH)₄⁻ → CrO₄²⁻ + 3e⁻

5. Equalize the number of electrons in both half-reactions:

Multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2 to get 6 electrons in each half-reaction:

3(ClO⁻ + 2e⁻ → Cl⁻)

2(Cr(OH)₄⁻ → CrO₄²⁻ + 3e⁻)

This results in:

3ClO⁻ + 6e⁻ → 3Cl⁻

2Cr(OH)₄⁻ → 2CrO₄²⁻ + 6e⁻

6. Add both half-reactions and cancel out the electrons:

3ClO⁻ + 6e⁻ + 2Cr(OH)₄⁻ → 3Cl⁻ + 2CrO₄²⁻ + 6e⁻

The electrons cancel each other out:

3ClO⁻ + 2Cr(OH)₄⁻ → 3Cl⁻ + 2CrO₄²⁻

7. Balance the remaining atoms using hydroxide (OH⁻) and water (H₂O) molecules:

The oxygen atoms are already balanced, but the hydrogen atoms are not balanced. Add 8 OH⁻ ions to the right side of the equation:

3ClO⁻ + 2Cr(OH)₄⁻ → 3Cl⁻ + 2CrO₄²⁻ + 8OH⁻

8. Now the final balanced redox reaction in basic solution is:

3ClO⁻(aq) + 2Cr(OH)₄⁻(aq) → 3Cl⁻(aq) + 2CrO₄²⁻(aq) + 8OH⁻(aq)

Step 7:

The oxygen atoms are already balanced, but the hydrogen atoms are not balanced. There are 8 hydrogen atoms on the left side of the equation (from Cr(OH)₄⁻). To balance the hydrogens, add 4 water (H₂O) molecules to the right side of the equation:

3ClO⁻ + 2Cr(OH)₄⁻ → 3Cl⁻ + 2CrO₄²⁻ + 4H₂O

Now balance the hydroxide ions:

There are 8 oxygen atoms in the form of hydroxide ions (OH⁻) on the left side of the equation. Since there are 4 water molecules on the right side, we need to add 4 hydroxide ions (OH⁻) to the right side to balance the oxygen atoms:

3ClO⁻ + 2Cr(OH)₄⁻ → 3Cl⁻ + 2CrO₄²⁻ + 4H₂O + 4OH⁻

Now the final balanced redox reaction in basic solution is:

3ClO⁻(aq) + 2Cr(OH)₄⁻(aq) → 3Cl⁻(aq) + 2CrO₄²⁻(aq) + 4H₂O(l) + 4OH⁻(aq)

Answer 2

Balancing ClO^-:

ClO⁻(aq) → Cl⁻(aq)

ClO⁻(aq) → Cl⁻(aq) + H₂O (l)

ClO⁻(aq) + 2H⁺ + 2e⁻→ Cl⁻(aq) + H₂O (l)

For basic solution,

ClO⁻(aq) + 2H⁺ + 2e⁻→ Cl⁻(aq) + H₂O (l) and add OH^-

ClO⁻(aq) + H₂O + 2e⁻→ Cl⁻(aq) + 2OH^-(l)

Now balance this part:

Cr(OH)₄^- (aq) → CrO2₄⁻(aq)

Cr(OH)₄^- (aq) + 4OH⁻(aq)→ CrO₂^4-(aq) + 4H₂O(l) + 3e⁻

3ClO⁻(aq) + 3H₂O(l) + 6e⁻→ 3Cl⁻(aq) + 6OH^-(l) multiply by 3

2Cr(OH)₄^-(aq) + 8OH⁻(aq)→ 2CrO₂^4-(aq) + 8H₂O(l) + 6e⁻ multiply by 2

Add the above two equations, you get:

3ClO⁻(aq) + 2Cr(OH)₄⁻(aq) + 2OH^-(aq) → 3Cl⁻(aq) + 2CrO₄²⁻(aq) + 5H₂O (l)