BALANCING AN EQUATION FOR A REACTION IN BASE Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with chromite ion [Cr(OH)4^-] in basic solution to yield chromate ion (CrO4^2- ) and chloride ion. The net ionic equation is

Question

BALANCING AN EQUATION FOR A REACTION IN BASE

Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with chromite ion [latex][Cr(OH)_{4}^{-}][/latex] in basic solution to yield chromate ion [latex](CrO_{4}^{2-})[/latex] and chloride ion. The net ionic equation is

[latex]ClO^{-}(aq) + Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + Cl^{-}(aq)[/latex] Unbalanced

Balance the equation using the half-reaction method.

STRATEGY
Follow the steps outlined in Figure 4.4.

Accepted Answer

Steps 1 and 2. The unbalanced net ionic equation shows that chromium is oxidized (from +3 to +6) and chlorine is reduced (from +1 to -1). Thus, we can write the following half-reactions:

Oxidation half-reaction: [latex]Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq)[/latex]

Reduction half-reaction: [latex]ClO^{-}(aq) → Cl^{-}(aq)[/latex]

Step 3. The half-reactions are already balanced for atoms other than O and H.

Step 4. Balance both half-reactions for O by adding [latex]H_{2}O[/latex] to the sides with less O, and then balance both for H by adding [latex]H^{+}[/latex] to the sides with less H:

Oxidation: [latex]Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + 4 H^{+}(aq)[/latex]

Reduction: [latex]ClO^{-}(aq) + 2 H^{+}(aq) → Cl^{-}(aq) + H_{2}O(l)[/latex]

Step 5. Balance both half-reactions for charge by adding electrons to the sides with the greater positive charge:

Oxidation: [latex]Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + 4 H^{+}(aq) + 3 e^{-}[/latex]

Reduction: [latex]ClO^{-}(aq) + 2 H^{+}(aq) + 2 e^{-} → Cl^{-}(aq) + H_{2}O(l)[/latex]

Next, multiply the half-reactions by factors that make the electron count in each the same. The oxidation half-reaction must be multiplied by 2, and the reduction half-reaction must be multiplied by 3 to give 6 [latex]e^{-}[/latex] in both:

Oxidation: 2 × [latex][Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + 4 H^{+}(aq) + 3 e^{-}][/latex]

or 2 [latex]Cr(OH)_{4}^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 8 H^{+}(aq) + 6 e^{-}[/latex]

Reduction: 3 × [latex][ClO^{-}(aq) + 2 H^{+}(aq) + 2 e^{-} → Cl^{-}(aq) + H_{2}O(l)][/latex]

or 3 [latex]ClO^{-}(aq) + 6 H^{+}(aq) + 6 e^{-} → 3 Cl^{-}(aq) + 3 H_{2}O(l)[/latex]

Step 6. Add the balanced half-reactions:

2 [latex]Cr(OH)_{4}^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 8 H^{+}(aq) + 6 e^{-}[/latex]

[latex]\underline{3 ClO^{-}(aq) + 6 H^{+}(aq) + 6 e^{-} → 3 Cl^{-}(aq) + 3 H_{2}O(l)}[/latex]

[latex]2 Cr(OH)_{4}^{-}(aq) + 3 ClO^{-}(aq) + 6 H^{+}(aq) + 6 e^{-} → 2 CrO_{4}^{2-}(aq) + 3 Cl^{-}(aq) + 3 H_{2}O(l) + 8 H^{+}(aq) + 6 e^{-}[/latex]

Now, cancel the species that appear on both sides of the equation:

[latex]2 Cr(OH)_{4}^{-}(aq) + 3 C1O^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 3 C1^{-}(aq) + 3 H_{2}O(l) + 2 H^{+}(aq)[/latex]

Finally, since we know that the reaction takes place in basic solution, we must add 2 [latex]OH^{-}[/latex] ions to both sides of the equation to neutralize the [latex]2 H^{+}[/latex] ions on the right, giving 2 additional [latex]H_{2}O[/latex]. The final net ionic equation, balanced for both atoms and charge, is
[latex]2 Cr(OH)_{4}^{-}(aq) + 3 ClO^{-}(aq) + 2 OH^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 3 C1^{-}(aq) + 5 H_{2}O(l)[/latex]

Charge: (2 × -1) + (3 × -1) + (2 × -1) = -7 Charge: (2 × -2) + (3 × -1) = -7

Question 4.12: BALANCING AN EQUATION FOR A REACTION IN BASE Aqueous sodium ...

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