Question 4.12: BALANCING AN EQUATION FOR A REACTION IN BASE Aqueous sodium ...
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 (CrO_{4}^{2-}) and chloride ion. The net ionic equation is
ClO^{-}(aq) + Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + Cl^{-}(aq) Unbalanced
Balance the equation using the half-reaction method.
STRATEGY
Follow the steps outlined in Figure 4.4.
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: Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq)
Reduction half-reaction: ClO^{-}(aq) → Cl^{-}(aq)
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 H_{2}O to the sides with less O, and then balance both for H by adding H^{+} to the sides with less H:
Oxidation: Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + 4 H^{+}(aq)
Reduction: ClO^{-}(aq) + 2 H^{+}(aq) → Cl^{-}(aq) + H_{2}O(l)
Step 5. Balance both half-reactions for charge by adding electrons to the sides with the greater positive charge:
Oxidation: Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + 4 H^{+}(aq) + 3 e^{-}
Reduction: ClO^{-}(aq) + 2 H^{+}(aq) + 2 e^{-} → Cl^{-}(aq) + H_{2}O(l)
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 e^{-} in both:
Oxidation: 2 × [Cr(OH)_{4}^{-}(aq) → CrO_{4}^{2-}(aq) + 4 H^{+}(aq) + 3 e^{-}]
or 2 Cr(OH)_{4}^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 8 H^{+}(aq) + 6 e^{-}
Reduction: 3 × [ClO^{-}(aq) + 2 H^{+}(aq) + 2 e^{-} → Cl^{-}(aq) + H_{2}O(l)]
or 3 ClO^{-}(aq) + 6 H^{+}(aq) + 6 e^{-} → 3 Cl^{-}(aq) + 3 H_{2}O(l)
Step 6. Add the balanced half-reactions:
2 Cr(OH)_{4}^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 8 H^{+}(aq) + 6 e^{-}
\underline{3 ClO^{-}(aq) + 6 H^{+}(aq) + 6 e^{-} → 3 Cl^{-}(aq) + 3 H_{2}O(l)}
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^{-}
Now, cancel the species that appear on both sides of the equation:
2 Cr(OH)_{4}^{-}(aq) + 3 C1O^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 3 C1^{-}(aq) + 3 H_{2}O(l) + 2 H^{+}(aq)Finally, since we know that the reaction takes place in basic solution, we must add 2 OH^{-} ions to both sides of the equation to neutralize the 2 H^{+} ions on the right, giving 2 additional H_{2}O. The final net ionic equation, balanced for both atoms and charge, is
2 Cr(OH)_{4}^{-}(aq) + 3 ClO^{-}(aq) + 2 OH^{-}(aq) → 2 CrO_{4}^{2-}(aq) + 3 C1^{-}(aq) + 5 H_{2}O(l)
Charge: (2 × -1) + (3 × -1) + (2 × -1) = -7 Charge: (2 × -2) + (3 × -1) = -7