Balancing a Redox Equation for a Reaction in Basic Solution Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with [Cr(OH)4^-]in basic solution to yield (CrO4^2-) and chloride ion. The net ionic equation is ClO^- (aq) + Cr(OH)4^-(aq) → CrO4^2-(aq)+Cl^-1

Question

Balancing a Redox Equation for a Reaction in Basic Solution
Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with [latex][Cr(OH)^{-}_{4} ][/latex] in basic solution to yield [latex](CrO^{2-}_{4})[/latex] and chloride ion. The net ionic equation is

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

Balance the equation using the half-reaction method.

STRATEGY
Follow the steps outlined in Figure 19.1, and add [latex]OH^{-}[/latex] ions as a final step to neutralize any [latex]H^{+}[/latex] ions that appear in the equation.

Accepted Answer

Step 1. Write the unbalanced net ionic equation.

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

Step 2. Decide which atoms are oxidized and which are reduced, and write the two unbalanced half-reactions. 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^{2-}_{4} (aq)[/latex]
Reduction half-reaction [latex]ClO^{-} (aq) → Cl^{-}(aq)[/latex]

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

Step 4. Balance each half-reaction for O by adding [latex]H_{2}O[/latex] to the side with less O, and balance for H by adding [latex]H^{+}[/latex] to the side with less H.

Oxidation [latex] Cr(OH)^{-}_{4}(aq) → CrO^{2-}_{4}(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 [latex](e ^{-} )[/latex] to the side with the greater positive charge.

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

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

Step 6. Multiply the half-reactions by integers to make the electron count the same in both half-reactions. 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 [latex] 2 × [Cr(OH)]^{-}_{4}(aq) → CrO^{2-}_{4}(aq) + 4 H^{+}(aq) + 3 e^{-}][/latex]
or [latex]2 Cr(OH)^{-}_{4}(aq) → 2 CrO^{2-}_{4}(aq) + 8 H^{+}(aq) + 6e^{-}[/latex]

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

Step 7. Add the two half-reactions together, and cancel electrons and other species that occur on both sides of the equation. The electrons must always cancel.

[latex]\begin{matrix} 2 Cr(OH)^{-}_{4}(aq) \longrightarrow 2 CrO^{2-}_{4}(aq) + 8 H^{+}(aq) + \cancel{6 e^{-}} \ 3 ClO^{-}(aq) + 6 H^{+}(aq) + \cancel{6 e^{-}} \longrightarrow 3 Cl^{-}(aq) + 3 H_{2}O(l) \ \hline 2 Cr(OH)^{-}_{4}(aq)+3ClO^{-}(aq)+6H^{+}(aq) \longrightarrow 2CrO^{2-}_{4}(aq)+Cl^{-}(aq)+3H_{2}O(l)+8H^{+}(aq) \end{matrix} [/latex]

Now, cancel the other species that appear on both sides of the equation. Since there are [latex]8 H^{+}[/latex] in the products and [latex]6 H^{+}[/latex] in the reactants, the result after cancellation is [latex]2 H^{+}[/latex]

[latex]2 Cr(OH)^{-}_{4}(aq) + 3 ClO^{-}(aq) → 2 CrO^{2-}_{4}(aq) + 3 Cl^{-}(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 [latex]2 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]

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

[latex]2 OH^{-}(aq) + 2 Cr(OH)^{-}_{4}(aq) + 3 ClO^{-}(aq) → 2 CrO^{2-}_{4}(aq) + 3 Cl^{-}(aq)+ 3 H_{2}O(l) + 2 H_{2}O(l)[/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^{2-}_{4}(aq)+3 Cl^{-}(aq)+5 H_{2}O(l)[/latex]

[latex]\begin{matrix} ext{Charge: }(2 × -1) + (3 × -1) + (2 × -1) = -7 & ext{ Charge: } (2 × -2) + (3 × -1) = -7 \ Cr = 2 & Cr = 2 \ Cl = 3 & Cl = 3 \ O = 13 & O = 13 \ H = 10 & H = 10 \end{matrix}[/latex]

Question 19.2: Balancing a Redox Equation for a Reaction in Basic Solution ...

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