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

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:

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