Question

Complete and balance the following half-reaction in basic solution

Cr(OH)3(s) -> CrO4^2-(aq)

Cr(OH)3 + H2O ==> CrO42-

Cr(OH)3 + H2O + 5OH- ==> CrO42- + 5H2O

Cr(OH)3 + H2O + 5OH- ==> CrO42- + 5H2O + 3e-

Answer 1

The half-reaction Cr(OH)₃ to CrO₄²- in basic solution is balanced by adding water, hydroxide ions, and electrons, and then simplifying to show the conversion of chromium hydroxide to chromate ion.

Step-by-step explanation:

To balance the given half-reaction in a basic solution, we need to follow a systematic approach. We start with the unbalanced half-reaction:

Cr(OH)₃(s) → CrO₄²--(aq)

First, we need to balance the chromium atoms. There is one chromium atom on each side of the reaction, so they are already balanced. Next, we balance the oxygen atoms by adding water molecules:

Cr(OH)₃ + 3H₂O →CrO₄²-- + 5OH-

Now we need to balance the hydrogen atoms by adding OH- ions to the most positive side:

Cr(OH)₃ + 3H₂O + 5OH- →CrO₄²-- + 8H₂O

To balance the charges, we add electrons to the least positive side:

Cr(OH)3 + 3H₂O + 5OH- + 3e- → CrO₄₂- + 8H₂O

Last, we cancel out the water molecules that appear on both sides to simplify the equation:

Cr(OH)₃ + 5OH- + 3e- → CrO₄²-- + 4H₂O

The balanced half-reaction shows the conversion of solid chromium hydroxide to aqueous chromate ion in a basic solution. This demonstrates the change in oxidation state and charge balance that occurs during the reaction.