For a particular redox reaction, Cr is oxidized to CrO4^2- and Cu^2+ is reduced to Cu+ . Complete and balance the equation for this reaction in basic solution. The phases are optional.

Answer 1

Start out by writing the half reactions for reduction and oxidation.

Cr→ CrO₄^2- oxidation (notice that the oxidation number of Cr goes from 0 to +6)

Cu²⁺→ Cu⁺ reduction (notice that the oxidation number of copper goes from +2 to +1)

Next determine where, and how many electrons are added or removed in the reduction and oxidation reactions respectively. For copper, it is straightforward. If the charge went from +2 to +1, clearly, and electron was added. Represent that reaction like this:

Cu²⁺ + 1 e^- → Cu⁺

For the oxidation of chromium, in basic solution, you add water to balance out the oxygens, but do so on the same side where the oxygens are (yes, I know this may seem counterintuitive at first, but read on...).

After you have done so, you balance the oxygens and hydrogens on the opposite side of the equation. Once you have done that, you balance the charges by adding electrons to the appropriate side of the equation. Completing those three steps results in this half reaction:

8OH^- + Cr → CrO₄^2- + 4H₂O + 6e^- now multiply the reduction reaction by 6 so that it gains 6e^-

6Cu²⁺ + 6e^- →6 Cu⁺ Add the two half reactions. The 6e^- cancel out since they are on opposite

sides of the equation. Final balanced reaction is:

8OH^- + Cr + 6Cu²⁺ → CrO₄^2- + 4H₂O +6Cu⁺

When you complete balancing a redox reaction, always check mass and charge balance, that is, make sure the numbers of each element on the left side of the arrow are equal their numbers on the right side. Then check that total charge on the left side of the equation is equal to total charge on the right. In this case, on the left, you have 12+ and 8-, giving you a net charge of 4+. On the right, you have 6+ and 2-, giving a net charge of 4+. I will leave you to determine if the elements are balanced.