In a particular redox reaction, Cr is oxidized to CrO42– and Cu2 is reduced to Cu . Complete and balance the equation for this reaction in acidic solution. Phases are optional.
Cr + Cu2+ = CrO42- + Cu+
complete and balance the reaction in an acidic solution. I know the half reactions are
Cr = CrO4^2-
Cu^2+ = Cu^+
To balance the redox equation, we need to ensure that the numbers of atoms on both sides of the equation are equal and that the charges are balanced. Here's how you can balance the given equation for the redox reaction in acidic solution step by step:
1. Write down the skeleton equation:
Cr + Cu2+ → CrO42- + Cu
2. Split the reaction into two half-reactions, one for the oxidation and one for the reduction:
Oxidation half-reaction: Cr → CrO42-
Reduction half-reaction: Cu2+ → Cu
3. Balance the atoms in each half-reaction:
For the oxidation half-reaction, there is already one Cr atom on each side. However, there are four oxygen (O) atoms on the right side, so we need to balance the oxygen atoms by adding four water (H2O) molecules to the left side:
Cr → CrO42- + 4H2O
For the reduction half-reaction, there is already one Cu atom on each side. To balance the charge, we need to add two electrons (e-) to the left side:
Cu2+ + 2e- → Cu
4. Balance the charges in each half-reaction:
In the oxidation half-reaction, the left side has a charge of 0, while the right side has a charge of -2. To balance the charges, we add six hydrogen ions (H+) to the left side:
Cr + 6H+ → CrO42- + 4H2O
In the reduction half-reaction, the left side has a charge of +2, while the right side has a charge of 0. The charges are already balanced.
5. Balance the number of electrons in each half-reaction:
In the oxidation half-reaction, there are ten electrons on the right side due to the four Cr-O bonds. To balance the electrons, we need to multiply the reduction half-reaction by ten:
10Cu2+ + 10e- → 10Cu
6. Combine the half-reactions:
To combine the half-reactions, we need to make the number of electrons in each half-reaction equal. Multiply the oxidation half-reaction by ten to match the number of electrons in the reduction half-reaction:
10Cr + 60H+ → 10CrO42- + 40H2O
10Cu2+ + 10e- → 10Cu
7. Add the two balanced half-reactions together:
Combine the equations above, and eliminate any species that appear on both sides of the equation. Cancel out the electrons, and simplify the equation if necessary:
10Cr + 60H+ + 10Cu2+ → 10CrO42- + 40H2O + 10Cu
Thus, the balanced redox equation for the given reaction in acidified solution is:
10Cr + 60H+ + 10Cu2+ → 10CrO42- + 40H2O + 10Cu
What is your problem with this?
Here is a site that explains how to do it. Memorize the set procedure.
http://www.chemteam.info/Redox/Redox.html
Post any question you have about the procedure.