In a particular redox reaction, Cr is oxidized to CrO4^2– and Cu2^+ is reduced to Cu^+ . Complete and balance the equation for this reaction in acidic solution.
Cr + Cu^2+ ---> CrO4^2-+ Cu^+
i started off by doing this
2 x SO3^2- ---> SO4^2- + 2e
2 x Cu^2+ +2e ---> Cu^+
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2SO3^2- ---> 2SO4^2- +4e
2Cu^2+ + 4e ----> 2Cu^+
the 4e cancel out and we got
2SO3^2- +2Cu^+ ----> 2SO4^2- + 2Cu^+
Please tell me if i did the process the wrong way. The prof hasn't started this topic but i just wanted to try it out.
I can tell you with some assurance that you will never balance chromate by using sulfite to sulfate. Here is a web site that will help you.
http://www.chemteam.info/Redox/Redox.html
oh no i messed up ! i wrote the wrong elements -_-'' im sorry.And thanks for the website!
You're on the right track! However, there are a few errors in your process and final equation.
To balance the redox reaction, we need to ensure that both the number of atoms and the charge are balanced on both sides of the equation. Here's the correct step-by-step process:
1. Split the reaction into two half-reactions, one for the oxidation and one for the reduction:
Oxidation half-reaction: Cr → CrO4^2-
Reduction half-reaction: Cu^2+ → Cu^+
2. Balance the atoms involved in each half-reaction, excluding the oxygen and hydrogen:
Oxidation half-reaction: Cr → CrO4^2-
Reduction half-reaction: Cu^2+ → Cu^+
3. Balance the number of oxygen atoms by adding water (H2O):
Oxidation half-reaction: Cr → CrO4^2-
Reduction half-reaction: Cu^2+ + H2O → Cu^+
4. Balance the charges by adding electrons (e^-) to the side that needs them:
Oxidation half-reaction: Cr → CrO4^2- + 4e^-
Reduction half-reaction: Cu^2+ + H2O → Cu^+ + 2e^-
5. Multiply each half-reaction by a coefficient to make the number of electrons in each equal, and add the half-reactions together:
2Cr + 7H2O → 2CrO4^2- + 8e^-
3Cu^2+ + 3H2O → 3Cu^+ + 6e^-
(Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4 to balance the electrons.)
6. Cancel out any common species on both sides of the equation:
6Cr + 7H2O + 3Cu^2+ → 2CrO4^2- + 3Cu^+ + 6H^+
Finally, check that the atoms and charges are balanced on both sides of the equation.
The balanced equation for the redox reaction in acidic solution is:
6Cr + 7H2O + 3Cu^2+ → 2CrO4^2- + 3Cu^+ + 6H^+
Your approach of balancing the redox reaction is correct, but there is a mistake in your final balanced equation.
To balance a redox reaction in acidic solution, you need to follow these steps:
1. Write the unbalanced equation for the reaction:
Cr + Cu^2+ ⟶ CrO4^2- + Cu^+
2. Separate the reaction into two half-reactions, one for the oxidation and one for the reduction:
Oxidation half-reaction: Cr ⟶ CrO4^2-
Reduction half-reaction: Cu^2+ ⟶ Cu^+
3. Balance the atoms other than H and O in each half-reaction. Begin by balancing the atoms undergoing oxidation or reduction:
Oxidation half-reaction: Cr ⟶ CrO4^2-
Reduction half-reaction: Cu^2+ ⟶ Cu^+
Since Cr is being oxidized, place a 4 in front of Cr to balance the number of Cr atoms:
4Cr ⟶ CrO4^2-
Since Cu^2+ is being reduced, balance the Cu atoms by placing a 2 in front of Cu^+:
Cu^2+ ⟶ 2Cu^+
4. Balance the charge in each half-reaction by adding electrons (e^-):
Oxidation half-reaction: 4Cr ⟶ CrO4^2- + 4e^-
Reduction half-reaction: Cu^2+ + 2e^- ⟶ 2Cu^+
5. Multiply the half-reactions by coefficients to make the number of electrons equal in both half-reactions (if necessary):
Multiply the oxidation half-reaction by 2:
8Cr ⟶ 2CrO4^2- + 8e^-
6. Combine the half-reactions and cancel out the electrons:
8Cr + 3Cu^2+ ⟶ 2CrO4^2- + 6Cu^+
Thus, the balanced equation for the redox reaction in acidic solution is:
8Cr + 3Cu^2+ ⟶ 2CrO4^2- + 6Cu^+