Balance the reaction using half reactions under BASIC conditions:
Al + CrO4^-2 ---> Al(OH)3 + Cr(OH)3
I tried to solve this question and ended up with this:
Here are the two half equations
Al ---> Al(OH)3
CrO4^-2 ---> Cr(OH)3
For Al ---> Al(OH)3, I did
3OH^- + 3H2O + Al ---> Al(OH)3 + 3H^+ + 3OH^-
3OH^- + 3H2O + Al ---> Al(OH)3 + 3H2O
3OH^- + Al ---> Al(OH)3
3e^- + 3OH^- + Al ---> Al(OH)3
For CrO4^-2 ---> Cr(OH)3 I did:
5OH^- + 5H^+ CrO4^-2 ---> Cr(OH)3 + H2O + 5OH^-
5H2O + CrO4^-2 ---> Cr(OH)3 + H2O + 5OH^-
4H2O + CrO4^-2 ---> Cr(OH)3 + 5OH^-
3e^- + 4H2O + CrO4^-2 ---> Cr(OH)3 + 5OH^-
After combining the reactions I got:
3e^- + 3OH^- + Al + Cr(OH)3 + 5OH^- ---> Al(OH)3 + 3e^- + 4H2O + CrO4^-2
simplified it and got:
8OH^- + Al + Cr(OH)3 ---> Al(OH)3 + 4H2O + CrO4^-2
Is this correct?
The Al is not right.
Al atoms balance.
H atoms balance
O atoms balance.
Charge does not balance. You have 6- on the left and zero on the right. You can fix that by placing the 3e on the right.
The CrO4^2- half reaction is balanced.
The total equation is not balanced.
Al is ok
Cr is ok
O is ok
H is ok
charge does not balance. You have 8- on the left and 2- on the right. I think you just didn't add the two equations right. You somehow reversed the 5OH and 4H2O. Each belongs on the other side.
If you add the two equations you get
4H2O + CrO4^2- + 3OH^- + Al ==> Al(OH)3 + Cr(OH)3 + 5OH^- , then you adjust the OH^- on each side by removing the 3 OH^- on the left and reducing the 5OH^- on the right to 2 OH^-
Your overall approach is correct, but there are a couple of errors and areas that need clarification. Let's go through the process step by step to arrive at the correctly balanced equation.
First, let's write the half-reactions:
1. Reduction half-reaction (Al to Al(OH)3):
Al → Al(OH)3
2. Oxidation half-reaction (CrO4^-2 to Cr(OH)3):
CrO4^-2 → Cr(OH)3
Next, let's balance the individual half-reactions:
Balancing the reduction half-reaction:
Al → Al(OH)3
Since the Al(OH)3 has one aluminum atom, we balance it by adding three water molecules to the right side:
Al + 3H2O → Al(OH)3
Now, let's balance the oxidation half-reaction:
CrO4^-2 → Cr(OH)3
Since the Cr(OH)3 has one chromium atom, we need to balance the number of chromium atoms on both sides. We can do this by adding four hydroxide ions (OH^-) to the left side:
CrO4^-2 + 4OH^- → Cr(OH)3
At this point, the individual half-reactions are balanced. Now, we need to balance the number of electrons transferred between the two half-reactions to ensure charge balance.
To do this, we look at the change in oxidation states. In the reduction half-reaction, the oxidation state of aluminum changes from 0 to +3, meaning it lost three electrons. In the oxidation half-reaction, the oxidation state of chromium changes from +6 to +3, meaning it gained three electrons.
Since the number of electrons transferred should be the same in both reactions, we multiply the reduction half-reaction by three:
3Al + 9H2O → 3Al(OH)3
Now, let's combine the two balanced half-reactions:
3Al + 9H2O + CrO4^-2 + 4OH^- → 3Al(OH)3 + Cr(OH)3
Finally, we simplify the equation by canceling out the common species on both sides:
3Al + 3H2O + CrO4^-2 + 4OH^- → 3Al(OH)3 + Cr(OH)3
This is the correctly balanced equation in basic conditions. Note that there are no electrons shown explicitly because they cancel out when combining the two half-reactions.
Therefore, the balanced reaction is:
3Al + 3H2O + CrO4^-2 + 4OH^- → 3Al(OH)3 + Cr(OH)3
To balance the given reaction under basic conditions, you need to follow these steps:
1. Write the unbalanced equation: Al + CrO4^-2 ---> Al(OH)3 + Cr(OH)3
2. Separate the reaction into two half-reactions, one for the reduction and one for the oxidation.
Reduction half-reaction: CrO4^-2 ---> Cr(OH)3
Oxidation half-reaction: Al ---> Al(OH)3
3. Balance the atoms other than hydrogen and oxygen in each half-reaction separately.
For the reduction half-reaction: CrO4^-2 ---> Cr(OH)3, you correctly balanced the atoms.
For the oxidation half-reaction: Al ---> Al(OH)3, you should include water molecules in the reactants and products to balance the oxygens:
Al + 3H2O ---> Al(OH)3 + 3e^-
Additionally, you need to balance the charges by adding hydroxide ions (OH^-) to the side where the charge is not balanced. In this case, add 3 OH^- ions to the right side:
Al + 3H2O + 3OH^- ---> Al(OH)3 + 3e^- + 3OH^-
4. Balance the electrons in each half-reaction by multiplying one or both of the half-reactions by a suitable number. In this case, multiplying the reduction half-reaction by 3 cancels out the electrons:
3CrO4^-2 ---> 3Cr(OH)3
5. Combine the two half-reactions by adding them together, canceling out common species on both sides.
3CrO4^-2 + Al + 3H2O + 3OH^- ---> 3Cr(OH)3 + Al(OH)3 + 3e^- + 3OH^-
Simplify the equation by canceling out the excess OH^- on both sides:
CrO4^-2 + Al + H2O ---> Cr(OH)3 + Al(OH)3 + e^-
Therefore, the balanced equation for the reaction under basic conditions is:
CrO4^-2 + Al + H2O ---> Cr(OH)3 + Al(OH)3
Please note that the charges, water molecules, and OH^- ions need to be balanced on both sides of the equation to achieve a correctly balanced reaction under basic conditions.