Balance the reaction using half reactions under BASIC conditions:
Al + CrO4^-2 ---> Al(OH)3 + Cr(OH)3
I'm not sure how to do this question.
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http://www.chemteam.info/Redox/Redox.html
To balance the given reaction under basic conditions, we need to follow a few steps:
Step 1: Split the overall reaction into two half reactions, one for each element involved.
Al + CrO4^-2 ---> Al(OH)3 + Cr(OH)3
Half-reaction 1: Al ---> Al(OH)3
Half-reaction 2: CrO4^-2 ---> Cr(OH)3
Step 2: Balance the elements other than hydrogen and oxygen in each half reaction.
In Half-reaction 1:
The number of Al atoms on the left side is equal to the number of Al atoms on the right side, so the elements (except for H and O) are already balanced.
In Half-reaction 2:
To balance the Cr atoms, we need to add a coefficient of 2 in front of Cr(OH)3 on the right side.
CrO4^-2 ---> 2Cr(OH)3
Step 3: Balance oxygen atoms by adding water (H2O) molecules to the side of each half reaction that needs oxygen.
In Half-reaction 1:
There are no oxygen atoms in Al(OH)3.
In Half-reaction 2:
To balance oxygen atoms, add 4 H2O molecules to the left side.
CrO4^-2 + 4H2O ---> 2Cr(OH)3
Step 4: Balance hydrogen atoms by adding hydrogen ions (H+) to the side that needs hydrogen.
In Half-reaction 1:
Add 3 H+ ions to the left side.
3H+ + Al ---> Al(OH)3
In Half-reaction 2:
There are no hydrogen atoms in Cr(OH)3.
Step 5: Balance the overall charge by adding electrons to the side that needs it.
In Half-reaction 1:
The left side has a charge of +3 from the H+ ions, and the right side has no charge. To balance the charges, add 3 electrons on the left side.
3H+ + Al ---> Al(OH)3 + 3e-
In Half-reaction 2:
We need to balance the charge. Since CrO4^-2 has a charge of -2 and Cr(OH)3 has a charge of 0, we need to add 2 electrons on the left side.
CrO4^-2 + 4H2O + 2e- ---> 2Cr(OH)3
Step 6: Multiply each half reaction by the necessary coefficients so that the number of electrons gained in the oxidation half-reaction is equal to the number of electrons lost in the reduction half-reaction.
To do this, multiply the first half reaction by 2 and the second half reaction by 3.
6H+ + 2Al ---> 2Al(OH)3 + 6e-
3CrO4^-2 + 12H2O + 6e- ---> 6Cr(OH)3
Step 7: Combine the two half reactions and cancel out any common terms on both sides.
6H+ + 2Al + 3CrO4^-2 + 12H2O ---> 2Al(OH)3 + 6e- + 6Cr(OH)3
By canceling out the common terms, the balanced overall reaction becomes:
2Al + 3CrO4^-2 + 12H2O ---> 2Al(OH)3 + 6Cr(OH)3
This is the balanced equation for the given reaction under basic conditions.