Balance the following in basic solution: Al(s) + CrO42¯(aq) → Al(OH)3(s) + Cr(OH)4¯(aq)
Can someone please help me do this. I having a hard time, especially determining the charge of Al(OH)3. Thanks to anyone who helps
The charge of Al(OH)3 is zero as is the charge of all compounds.
In the equation Al has changed from an oxidation state of zero to an oxidation state of +3 in Al(OH)3.
To balance the given chemical equation in basic solution, follow these steps:
Step 1: Write the unbalanced equation:
Al(s) + CrO4^2-(aq) → Al(OH)3(s) + Cr(OH)4^-(aq)
Step 2: Balance the atoms except for hydrogen and oxygen. Start with aluminum (Al), then balance chromium (Cr), and finally balance oxygen (O):
Al(s) + CrO4^2-(aq) → Al(OH)3(s) + Cr(OH)4^-(aq)
Step 3: Balance the hydrogen atoms by adding enough H2O molecules to the opposite side of the equation to equalize the number of hydrogen atoms:
Al(s) + CrO4^2-(aq) + 4 H2O(l) → Al(OH)3(s) + Cr(OH)4^-(aq)
Step 4: Balance the oxygen atoms by adding OH- ions to the opposite side of the equation. The number of OH- ions added should equal the number of missing O atoms from the previous step:
Al(s) + CrO4^2-(aq) + 4 H2O(l) → Al(OH)3(s) + Cr(OH)4^-(aq) + 7 OH-(aq)
Step 5: Determine the charge of Al(OH)3. The charge of a compound can be determined based on the charges of its constituent ions. Here, Al(OH)3 consists of Al^3+ cation and OH- anions. Since the compound is neutral, the total charge must balance out to zero. Given that OH- carries a charge of -1, the charge on Al^3+ must be +3. Therefore, Al(OH)3 has a charge of 0.
The balanced equation in basic solution is:
2 Al(s) + 3 CrO4^2-(aq) + 8 H2O(l) → 2 Al(OH)3(s) + 3 Cr(OH)4^-(aq)