How do i solve this half reaction with Mg(OH)2(s) --> Mg(s)?
Complete and balance the displayed half reaction in basic aqueous solution taking into account the electrons.
Mg(OH)2(s) → Mg(s)
Mg(OH)2 +2e ==> Mg + 2OH^-
To solve the given half reaction, we need to balance the number of atoms on both sides of the equation, including the oxygen and hydrogen atoms. Here's how you can do it step by step:
1. Start by writing the given half reaction equation:
Mg(OH)2(s) → Mg(s)
2. Identify the elements and their oxidation states on both sides of the equation. In this case, we have magnesium (Mg) with an oxidation state of +2 on both sides.
3. Balance the atoms other than hydrogen and oxygen. In this case, we only have Mg as the only atom other than hydrogen and oxygen. Since Mg is already balanced, we can move to the next step.
4. Balance the hydrogen atoms by adding water (H2O) molecules to the side that needs more hydrogens. In this case, there are 2 hydrogens on the left side and 0 hydrogens on the right side. So, we need to add 2 water molecules on the right side:
Mg(OH)2(s) → Mg(s) + 2H2O
5. Balance the oxygen atoms by adding hydroxide ions (OH-) to the side that needs more oxygens. To balance the oxygen, we add 2 OH- ions on the left side:
Mg(OH)2(s) + 2OH- → Mg(s) + 2H2O
6. Finally, balance the charges by adding the appropriate number of electrons (e-) to the side that needs it. In this case, we need 2 electrons on the left side to balance the charge:
Mg(OH)2(s) + 2OH- + 2e- → Mg(s) + 2H2O
Now, the half reaction equation is balanced, taking into account the electrons and the presence of hydroxide ions in basic aqueous solution.