Balance the following half-reaction under basic conditions.
NO3^− → NH3
Why, hello there! Balancing equations, you say? Well, let's dive right in and balance that half-reaction under basic conditions.
First, let's start by balancing the nitrogen atoms. On the left side, we have just one N atom in NO3^-, while on the right side, we have only one N atom in NH3. So far, so good!
Next, let's balance the oxygen atoms. On the left side, we have 3 oxygen atoms in NO3^-, but on the right side, there are none. We can add water (H2O) to the right side to balance the oxygens.
NO3^- ---> NH3 + H2O
Now, we have 3 oxygens on both sides of the equation.
Lastly, let's balance the hydrogens. On the left side, there are none, while on the right side, we have 3 hydrogens in NH3 and 1 hydrogen in H2O. To balance the hydrogens, we can add OH^- to the left side of the equation.
NO3^- + OH^- ---> NH3 + H2O
And voila! We have balanced the half-reaction under basic conditions. Keep up the great work, you chemistry wizard!
To balance the half-reaction under basic conditions, follow these steps:
Step 1: Determine the elements that are being oxidized and reduced.
In the given half-reaction:
NO3^- is being reduced (changing from a higher oxidation state to a lower oxidation state) to NH3.
Step 2: Balance the non-oxygen and non-hydrogen atoms.
Count the number of nitrogen atoms on both sides of the equation:
On the left side: 1 nitrogen atom (NO3^-)
On the right side: 1 nitrogen atom (NH3)
Step 3: Balance the oxygen atoms.
On the left side, there are 3 oxygen atoms in the nitrate ion. To balance this on the right side, we need to add water (H2O) molecules.
NO3^- → NH3 + H2O
Now, there are 3 oxygen atoms on the right side, which balances the equation.
Step 4: Balance the hydrogen atoms.
On the right side, there are 3 hydrogen atoms from the ammonia molecule.
To balance this, we need to add hydroxide ions (OH^-) to the left side of the equation. Each hydroxide ion contributes one hydrogen atom.
NO3^- + 3OH^- → NH3 + H2O
Now, there are 3 hydrogen atoms on both sides of the equation.
Step 5: Balance the charges.
On the left side, the nitrate ion (NO3^-) has a charge of -1.
On the right side, ammonia (NH3) has no charge.
To balance the charges, we need to add electrons (e^-) to the left side of the equation. Each electron has a charge of -1.
2e^- + NO3^- + 3OH^- → NH3 + H2O
Now, the charges are balanced.
Step 6: Check the final equation.
Count the number of atoms on both sides of the equation to ensure that they are balanced:
On the left side: 1 nitrogen atom, 6 oxygen atoms, 3 hydrogen atoms, and 2 electrons
On the right side: 1 nitrogen atom, 4 hydrogen atoms, and 1 oxygen atom
Since all the atoms are balanced, the final balanced half-reaction equation under basic conditions is:
2e^- + NO3^- + 3OH^- → NH3 + H2O
To balance the given half-reaction, NO3^- → NH3, under basic conditions, you need to follow these steps:
Step 1: Write the unbalanced half-reaction:
NO3^- → NH3
Step 2: Balance the atoms other than hydrogen and oxygen:
In this case, we only have nitrogen (N) on both sides. Therefore, there is no need to balance any other atoms.
Step 3: Balance oxygen by adding water (H2O):
The number of oxygen atoms on the reactant side is 3 (NO3^-), while the product side has none. To balance the oxygen, you need to add water molecules (H2O) to the product side.
Thus, the balanced equation becomes:
NO3^- → NH3 + H2O
Step 4: Balance hydrogen (H) by adding hydrogen ions (H+):
Next, you need to balance the hydrogen atoms. Since we are under basic conditions, you add H2O to balance the hydrogen atoms on the product side.
Now the balanced equation is:
NO3^- + H2O → NH3 + OH^-
Step 5: Balance the charges by adding electrons (e^-):
In this case, the charge of the reactant (NO3^-) is -1, while the product side has hydroxide ions (OH^-) with a charge of -1. The charge is already balanced.
Finally, the fully balanced equation under basic conditions is:
NO3^- + H2O → NH3 + OH^-
Please don't change screen names. Refer to your post above under anonymous. For help on balancing redox equation see this link.
http://www.chemteam.info/Redox/Redox.html