Balance the following in basic solution.

NO2− + H2 → N2O

Find Oxidation reaction, Reduction reaction, and Net reaction

To balance the given chemical equation in basic solution, we will follow these steps:

Step 1: Write the unbalanced equation:
NO2− + H2 → N2O

Step 2: Separate the reaction into half-reactions, one for the oxidation and one for the reduction.

Oxidation half-reaction: NO2− → N2O (Assume oxygen is balancing with water)

Reduction half-reaction: H2 → N2O (Assume hydrogen is balancing with water)

Step 3: Balance the atoms other than hydrogen and oxygen in each half-reaction.

Oxidation half-reaction: 2NO2− → 2N2O

Reduction half-reaction: 2H2 → 4H2O

Step 4: Balance the oxygen atoms by adding water to the side that needs more oxygen.

Oxidation half-reaction: 2NO2− → 2N2O + 2OH−

Reduction half-reaction: 2H2 + 2OH− → 4H2O

Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen.

Oxidation half-reaction: 2NO2− + 4H+ → 2N2O + 2OH−

Reduction half-reaction: 2H2 + 2OH− → 4H2O

Step 6: Balance the charges by adding electrons (e-) to the side that has a higher positive charge.

Oxidation half-reaction: 2NO2− + 4H+ + 6e- → 2N2O + 2OH−

Reduction half-reaction: 2H2 + 2OH− → 4H2O + 4e-

Step 7: Multiply the half-reactions by appropriate coefficients to make the number of electrons in both reactions equal.

Oxidation half-reaction: 2NO2− + 4H+ + 6e- → 2N2O + 2OH−

Reduction half-reaction: 2H2 + 2OH− + 4e- → 4H2O + 4e-

Step 8: Add the two half-reactions together and cancel out any common species.

Oxidation half-reaction: 2NO2− + 4H+ + 6e- → 2N2O + 2OH−

Reduction half-reaction: 2H2 + 2OH− + 4e- → 4H2O + 4e-

Net reaction: 2NO2− + 2H2 → 2N2O + 4H2O

To balance the given equation in basic solution, we need to make sure that the number of atoms on both sides of the equation is equal and that the charges are balanced. Here's how you can balance it step by step:

1. Start by balancing the atoms other than hydrogen and oxygen. In this case, we have nitrogen (N) on both sides, so that's already balanced.

NO2− + H2 → N2O

2. Balance oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen atoms. Since the nitrogen dioxide ion (NO2-) has one oxygen atom, we need to add a water molecule (H2O) to balance the oxygen atoms on the left-hand side.

NO2− + H2O + H2 → N2O

3. Now, let's balance the hydrogen atoms. On the left-hand side, there are two hydrogen atoms in the H2 molecule, while on the right-hand side, there are four hydrogen atoms in the water (H2O) and hydrogen (H2) molecules. To balance the hydrogens, we need to add two hydroxide ions (OH-) to the left-hand side.

NO2− + H2O + H2 + 2OH- → N2O

4. Now, examine the charges. On the left-hand side, the nitrite ion (NO2-) has a charge of -1, while the hydroxide ions (OH-) have a charge of -1 each. So, the total charge on the left-hand side is -1 -1 = -2. On the right-hand side, the nitrous oxide (N2O) is neutral (charge of 0). To balance the charges, we need to add two electrons (e-) to the left-hand side.

NO2− + H2O + H2 + 2OH- + 2e- → N2O

Now, we have successfully balanced the equation in basic solution.

Oxidation reaction: NO2- → N2O + 2e-
Reduction reaction: H2 + 2OH- → 2H2O + 2e-
Net reaction: NO2- + H2 + 2OH- → N2O + 2H2O

3H2O + 4e + 2NO2^- ==> N2O + 6OH^-

4OH^- + H2 ==> 2OH^- + 2e + 2H2O which becomes
2OH^- +H2 ==> 2e + 2H2O

For balanced equation multiply the 1st one by 1 and the last one by 2 and add the two. Check my work.