Balance the following redox reaction in acidic solution: H3PO4 + HNO2--> N2O4 + H3PO3

I'll help. P goes from +5 oxidation state on the left to +3 on the right. N goes from + 3 on the left to +4 (for each)N on the right. To make it easier to balance, I would start by multiplying HNO2 by 2 so the equation will read

H3PO4 + 2HNO2 ==> N2O4 + H3PO3
To make H balance add H2O to the appropriate side.

To balance the given redox reaction in acidic solution, we need to follow these steps:

Step 1: Break the reaction into two half-reactions:
H3PO4 → H3PO3 (Phosphorus reduction)
HNO2 → N2O4 (Nitrogen oxidation)

Step 2: Balance the atoms involved in the half-reactions (except for oxygen and hydrogen):
H3PO4 → H3PO3 (No changes needed)
HNO2 → 2NO2 (Doubled the number of nitrogen atoms)

Step 3: Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen:
HNO2 → 2NO2 + H2O

Step 4: Balance the hydrogen atoms by adding H⁺ ions (acidic solution):
HNO2 + H⁺ → 2NO2 + H2O

Step 5: Balance the charges by adding electrons (e⁻):
HNO2 + H⁺ + 2e⁻ → 2NO2 + H2O (Nitrogen oxidation)
H3PO4 → H3PO3 (Phosphorus reduction)

Step 6: Multiply the half-reactions by the appropriate factors to equalize the number of electrons in each half-reaction:
H3PO4 + 3H⁺ + 3e⁻ → H3PO3 (multiplied by 3)
HNO2 + H⁺ + 2e⁻ → 2NO2 + H2O

Step 7: Add the balanced half-reactions together and cancel out common terms:
3H3PO4 + 9H⁺ + 9e⁻ + HNO2 + H⁺ + 2e⁻ → 3H3PO3 + 2NO2 + H2O

Step 8: Simplify the equation:
3H3PO4 + 10H⁺ + 3HNO2 → 3H3PO3 + 2NO2 + H2O

So, the balanced redox reaction in acidic solution is:
3H3PO4 + 10H⁺ + 3HNO2 → 3H3PO3 + 2NO2 + H2O

To balance the given redox reaction, follow the steps below:

Step 1: Assign oxidation numbers to each element in the reaction.

H3PO4 + HNO2 → N2O4 + H3PO3

The oxidation number of hydrogen (H) is +1, oxygen (O) is -2, nitrogen (N) is +3, and phosphorus (P) is +5. The sum of the oxidation numbers should be zero for a neutral molecule and the overall charge for an ion.

Step 2: Identify the elements that undergo changes in oxidation numbers (oxidation and reduction).

In this reaction, the oxidation number of nitrogen goes from +3 in HNO2 to +4 in N2O4. It is undergoing an increase in oxidation number, so it is being oxidized.

The oxidation number of phosphorus goes from +5 in H3PO4 to +3 in H3PO3. It is undergoing a decrease in oxidation number, so it is being reduced.

Step 3: Split the reaction into half-reactions.

Since there is only one element undergoing oxidation (nitrogen) and one element undergoing reduction (phosphorus), we will split the reaction into two half-reactions:

Oxidation half-reaction:
HNO2 → N2O4

Reduction half-reaction:
H3PO4 → H3PO3

Step 4: Balance the atoms involved in the half-reactions (excluding hydrogen and oxygen).

Oxidation half-reaction:
HNO2 → N2O4 (Already balanced since there are equal atoms on both sides)

Reduction half-reaction:
H3PO4 → H3PO3 (Already balanced since there are equal atoms on both sides)

Step 5: Balance the oxygen atoms by adding water molecules (H2O).

Oxidation half-reaction:
HNO2 → N2O4 + H2O

Reduction half-reaction:
H3PO4 → H3PO3 + H2O

Step 6: Balance the hydrogen atoms by adding H+ ions.

Oxidation half-reaction:
HNO2 + 3H+ → N2O4 + H2O

Reduction half-reaction:
H3PO4 + 3H+ → H3PO3 + H2O

Step 7: Balance the charges by adding electrons (e-).

Oxidation half-reaction:
HNO2 + 3H+ + 2e- → N2O4 + H2O

Reduction half-reaction:
H3PO4 + 3H+ + 2e- → H3PO3 + H2O

Step 8: Equalize the number of electrons in both half-reactions.

Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 to equalize the number of electrons in both half-reactions.

2HNO2 + 6H+ + 4e- → 2N2O4 + 2H2O
3H3PO4 + 9H+ + 6e- → 3H3PO3 + 3H2O

Step 9: Combine the half-reactions.

Add the balanced oxidation half-reaction and the balanced reduction half-reaction together.

2HNO2 + 6H+ + 4e- + 3H3PO4 + 9H+ + 6e- → 2N2O4 + 2H2O + 3H3PO3 + 3H2O

Simplify and cancel out common elements:

2HNO2 + 9H+ + 3H3PO4 → 2N2O4 + 6H2O + 3H3PO3

Step 10: Verify the balance of atoms and charges.

Count the atoms on both sides of the equation to ensure they are equal. In this case, there are 2 nitrogens (N), 12 hydrogens (H), and 16 oxygens (O) on both sides.

Finally, check the charge by counting the total charge on both sides. In this case, the charges are balanced with a total charge of zero on both sides.

Therefore, the balanced redox reaction in acidic solution is:

2HNO2 + 9H+ + 3H3PO4 → 2N2O4 + 6H2O + 3H3PO3