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

I balanced this below (or gave information to balance it).

To balance a redox reaction, we need to make sure that the number of atoms and charges are the same on both sides of the equation. Here's how to balance the given redox reaction in acidic solution:

Step 1: Assign oxidation states to each element in the reaction. In this case, the oxidation state of H in H3PO4 is +1, H in HNO2 is +1, N in HNO2 is +3, O in HNO2 is -2, P in H3PO4 is +5, and O in H3PO4 is -2.

Step 2: Identify the elements undergoing oxidation and reduction. In this reaction, the oxidation state of N is changing from +3 to +4, so it is undergoing oxidation, and the oxidation state of P is changing from +5 to +3, so it is undergoing reduction.

Step 3: Write two separate half-reactions, one each for oxidation and reduction.

Oxidation half-reaction: HNO2 --> N2O4

Reduction half-reaction: H3PO4 --> H3PO3

Step 4: Balance the atoms other than H and O in both half-reactions. Start with the element that is changing oxidation state.

Balancing oxidation half-reaction:
We have 2 N on the product side, so we need to put a 2 in front of the NO2 to balance the nitrogen atoms.

2HNO2 --> N2O4

Balancing reduction half-reaction:
We have 4 H on the product side, so we need to put a 4 in front of the H3PO3 to balance the hydrogen atoms.

H3PO4 --> 4H3PO3

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

In the oxidation half-reaction, there are 4 oxygen atoms on the left side, but only 2 on the right side. Adding 2 H2O molecules to the right side balances the oxygen atoms.

2HNO2 --> N2O4 + 2H2O

In the reduction half-reaction, there are 12 oxygen atoms on the left side, but only 6 on the right side. Adding 6 H2O molecules to the right side balances the oxygen atoms.

H3PO4 + 6H2O --> 4H3PO3

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

In the oxidation half-reaction, there are 4 hydrogen atoms on both sides.

2HNO2 + 2H+ --> N2O4 + 2H2O

In the reduction half-reaction, there are 12 hydrogen atoms on the left side, but only 24 on the right side. Adding 24 H+ ions to the left side balances the hydrogen atoms.

H3PO4 + 6H2O + 24H+ --> 4H3PO3

Step 7: Balance the charges by adding electrons (e-) to the side with the higher positive charge.

In the oxidation half-reaction, the total charge on the left side is 0, while on the right side, it is +2 due to H+ ions. We need to add 2 electrons to the left side to balance the charges.

2HNO2 + 2H+ + 2e- --> N2O4 + 2H2O

In the reduction half-reaction, the total charge on the left side is +1 due to H3PO4, while on the right side, it is 0. We need to add 1 electron to the left side to balance the charges.

H3PO4 + 6H2O + 24H+ + e- --> 4H3PO3

Step 8: Multiply the half-reactions by the appropriate coefficients to equalize the number of electrons.

Multiply the oxidation half-reaction by 1 and the reduction half-reaction by 2 to balance the electrons.

2HNO2 + 2H+ + 2e- --> N2O4 + 2H2O
2H3PO4 + 12H2O + 48H+ + 2e- --> 8H3PO3

Step 9: Combine the half-reactions and cancel out any common species.

Adding the two balanced half-reactions together gives the balanced redox reaction:

2HNO2 + 2H+ + 2e- + 2H3PO4 + 12H2O + 48H+ + 2e- --> N2O4 + 2H2O + 8H3PO3

Simplifying the equation gives the final balanced redox reaction:

2HNO2 + 2H3PO4 + 12H2O --> N2O4 + 8H3PO3