Balance the following redox equation, identifying the element oxidized and the element reduced. show all of the work used to solve the problem.

hno3 + p yields h3po4 + no

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N is reduced; P is oxidized;

5HNO₃ + 2H₂O + 3P ⟶ 5NO + 3H₃PO₄  

Explanation:

Probably the easiest way to balance the equation is to use the ion-electron method.

Step 1. Write the skeleton equation

HNO₃ + P ⟶ H₃PO₄ + NO

Step 2. Separate into two half-reactions.

HNO₃ ⟶ NO

P ⟶ H₃PO₄  

Step 3.  Balance all atoms other than H and O

Done

Step 4. Balance O by adding H₂O molecules to the deficient side.

HNO₃ ⟶ NO + 2H₂O

4H₂O + P ⟶ H₃PO₄  

Step 5. Balance H by adding H⁺ ions to the deficient side.

3H⁺ + HNO₃ ⟶ NO + 2H₂O

4H₂O + P ⟶ H₃PO₄ + 5H⁺

Step 6. Balance charge by adding electrons to the deficient side.

3H⁺ + HNO₃ + 3e⁻ ⟶ NO + 2H₂O

4H₂O + P ⟶ H₃PO₄ + 5H⁺ + 5e⁻

The N atom is reduced because it gains electrons. The P atom is oxidized because it loses electrons.

Step 7. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [3H⁺ + HNO₃ + 3e⁻ ⟶ NO + 2H₂O]

3 × [4H₂O + P ⟶ H₃PO₄ + 5H⁺ + 5e⁻]

Step 8. Add the two half-reactions.

15H⁺ + 5HNO₃ + 15e⁻ ⟶ 5NO + 10H₂O

12H₂O + 3P ⟶ 3H₃PO₄ +15H⁺ + 15e⁻  

15H⁺ + 5HNO₃ + 15e⁻ + 12H₂O + 3P ⟶ 5NO + 10H₂O + 3H₃PO₄ + 15H⁺ + 15e⁻

Step 9. Cancel species that occur on each side of the equation

15H⁺ + 5HNO₃ + 15e⁻ + 2(12)H₂O + 3P ⟶ 5NO + 10H₂O + 3H₃PO₄ + 15H⁺ + 15e⁻

becomes

5HNO₃ + 2H₂O + 3P ⟶ 5NO + 3H₃PO₄  

Step 10. Check that all atoms are balanced.

Step 11. Check that charge is balanced

Everything checks. The balanced equation is

5HNO₃ + 2H₂O + 3P ⟶ 5NO + 3H₃PO₄