Answer:
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.
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{H} & 9 & 9\\\text{N} & 5 & 5\\\text{O} & 17 & 17\\\text{P} & 3 & 3\\\end{array}[/tex]
Step 11. Check that charge is balanced
[tex]\begin{array}{cc}\textbf{On the left} & \textbf{On the right}\\0 & 0\\\end{array}[/tex]
Everything checks. The balanced equation is
5HNO₃ + 2H₂O + 3P ⟶ 5NO + 3H₃PO₄
