Question

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

Will make brainliest!

Answer 1

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.

$\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{H} & 9 & 9\\\text{N} & 5 & 5\\\text{O} & 17 & 17\\\text{P} & 3 & 3\\\end{array}$

Step 11. Check that charge is balanced

$\begin{array}{cc}\textbf{On the left} & \textbf{On the right}\\0 & 0\\\end{array}$

Everything checks. The balanced equation is

5HNO₃ + 2H₂O + 3P ⟶ 5NO + 3H₃PO₄