Question

12 hclo4 + p4o10 → 6 cl2o7 + 4 h3po4 Уравняйте при помощи электронно-ионных схем

Answer 1

12HClO4+P4O10=6Cl2O7+4H3PO4

Answer 2

12 HClO4 + P4O10 → 6 Cl2O7 + 4 H3PO4

To balance the equation using the electron-ion method, follow these steps:

1. Separate the equation into half-reactions for oxidation and reduction. Oxidation half-reaction: P4O10 → 4 PO4^3- (Gain 12 electrons)

   Reduction half-reaction: 12 HClO4 + 12 e^- → 6 Cl2O7 + 12 H^+ (Loss 12 electrons)

2. Balance the atoms except for hydrogen and oxygen in each half-reaction. Oxidation half-reaction: P4O10 + 12 H2O → 4 PO4^3- + 24 H^+ Reduction half-reaction: 12 HClO4 + 12 e^- → 6 Cl2O7 + 12 H^+

3. Balance the oxygen atoms by adding water molecules. Oxidation half-reaction: P4O10 + 12 H2O → 4 PO4^3- + 24 H^+ Reduction half-reaction: 12 HClO4 + 12 e^- + 12 H2O → 6 Cl2O7 + 12 H^+

4. Balance the hydrogen atoms by adding hydrogen ions (H+). Oxidation half-reaction: P4O10 + 12 H2O → 4 PO4^3- + 24 H^+ Reduction half-reaction: 12 HClO4 + 12 e^- + 12 H2O → 6 Cl2O7 + 12 H^+

5. Balance the charges by adding electrons to the side that needs them. Oxidation half-reaction: P4O10 + 12 H2O + 12 e^- → 4 PO4^3- + 24 H^+ Reduction half-reaction: 12 HClO4 + 12 e^- + 12 H2O → 6 Cl2O7 + 12 H^+

6. Multiply each half-reaction by the appropriate factor to equalize the number of electrons. Oxidation half-reaction: 4(P4O10 + 12 H2O + 12 e^- → 4 PO4^3- + 24 H^+) Reduction half-reaction: 3(12 HClO4 + 12 e^- + 12 H2O → 6 Cl2O7 + 12 H^+)

The balanced equation is: 4P4O10 + 12H2O + 12HClO4 → 4PO4^3- + 24H^+ + 6Cl2O7