How did they go from: 2As + 5H2O → As2O5 + 10e- + 10H+

Question

To: 6As + 15H2O → 3As2O5 + 30e- + 30H+
When balancing a redox reaction in acidic conditions using the oxidation number method?

Answer 1

To balance the redox reaction in acidic conditions using the oxidation number method, we follow these steps:

1. Write out the unbalanced equation and assign oxidation numbers to all elements in the equation.

2As + 5H2O → As2O5 + 10e- + 10H+

2. Identify the element undergoing oxidation and reduction, and write half-reactions for each process. In this case, As is being oxidized and H2O is being reduced.

Oxidation: As → As2O5
Reduction: 5H2O → 10e- + 10H+

3. Balance the atoms of each half-reaction, except for H and O.

Oxidation: 2As → As2O5
Reduction: 5H2O → 10e- + 10H+

4. Balance the O atoms by adding H2O molecules to the appropriate side of the equation.

Oxidation: 2As + 5H2O → As2O5 + 10e- + 10H+
Reduction: 5H2O + 10e- + 10H+ → 5H2O + 10H+

5. Balance the H atoms by adding H+ ions to the appropriate side of the equation.

Oxidation: 2As + 5H2O → As2O5 + 10e- + 10H+
Reduction: 5H2O + 10e- + 10H+ → 5H2O + 10H+ + 10e-

6. Multiply each half-reaction by a factor that will make the number of electrons equal for both half-reactions. In this case, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

Oxidation: 10As + 25H2O → 5As2O5 + 50e- + 50H+
Reduction: 10H2O + 20e- + 20H+ → 10H2O + 20H+ + 20e-

7. Add the half-reactions together and cancel out any common terms.

10As + 25H2O + 10H2O + 20e- + 20H+ → 5As2O5 + 50e- + 50H+ + 10H2O + 20H+

8. Simplify and write the balanced equation.

6As + 15H2O → 3As2O5 + 30e- + 30H+