How did they go from: 2As + 5H2O → As2O5 + 10e- + 10H+
To: 6As + 15H2O → 3As2O5 + 30e- + 30H+
When balancing a redox reaction in acidic conditions using the oxidation number method?
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+
To balance the given redox reaction in acidic conditions using the oxidation number method, follow these steps:
Step 1: Assign oxidation numbers to all elements in the reaction.
2As + 5H2O → As2O5 + 10e- + 10H+
In this reaction, the oxidation state of As changes from 0 to +5, and the oxidation state of H changes from +1 to +.
Step 2: Separate the reaction into two half-reactions, oxidation and reduction.
Oxidation half-reaction: 2As → As2O5
Reduction half-reaction: 5H2O + 10e- + 10H+ → 5H2O
In the reduction half-reaction, you need to balance hydrogen by adding H+ ions, and balance the charge by adding electrons.
Step 3: Balance the atoms other than hydrogen and oxygen in the half-reactions.
Oxidation half-reaction: 2As → As2O5
Reduction half-reaction: 5H2O + 10e- + 10H+ → 30H2O
Since the number of oxygen atoms on the reactant side of the oxidation half-reaction is 2, and the product side is 5, balance the oxygen atoms by adding water molecules to the reactant side.
Step 4: Balance the hydrogen atoms in the half-reactions by adding H+ ions.
Oxidation half-reaction: 2As + 5H2O → As2O5 + 10H+
Reduction half-reaction: 5H2O + 10e- + 10H+ → 30H2O + 10H+
Now, the hydrogen atoms are balanced.
Step 5: Balance the charge in the half-reactions by adding electrons.
Oxidation half-reaction: 2As → As2O5 + 10e-
Reduction half-reaction: 5H2O + 10e- + 10H+ → 30H2O + 10e-
The number of electrons in the oxidation half-reaction is 10, so add 10 electrons to the reactant side of the oxidation half-reaction.
Step 6: Multiply the half-reactions by appropriate numbers to equalize the number of electrons in both half-reactions.
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the number of electrons.
3(2As → As2O5 + 10e-) <-- oxidation half-reaction
2(5H2O + 10e- + 10H+ → 30H2O + 10e-) <-- reduction half-reaction
This gives us:
6As + 15H2O → 3As2O5 + 30e- + 30H+
Now, the equation is balanced in terms of both atoms and charges.