Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (Use the lowest possible coefficients.

Question

Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (Use the lowest possible coefficients.

(a) H2O2(l) + Pb(s) → H2O(l) + Pb2+(aq)
(b) O2(g) + Mn(s) → H2O(l) + Mn2+(aq)

Answer 1

Here is a site that provide a lot of information on the redox equations including the half rxn method.

http://www.chemteam.info/Redox/Redox.html

I shall be happy to answer in specific questions you have to help get you started.
The half rxns for #1 are
H2O2 ==> H2O and
Pb ==> Pb^2+

Answer 2

To balance oxidation-reduction reactions in acidic solution using the half-reaction method, follow these steps:

Step 1: Identify the oxidation and reduction half-reactions.
- Oxidation: The species that loses electrons (increases in oxidation state).
- Reduction: The species that gains electrons (decreases in oxidation state).

(a) H2O2(l) + Pb(s) → H2O(l) + Pb2+(aq)

In this reaction, H2O2 is reduced to H2O, and Pb is oxidized to Pb2+.

The oxidation half-reaction: Pb(s) → Pb2+(aq)
The reduction half-reaction: H2O2(l) → H2O(l)

(b) O2(g) + Mn(s) → H2O(l) + Mn2+(aq)

In this reaction, O2 is reduced to H2O, and Mn is oxidized to Mn2+.

The oxidation half-reaction: Mn(s) → Mn2+(aq)
The reduction half-reaction: O2(g) → H2O(l)

Step 2: Balance the atoms in each half-reaction, except oxygen and hydrogen.
Start by balancing the atoms other than oxygen and hydrogen.

(a) Oxidation half-reaction: Pb(s) → Pb2+(aq)
This reaction is already balanced.

Reduction half-reaction: H2O2(l) → H2O(l)
To balance this half-reaction, we need to balance oxygen. We can add H2O to the reactant side and H+ (from the acid solution) on the product side. Start by adding H2O:

H2O2(l) → 2H2O(l)

Now, balance hydrogen by adding H+ on the product side:

H2O2(l) + 2H+(aq) → 2H2O(l)

(b) Oxidation half-reaction: Mn(s) → Mn2+(aq)
This reaction is already balanced.

Reduction half-reaction: O2(g) → H2O(l)
To balance this half-reaction, we need to balance oxygen. Add H2O to the product side:

O2(g) + H2O(l) → 2H2O(l)

Step 3: Balance the charge in each half-reaction.
(a) Oxidation half-reaction: Pb(s) → Pb2+(aq)
To balance the charge, add two electrons (2e-) to the left-hand side:

Pb(s) → Pb2+(aq) + 2e-

Reduction half-reaction: H2O2(l) + 2H+(aq) → 2H2O(l)
To balance the charge, add two electrons (2e-) to the right-hand side:

H2O2(l) + 2H+(aq) + 2e- → 2H2O(l)

(b) Oxidation half-reaction: Mn(s) → Mn2+(aq)
To balance the charge, add two electrons (2e-) to the right-hand side:

Mn(s) → Mn2+(aq) + 2e-

Reduction half-reaction: O2(g) + 4H+(aq) + 4e- → 4H2O(l)
To balance the charge, add four electrons (4e-) to the left-hand side:

O2(g) + 4H+(aq) + 4e- → 4H2O(l)

Step 4: Multiply the half-reactions to equalize electron transfer.
To make the number of transferred electrons equal in both half-reactions, multiply each equation by the necessary coefficients.

(a) Pb(s) → Pb2+(aq) + 2e-
H2O2(l) + 2H+(aq) + 2e- → 2H2O(l)

Multiply the oxidation half-reaction by 2:
2Pb(s) → 2Pb2+(aq) + 4e-
Now the number of electrons is equal in both half-reactions.

(b) Mn(s) → Mn2+(aq) + 2e-
2O2(g) + 8H+(aq) + 8e- → 8H2O(l)

Multiply the oxidation half-reaction by 4:
4Mn(s) → 4Mn2+(aq) + 8e-
Again, the number of electrons is equal in both half-reactions.

Step 5: Add the balanced half-reactions together.
(a) 2Pb(s) + H2O2(l) + 2H+(aq) → 2Pb2+(aq) + 2H2O(l)

(b) 4Mn(s) + 2O2(g) + 8H+(aq) → 4Mn2+(aq) + 8H2O(l)

Now the equation is balanced for both mass and charge in acidic solution.