Question

GUYS ITS DUE IN 2 HOURS AND I CANT DO HALF REACTIONS HELLLLP PLEASE ap chem
a) Identify the oxidation and reduction half-reactions for the equation below and describe the number of electrons that are involved in the reaction.
2Ag+(aq) + Pb(s) → 2Ag(s) + Pb2+(aq)

b) Explain the reaction that would occur in a galvanic cell between solutions of Al3+ and Pb2+. Identify the oxidation and reduction reactions and the location of each reaction. Then, use the standard reduction potential tables to determine the cell potential.

Answer 1

To analyze this redox reaction, let's break it down into its oxidation and reduction half-reactions and determine the number of electrons involved.

1. Identifying the half-reactions:

Reduction half-reaction:
Ag+ (aq) + e- → Ag(s)

Oxidation half-reaction:
Pb(s) → Pb2+ (aq) + 2e-

2. Number of electrons involved:

In the reduction half-reaction:
- Each Ag+ ion gains one electron to become Ag(s)
- There are two Ag+ ions in the overall equation
- Therefore, 2 electrons are gained in total

In the oxidation half-reaction:
- Each Pb atom loses two electrons to become Pb2+
- There is one Pb atom in the overall equation
- Therefore, 2 electrons are lost in total

The number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction, which is 2 electrons.

In summary:
- Reduction: 2Ag+ (aq) + 2e- → 2Ag(s)
- Oxidation: Pb(s) → Pb2+ (aq) + 2e-
- Total number of electrons transferred: 2 electrons

Answer 2

In a galvanic cell between solutions of Al3+ and Pb2+, the following reaction would occur:

2Al(s) + 3Pb2+(aq) → 2Al3+(aq) + 3Pb(s)

The half-reactions and their locations are:

Oxidation (occurs at the anode):
Al(s) → Al3+(aq) + 3e-

Reduction (occurs at the cathode):
Pb2+(aq) + 2e- → Pb(s)

To determine the cell potential, we need to use the standard reduction potentials:

Al3+(aq) + 3e- → Al(s) E° = -1.66 V
Pb2+(aq) + 2e- → Pb(s) E° = -0.13 V

The cell potential (E°cell) is calculated by subtracting the oxidation potential from the reduction potential:

E°cell = E°reduction - E°oxidation
E°cell = E°(Pb2+/Pb) - E°(Al3+/Al)
E°cell = (-0.13 V) - (-1.66 V) = 1.53 V

Therefore, the standard cell potential for this galvanic cell is 1.53 V[1][2].

Answer 3

TYSM!