What is the reaction that would occur in a galvanic cell between solutions of Al3+ and Pb2, where would oxidation and reduction take place, and what would the cell potentials be?
Aluminum is much more prone to oxidation, so it will lose electrons (oxidation).
Half cell potentials would be
Al>>Al^3+3e -1.66 is reduction potential
Pb>>Pb^2 + 2e -.13 is reduction potential
If Eo means oxidation potential (-reduction potential)
Cell potential = EoCell= EoRed,Cathode−EoRed,Anode
then
Cell potential= 1.66-(-1.3)=1.79 volts.
check my thinking.
bobpursley, I think the calculation is wrong.
Al is oxidation, Pb is reduction. Ecell= E reduction (Pb) - E oxidation
So the cell potential will be Ecell = -0.13 - (+1.68) = -1.81 V
The reaction that would occur in a galvanic cell between solutions of Al3+ and Pb2+ can be represented as follows:
Al3+ + 3e- -> Al (Reduction half-reaction)
Pb2+ -> Pb + 2e- (Oxidation half-reaction)
In this reaction, aluminum ion (Al3+) is being reduced to aluminum (Al), while lead ion (Pb2+) is being oxidized to lead (Pb).
To determine the cell potential, we need to know the standard reduction potentials of the half-reactions involved. These values can be found in tables or reference sources.
The standard reduction potential of the aluminum half-reaction is -1.66 V, and the standard reduction potential of the lead half-reaction is -0.13 V.
To calculate the cell potential, we subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction.
Cell potential = E(red) - E(ox)
= -1.66 V - (-0.13 V)
= -1.53 V
The cell potential for this reaction is -1.53 V, indicating that it is a spontaneous reaction.
To determine the reaction, oxidation and reduction sites, as well as the cell potentials in a galvanic cell, we need to consider the reduction potentials of the species involved.
1. Look up the reduction potentials: The first step is to find the reduction potentials (also known as standard electrode potentials) of Al and Pb. These values can be found in tables or online resources. Let's assume:
Reduction potential of Al³⁺ (Al3+) → Al(s): E°(Al³⁺/Al) = -1.66 V
Reduction potential of Pb²⁺ (Pb2+) → Pb(s): E°(Pb²⁺/Pb) = -0.13 V
2. Identify the oxidation and reduction reactions: In a galvanic cell, oxidation occurs at the anode, and reduction occurs at the cathode. The species with the higher (less negative) reduction potential gets reduced, while the one with the lower (more negative) reduction potential gets oxidized.
In this case, Al³⁺ (Al3+) has a more negative reduction potential compared to Pb²⁺ (Pb2+). Therefore, Al will be oxidized at the anode, and Pb²⁺ (Pb2+) will be reduced at the cathode.
Oxidation half-reaction: Al(s) → Al³⁺(aq) + 3e⁻
Reduction half-reaction: Pb²⁺(aq) + 2e⁻ → Pb(s)
3. Calculate the overall cell potential: The cell potential (Ecell) can be determined by subtracting the reduction potential of the oxidation reaction from the reduction potential of the reduction reaction.
Ecell = E°(cathode) - E°(anode)
Ecell = E°(Pb²⁺/Pb) - E°(Al³⁺/Al)
Ecell = (-0.13 V) - (-1.66 V)
Ecell = +1.53 V
Therefore, the reaction in the galvanic cell between Al³⁺ and Pb²⁺ would be:
Anode (oxidation): Al(s) → Al³⁺(aq) + 3e⁻
Cathode (reduction): Pb²⁺(aq) + 2e⁻ → Pb(s)
The oxidation would take place at the aluminum electrode (anode), and the reduction would occur at the lead electrode (cathode). The cell potential (Ecell) would be +1.53 V.