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.
Would the reaction for the two metals need something to bond with that doesn't really have an effect, like NO3?
To determine the reaction that would occur in a galvanic cell between solutions of Al3+ and Pb2+, we first need to identify the oxidation and reduction reactions and the location of each reaction.
In this case, aluminum ion (Al3+) would be the reducing agent and it will undergo oxidation, losing electrons to form aluminum metal (Al). The oxidation reaction can be written as:
Al3+ (aq) - 3e- --> Al (s) (1)
On the other hand, lead ion (Pb2+) would act as the oxidizing agent and gain electrons to form solid lead (Pb). The reduction reaction can be written as:
Pb2+ (aq) + 2e- --> Pb (s) (2)
Now, let's discuss the location of each reaction. In a galvanic cell, oxidation typically occurs at the anode (negative electrode), and reduction occurs at the cathode (positive electrode). Therefore, equation (1) representing the oxidation reaction would occur at the anode, while equation (2) representing the reduction reaction would occur at the cathode.
To determine the cell potential, we can make use of standard reduction potential tables. The standard reduction potential (E°) measures the tendency of a species to gain electrons and tells us whether the reaction is spontaneous or not.
Looking up the standard reduction potentials for Al3+ and Pb2+ in a table, we find that:
E°(Al3+ / Al) = -1.66 V
E°(Pb2+ / Pb) = -0.13 V
Since the cell potential (Ecell) is the difference between the reduction potentials at the two electrodes, we can calculate it by subtracting the reduction potential at the anode from that at the cathode:
Ecell = E°(Pb2+ / Pb) - E°(Al3+ / Al)
= -0.13 V - (-1.66 V)
= -0.13 V + 1.66 V
= 1.53 V
Therefore, the cell potential for the galvanic cell between Al3+ and Pb2+ solutions would be 1.53 volts.