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If the concentration of sn2+ in the cathode compartment is 1.50 m and the cell generates an emf of 0.24 v , what is the concentration of pb2+ in the anode compartment?

User McGrady
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2 Answers

5 votes

Final answer:

To determine the concentration of Pb2+ in the anode compartment, we can use the Nernst equation. By rearranging the Nernst equation and plugging in the given values, we can calculate the concentration of Pb2+ in the anode compartment.

Step-by-step explanation:

To determine the concentration of Pb2+ in the anode compartment, we need to use the Nernst equation. The Nernst equation relates the cell potential to the concentration of ions in each compartment of the cell. It is given by: Ecell = E°cell - (0.0592/n) × log([Pb2+]/[Sn2+]), where Ecell is the cell potential, E°cell is the standard cell potential, [Pb2+] is the concentration of Pb2+ in the anode compartment, [Sn2+] is the concentration of Sn2+ in the cathode compartment, and n is the number of electrons transferred in the balanced redox reaction.

In this case, the cell potential is 0.24 V, [Sn2+] is 1.50 M, and n is 2 (since it is a Sn2+/Pb2+ redox reaction). Therefore, the equation becomes: 0.24 = E°cell - (0.0592/2) × log([Pb2+]/1.50). We can rearrange this equation to solve for [Pb2+]: [Pb2+] = (10^((E°cell - 0.24) × 2 / 0.0592)) × 1.50.

User Kostmo
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5 votes

Answer:

4.8*10^-8M

Step-by-step explanation:

If the concentration of sn2+ in the cathode compartment is 1.50 m and the cell generates-example-1
User Wayneio
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