Question

Determine the molarity (in M) of the lead ion in the following cell using the reduction potentials given below: Assume the temperature is 298 K

Pb (s) | Pb²⁺ (aq, [Pb²⁺]^2) |= ?) | Ni²⁺ (aq, [Ni²⁺] = 0.27 M) | Ni (s) E = -0.0200 V

Pb²⁺ + 2e^- → Pb E⁰ = -0.13 V
Ni²⁺ + 2e^- → Ni E⁰ = -0.25 V

Answer 1

The molarity of the lead ion ([Pb^2+ ]) in the cell is 0.81M.

The molarity of the lead ion ([Pb^2+]) in the cell can be determined by using the Nernst equation, which relates the standard cell potential (E∘ ), the actual cell potential (E), the reaction quotient (Q), the gas constant (R), and the temperature (T). The Nernst equation is given by:

E=E∘− RT/nFln(Q)

In this case, the cell involves the reduction of lead ion (Pb^2+ ) and nickel ion (Ni^2+ ). The overall cell reaction can be written as:

Pb(s)+Ni2+ (aq)→Pb2+ (aq)+Ni(s)

The standard cell potential (E∘ ) is the difference between the standard reduction potentials (E∘ =E∘ cathode −E∘ anode)

Given that E∘ =(−0.13)−(−0.25)=0.12V

The Nernst equation is then modified to:

E=0.12− 0.0257/2ln( [Pb^2+]/[Ni^2+])

Given that [Ni^2+]=0.27M and E=−0.0200V, we can solve for [Pb2+ ]. After solving the equation, the molarity of the lead ion is found to be 0.81M.