A voltaic cell consists of a Pb /Pb^2+ half-cell and a Cu/Cu^2 half-cell at 298K. The initial concentrations of Pb^2+ and Cu^2 are 5.20×10−2 M and 1.40 M, respectively.

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To determine the potential of the voltaic cell, you need to know the standard reduction potentials for the reactions occurring at each electrode. The standard reduction potential (E°) values can be found in a reference table or in a chemistry textbook.

The half-reactions for the Pb/Pb^2+ and Cu/Cu^2+ cells are as follows:

Pb + 2e^- -> Pb^2+ (Reduction half-reaction for the Pb/Pb^2+ half-cell)
Cu^2+ + 2e^- -> Cu (Reduction half-reaction for the Cu/Cu^2+ half-cell)

Now, let's find the overall cell reaction by combining the two half-reactions:

Pb + 2e^- -> Pb^2+ (Oxidation half-reaction for the Pb/Pb^2+ half-cell)
Cu^2+ + 2e^- -> Cu (Reduction half-reaction for the Cu/Cu^2+ half-cell)

By adding these two half-reactions, we get the overall cell reaction:

Pb + Cu^2+ -> Pb^2+ + Cu

Now, use the Nernst equation to calculate the cell potential at 298K:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298K)
n is the number of electrons transferred in the balanced cell reaction
F is the Faraday constant (96485 C/mol)
Q is the reaction quotient, which can be determined using the concentrations of the species in the cell.

In this case, n = 2 because 2 electrons are transferred in the balanced cell reaction. We can substitute the values into the equation and solve for Ecell.

Note: To calculate Q, you need to take into account the stoichiometry of the balanced cell reaction and the concentrations of products and reactants.

Q = [Pb^2+] / [Cu^2+]

Now substitute the given initial concentrations:
[Pb^2+] = 5.20×10^-2 M
[Cu^2+] = 1.40 M

Calculate Q by substituting these concentration values into the equation.

Now you can calculate the cell potential (Ecell) by substituting the values into the Nernst equation.