Potential of a cell is 0.1776 volts

Pb|Pb2+(??M)||Pb2+(1.0M)|Pb

I know i have to use Nernst's equation .. i just have no clue how to attempt this :S

This is a concentration cell since the electrodes are the same but the molarities are different (we assume they are anyway).

Ecell = Eo-(0.0592/n)log Q
I would plug in 0.1776 for Ecell. Eo is always zero in a concn cell, n = 2 for this cell and solve for Q.
Q = (x/1M) and solve for x.

To find the concentration of Pb2+ at the cathode in the given cell, you can use the Nernst equation. The Nernst equation relates the potential of the cell (Ecell) to the standard potential (E°cell) and the concentrations of the species involved.

The Nernst equation is given as follows:

Ecell = E°cell - (RT / nF) * ln(Q)

Where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the ideal gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
n is the number of electrons involved in the balanced redox equation
F is the Faraday constant (96,485 C/mol)
ln(Q) is the natural logarithm of the reaction quotient (Q)

In our case, the cell is represented as follows:

Pb|Pb2+(??M)||Pb2+(1.0M)|Pb

The anode is represented by Pb|Pb2+(??M), where the concentration of Pb2+ is unknown. The cathode is represented by Pb2+(1.0M)|Pb.

First, you need to determine the reaction taking place at the electrodes. In this case, the reaction is:

Pb2+(??M) + 2e- -> Pb(Anode)
Pb2+(1.0M) + 2e- -> Pb(Cathode)

The number of electrons involved in both reactions is 2 (n = 2).

Next, substitute the known values into the Nernst equation:

Ecell = E°cell - (RT / nF) * ln(Q)

Given:
Ecell = 0.1776 V (potential of the cell)
R = 8.314 J/mol·K
T = temperature in Kelvin (should be specified in the question)
n = 2
F = 96,485 C/mol

Now you need to calculate Q, the reaction quotient. Q can be calculated using the concentrations of the species involved in the reaction.

Q = [Pb(Anode)] / [Pb(Cathode)]

In our case, [Pb(Anode)] is unknown and [Pb(Cathode)] is 1.0 M.

Now, you can substitute the known values into the Nernst equation and solve for [Pb(Anode)]:

0.1776 = E°cell - (RT / (2F)) * ln([Pb(Anode)] / 1.0)

Solve the equation for [Pb(Anode)], taking care to use the correct values for R, T, and F, as well as the sign conventions for the logarithm function.

NOTE: The answer will depend on the temperature since both E°cell and the gas constant R have temperature dependencies. Please provide the temperature so that the complete calculation can be done accurately.

To calculate the cell potential using Nernst's equation, you need to know the standard cell potential (E°) and the concentrations of the ions involved in the cell reaction. In this case, you have given the potential of the cell (0.1776 volts) and the concentrations of Pb2+ ions in both half-cells.

Nernst's equation is given as:

E = E° - (0.0592/n) * log(Q)

Where:
E = Cell potential
E° = Standard cell potential
n = Number of moles of electrons transferred in the balanced cell reaction
Q = Reaction quotient

First, let's identify the balanced cell reaction:

Pb + Pb2+ → Pb2+ + Pb

From the reaction, we can observe that two electrons (2e-) are transferred. Therefore, n = 2.

Next, we need to calculate the reaction quotient, Q. The reaction quotient is calculated using the concentrations of the participating species raised to the power of their stoichiometric coefficients in the balanced reaction.

Q = [Pb2+] / [Pb2+]

Given that the concentration of Pb2+ in the anode half-cell (left side) is ? M (unknown) and the concentration of Pb2+ in the cathode half-cell (right side) is 1.0 M, we can express Q as:

Q = [? M] / [1.0 M]

Since you have not provided the concentration of Pb2+ in the anode half-cell (? M), we cannot directly calculate the reaction quotient or determine the cell potential at this point.

you don't make sense