The measured potential of the following cell was 0.1776 volts:

Pb|Pb2+(??M)||Pb2+(1.0M)|Pb
Calculate the electrode potential of the unknown half-cell(anode).

Not sure how to approach this, considering the molarity is unknown. There isn't a lot of space on the page for the question, so I'm guessing it's a relatively simple solution.

Ecell = Eocell + (0.0592/#e)log(Pb^2+)/(Pb^2+)

Ecell = 0.1776
Eocell= I don't have my text handy but I think this is -0.126. You should verify that.
#e = 2
For the log term, the top number there is 1M and the lower number is x. You solve for x

To calculate the electrode potential of the unknown half-cell (anode), we can use the Nernst equation:

E = E° - (0.0592/n) * log(Q)

In this equation:
- E is the electrode potential of the unknown half-cell.
- E° is the standard electrode potential of the cell, which is given as 0.1776 V.
- 0.0592 is the value of the natural logarithm of 10 divided by the number of electrons transferred in the cell reaction (n). Since we don't have the complete reaction, we'll assume that 2 electrons are transferred.
- Q is the reaction quotient, which can be calculated using the concentrations of the species involved in the half-cell reaction.

Given that our half-cell reaction is:

Pb2+(?? M) + 2e- ⟶ Pb

We can see that the concentration of Pb is 1.0 M. Since we need to calculate the concentration of Pb2+ in the unknown half-cell, let's assume the concentration is x M.

Now, we can substitute the values into the Nernst equation:

0.1776 = E° - (0.0592/2) * log(x/(1.0^2)) [since the stoichiometry is 2:1]

Simplifying the equation:

0.1776 = E° - 0.0296 * log(x)

Next, let's isolate the term E°:

E° = 0.1776 + 0.0296 * log(x)

Now, you can calculate the electrode potential of the unknown half-cell (anode) by substituting the concentration value (x) and solving for E°.

To calculate the electrode potential of the unknown half-cell (anode), we can use the Nernst equation:

E = E° - (0.0592 V / n) * log(Q)

Where:
- E is the measured potential of the cell (0.1776 V in this case)
- E° is the standard electrode potential
- n is the number of moles of electrons transferred in the balanced redox reaction
- Q is the reaction quotient, which is calculated using the concentrations of the species involved in the half-reaction

In this case, the balanced redox reaction at the anode is:

Pb(s) → Pb2+(?? M) + 2e-

First, we need to determine the number of moles of electrons transferred (n). Since we have 2 electrons in the balanced redox reaction, n is equal to 2.

Now, let's calculate the reaction quotient (Q) using the concentrations of the species involved. Since the molarity of the Pb2+ solution is unknown, we'll represent it as x:

Q = [Pb2+(?? M)] / [Pb2+(1.0 M)]

Next, substitute the values into the Nernst equation and we get:

0.1776 V = E° - (0.0592 V / 2) * log([Pb2+(?? M)] / [Pb2+(1.0 M)])

Simplify the equation by multiplying and dividing by 2:

0.3552 V = 2E° - 0.0592 V * log([Pb2+(?? M)] / [Pb2+(1.0 M)])

Now, we need to determine the standard electrode potential (E°) for the half-reaction. You can refer to a standard electrode potential table or a chemistry textbook to find it. For the half-reaction Pb2+ + 2e- → Pb, E° = -0.13 V.

Substitute this value into the equation:

0.3552 V = 2(-0.13 V) - 0.0592 V * log([Pb2+(?? M)] / [Pb2+(1.0 M)])

Now, you can solve the equation for [Pb2+(?? M)] by rearranging the equation and using logarithmic properties. However, since you mentioned there isn't much space, I'm unable to continue solving the equation for you without knowing the actual values of the molar concentrations.