A concentration cell based on the following half reaction at 317 K

(Cu^2+) + (2e-) --> (Cu) SRP = 0.34 V

Has initial concentrations of 1.33 M Cu2+, 0.393 M Cu2+, and a potential of 0.01665 V at these conditions. After 4.1 hours the new potential of the cell is found to be 0.004008 V. What is the concentration of Cu2+ at the cathode at this new potential?

Please help

To find the concentration of Cu2+ at the cathode at the new potential, we need to use the Nernst equation:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:
- Ecell is the observed cell potential
- E°cell is the standard cell potential (SRP)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (317 K)
- n is the number of electrons transferred in the balanced equation (in this case, n = 2)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, calculated from the concentrations of the Cu2+ at the anode and cathode

First, let's find E°cell:
E°cell = 0.34 V (Given)

Next, let's find Q at the initial conditions:
Q = [Cu2+]cathode / [Cu2+]anode
= (1.33 M) / (0.393 M)
= 3.38

Now, let's solve for the ln(Q) term in the Nernst equation:

ln(Q) = ln(3.38) = 1.216

Now we can find Ecell at the initial conditions:

Ecell = E°cell - (RT/nF) * ln(Q)
= 0.34 V - [(8.314 J/(mol·K)) * (317 K)] / [(2 mol)(96,485 C/mol)] * 1.216
= 0.003907 V

Now, let's solve for Q at the new potential:
Ecell = E°cell - (RT/nF) * ln(Q)

Re-arranging the equation to solve for Q:

ln(Q) = (E°cell - Ecell) * (nF/RT)
Q = e^[(E°cell - Ecell) * (nF/RT)]

Plugging in the values:

Q = e^[(0.34 V - 0.004008 V) * (2 * 96,485 C/mol) / (8.314 J/(mol·K) * 317 K)]
= e^[-0.3358]

Using a scientific calculator, e^[-0.3358] is approximately 0.715

Now we can find the concentration of Cu2+ at the cathode at the new potential by rearranging the Q equation:

Q = [Cu2+]cathode / [Cu2+]anode

Since we know that [Cu2+]anode is 0.393 M, we can solve for [Cu2+]cathode:

0.715 = [Cu2+]cathode / 0.393 M

Solving for [Cu2+]cathode:

[Cu2+]cathode = 0.715 * 0.393 M
= 0.281 M

Therefore, the concentration of Cu2+ at the cathode at the new potential is 0.281 M.