need help please

Question

Calculate the potential of a concentration cell that uses the reaction below at 298 K using the Nernst equation. Enter only a numerical result. To enter a power of 10 such as 3.2 x 10^-5 use the format 3.2E-5

Cu2+(aq, 0.02 M) + Cu(s) → Cu2+(aq, 0.01 M) + Cu(s)

i tried so many times but ANS keep saying wrong

Answer 1

Nernst: E = Eº - RT/nF ln Q

Since this is at 298K, the RT/nF ln Q converts to 0.0592/n log Q

E = Eº - 0.0592/n log 0.01/0.02 (Q = [products]/[reactants]

E = 0 - 0.0592/2 log 0.5 (Eº for a concentration/electrolytic cell is zero)

E = -0.0296 x -0.302

E = 0.008910

E = 8.9E-3 V

Answer 2

Short answer:

E = Eº - (0.0592V/n) * logQ

= 0 - 0.0592V/2 * log(0.01M / 0.02M)

= 0.00891048787 V

Detailed answer:

Since the temperature is 298K, we can use the following form of the Nernst Equation:

E = Eº - (0.0592V/n) * logQ where E is the cell potential, Eº is the standard cell potential, Q is the reaction quotient, and n is the number of moles of electrons transferred in the reaction.

In a concentration cell, Eº = 0V. n = 2 since 2 moles of electrons are transferred in this reaction. Also, Q = 0.01M/0.02M = 0.5.

Hence, E = Eº - (0.0592V/n) * logQ = 0 - 0.0592V/2 * log(0.5) = 0.00891048787 V

Also, you might find the following KhanAcademy video (about ten minutes long) on concentration cells helpful: https://www.khanacademy.org/science/chemistry/oxidation-reduction/cell-potentials-under-nonstandard-conditions/v/concentration-cell

Hope this helps!