Final answer:
The cell potential for the given reaction is calculated using the Nernst equation, which takes into account the standard cell potential and the reaction quotient. In this case, the standard cell potential is negative, indicating that the reaction is not spontaneous under standard conditions. The calculated cell potential is the final answer.
Step-by-step explanation:
To calculate the cell potential for the given reaction, we need to use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
First, let's identify the half-reactions:
Anode (oxidation): Cu(s) → Cu²+(aq, 4.48 M) + 2e-
Cathode (reduction): Cu²+(aq, 0.0032 M) + 2e- → Cu(s)
The reaction is spontaneous when the cell potential (Ecell) is positive. Let's calculate the standard cell potential (E°cell) using the standard reduction potentials for the half-reactions:
E°cell = Er(cathode) - Er(anode)
Er(cathode) = 0.00 V (standard reduction potential of Cu²+(aq) + 2e- → Cu(s))
Er(anode) = +0.339 V (standard reduction potential of Cu(s) → Cu²+(aq) + 2e-)
E°cell = 0.00 V - 0.339 V = -0.339 V
Next, let's calculate the reaction quotient (Q):
Q = [Cu²+(aq, 4.48 M)] / [Cu²+(aq, 0.0032 M)]
Now we can substitute the values into the Nernst equation to find the cell potential (Ecell):
Ecell = -0.339 V - (0.0592/2) * log([Cu²+(aq, 4.48 M)] / [Cu²+(aq, 0.0032 M)])