Question #45017

A concentration cell is constructed at $298 \text{ K}$ using copper electrodes dipping into two aqueous solutions of copper(II) sulfate ($\ce{CuSO4}$):

$\ce{Cu(s) | CuSO4 (aq, c_1) || CuSO4 (aq, c_2) | Cu(s)}$

Due to significant ion pairing at higher concentrations, the salt does not behave ideally. The Van't Hoff factor ($i$) for the left half-cell ($c_1 = 0.05 \text{ M}$) is experimentally determined to be $1.80$, while for the right half-cell ($c_2 = 0.40 \text{ M}$) it is $1.40$.

Assuming the effective concentration of $\ce{Cu^{2+}}$ ions is determined by the degree of dissociation ($\alpha$) derived from the Van't Hoff factor, calculate the cell potential ($E_{\text{cell}}$) in millivolts (mV).

(Given: $\frac{2.303 RT}{F} = 0.06 \text{ V}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 7 = 0.84$)