[LHM]

Given the standard electrode potentials:

Cu²⁺+2e^- Cu(s) E^o=0.337V
Al³⁺+3e^- Al(s)   E°=-1.66V

The voltage of the cell Cu°|CuSO₄(0.5M)||Al₂(SO₄)₃|Al° would be
A) 0.00 V
B) 0.72 V
C) 2.01 V
D) 5.70 V

I tried doing E°=-0.337V-1.66V=-1.997V, and then: E=-1.997-8.31447*298.15/(96485*6)*ln(0.5³/0.1²)=-2.01 V. The answer is D though, what did I do wrong?

[Borek]

Even without switching calculator on C is the only viable answer. As you have calculated, standard potential for the cell is close to 2V. The potential difference due to dilution is around 60 mV per tenfold change in concentration. Standard solution is about 1M. You have two solutions - concentration of neither is more than 10 times larger/smaller than 1M, so the formal potential can't differ from the original one by more than 120 mV. Only C fits that range.

D is completely off.