Calculate the voltage of the following cell: Zn | Zn^{2+} (0.14 M ) || Cu^{2+} (0.30 M) | Cu?

Refer to: http://www2.ucdsb.on.ca/tiss/stretton/database/Standard_Reduction_Potentials.htm
Zn²⁺(aq) + 2e⁻ → Zn(s) … E° = -0.76 V
Cu²⁺(aq) + 2e⁻ → Cu(s) … E° = +0.34 V

Zn | Zn²⁺(0.14 M ) || Cu²⁺(0.30 M) | Cu
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) … E°(cell)
E°(cell) = E°(red) - E°(oxid) = (+0.34) - (-0.76) = 1.10 V

Nernst equation:
Voltage of the cell, E(cell)
= E° - [RT/(nF)] ln([Zn²⁺]/[Cu²⁺])
= 1.10 - [8.314 × 298 / 2 × 96485] ln(0.14/0.30) V
= 1.11 V