Redox reaction Zn(s) + Cu²⁺ (0.1M) → Zn²⁺ (1M) + Cu(s) takes place in a cell with E^o_cell = 1.1V then what is E_cell for the cell at 298K?  

1.07 V

E_cell = E^o_cell - \(\frac{0.059}{2}\)log\(\frac{[Zn^{2+}]}{[Cu^{2+}]}\)

E_cell = 1.1 - \(\frac{0.059}{2}\)log\(\frac{[1]}{[0.1]}\)  

E_cell = 1.1 - 0.0295 log10             [∴log10 = 1]

E_cell = 1.1 - 0.0295 = 1.0705 V