If the ratio of composition of oxidized and reduced species in electrochemical cell, is given as \(\frac{[O]}{[R]}\) = e2, the correct potential difference will be
1
E - E0 = \(\frac{2RT}{nF}\)
2
E - E0 = - \(\frac{2RT}{nF}\)
3
E - E0 = \(\frac{RT}{nF}\)
4
E - E0 = - \(\frac{RT}{nF}\)