Given the standard reduction potentials:
O2 + 4H+ + 4e- --> 2H2O E* = 1.23 V
Br2 + 2e- --> 2Br- E* = 1.08 V
2H+ + 2e- --> H2 E* = 0.00 V
Na+ + e- --> Na E* = -2.71 V
What products are formed in the electrolysis of 1 M NaBr in a solution with [H3O+] = 1 M?
(A) Na (s) and O2(g)
(B) Na (s) and Br2 (g)
(C) H2 (g) and Br (g)
(D) H2 (g) and O2 (g)
Since electrolysis, ΔG is not spontaneous and E cell has to be negative, right? So why can't it be B since adding the E of the half reaction result in -3.79 volts (non spontaneous)? (The correct above is (C))