34) and we have
H2O (l ) + e– ® ½H2(g) + OH–
(2 35)
At the anode the following oxidation reactions are possible:
Cl– (aq) ® ½ Cl2 (g) + e–
(
)
o
Ecell
 = 1 36 V
(2 36)
2H2O (l ) ® O2 (g) + 4H+(aq) + 4e–
(
)
o
Ecell
 = 1