2 H2O + 2e- -----> 2OH- + H2
Eo cell = -0.83 v at 25 C
How do I calculate reduction potential at different pH's?
pH = 7
H+ = 1.0x10-7
OH-=1.0x10-7
Q=(2OH)2(H2)
E Cell = -0.83v - 0.0257/2 x In [(2*1.0x10-7)2(1.0x10-7)]
This doesn't give me the right answer. How is this done?