Can someone help me with this question? I'm having my A2 this friday and would appreciate any help given. For part (i) why is the Ecell calculated as 0.40-(-0.83)= +1.23V? Based on the data booklet, 2H2O +2e ----> H2 + 2OH : E= -0.83V. isn't the electrode reaction for cathode change sign to +0.83V since it's reversed?
Turn on thread page Beta
- Thread Starter
- 17-10-2017 15:27
- 17-10-2017 21:43
No, you keep the standard potential as the sign it's given, it will become positive in the equation lhs-rhs = E