Quick Chem Question

[mattshen]

Hey, this question is from the 2010 Insight paper that im not too sure and hope you guys can clarify

Question 5
The nickel–cadmium cell is a rechargeable cell used in mobile phones, laptop computers and many other portable electronic devices.
The two relevant half-equations for the discharge reaction are
Cd(s) + 2OH–(aq) → Cd(OH)2(s) + 2e–
NiOOH(s) + H2O(l) + e– → Ni(OH)2(s) + OH–(aq)
a. Will the pH in the region immediately surrounding the anode increase or decrease as the cell discharges? Give a reason for your answer.

The answer says that pH will decrease because [OH] decreases
I'm not too sure but i thought pH was a measure for H+ ions, and since no H+ ions are involved, the pH should remain the same. :/
Thanks,

[vea]

[OH-][H+]=Kw (Kw at 25 degrees Celsius is 10^-14)
[H+]=Kw/[OH-]
So [H+] and [OH-] are inversely proportional.
When [OH-] decreases, [H+] increases, making the pH lower.

[mattshen]

hmmmmm okay
my friend suggested something that made more sense
Considering that the reaction is H2O <====> H+ + OH- 
When there is a decrease in OH- the system will counteract the decrease of OH by favouring a forward reaction and hence more H+ and OH- has increased (however H+ is still greater than OH-) resulting in a decrease in pH
You reckon this is correct?
Thanks

[b^3]

pH isn't related to the ratio of the two to each other (well it is, but we can only calculate this at 25 C). Your are partially correct. As OH^- decreases a forward reaction will be favoured, so more H₃O⁺ is produced. This means the [H₃O⁺] increases and as pH=-log([H₃O⁺]), ph will decrease. It is not the ratio that determines the pH, it is the concentration of H₃O⁺. I.e. this isn't correct ->

(however H+ is still greater than OH-) resulting in a decrease in pH

[mattshen]

ahhh yep thank you for correcting me