對於Cu系統
Cu2+ +2e-<==>Cu
equilibrium K = 1/[Cu2+]
由數學所知, ln1=0
foX= 1/[Cu2+] >1
ln X >0
so 0.02M 果個個E' 係細過standard condition 果個
所以佢會做anode(negative pole), loss of e-
係cathode(+ve) gain of e- of Cu2+ ->
我係CE人 AR ...
AL人 D野真係勁.. 完成唔明講緊咩 0== [[bao_29]]
答返問題
我係咁諗JA..
題目話左 In the setup, the electron flow in such a direction that the conc. of the Cu2+ ion in the two beaker become the same.
咁即係 X 放d Cu2+落水
Y吃左d Cu2+from水
咁先可以balance到個concentration