electrolyte既問題

In the solution of NaCl, NaCl is dissociated to give Na^+(aq) and Cl^-(aq) ions, and H2O is ionized to give H^+(aq) and OH^-(aq) ions. Therefore, the solution contains Na^+(aq), H^+(aq), Cl^-(aq) and OH^-(aq).

Anode (graphite) :
The anode is connected to the positive electrode of the cell. Therefore, Cl^-(aq) and OH^-(aq) ions migrate to the anode. Cl^-(aq) ions are preferentially discharged to give Cl2(g). OH^-(aq) ions are left.
2Cl^-(aq) → Cl2(g) + 2e^-

Cathode (graphite) :
The cathode is connected to the negative electrode of the cell. Therefore, Na^+(aq) and H^+(aq) ions migrate to the cathode. H^+(aq) ions are preferentially discharged to give H2(g). Na^+(aq) ions are left.

Conclusion :
Cl2(g) is formed at the anode, and H2(g) is formed at the cathode. Na^+(aq) and OH^-(aq) are left in the solution. In other words, the solution left is NaOH(aq).

根據我既理解
你指既aqueous sodium choloride應該係溶左落水既NaCl?
而根據我學過既知識
將海水電解左之後
負極會有hydrogen,因為hydrogen帶正電荷
正極就有chloride,因為chloride帶負電荷
而剩低既溶液就係NaOH(aq)

2008-03-08 14:23:47 補充：
正極上既係chlorine gas
負極就係hydrogen gas