Chem electrolysis

2008-03-30 3:18 am
我想問點解electrolysis of dilute NaCl soultion
around the cathode, H+ ions are discharged to form H2 gas 之後就會有excess amount of OH- ions near the cathode?

同樣點解 OH- ions are discharged to form O2之後又會有excess amount of H+ ions near the anode 呢?

回答 (2)

2008-03-30 7:43 am
✔ 最佳答案
In NaCl solution, NaCl gives Na+ and Cl- ions, and H2O is slightly ionized to give H+ and OH- ions. In the solutions, there are two types of cation (Na+ and H+) and two types of anion (Cl- and OH-).

When the NaCl solution is electrolysed, Na+ and H+ ions are attracted by the cathode. H+ ions are preferentially discharged because H+ ions are reduced more readily than Na+ ions.
2H+ + 2e- → H2
At the same time, Cl- and OH- ions are attracted by the anode. Cl- ions are preferentially discharged because Cl- ions are more concentrated than OH- ions.
2Cl- → Cl2 + 2e-

Before the electrolysis, the concentration and H+ ions and the concentration of OH- ions are equal. During the electrolysis, H+ ions are discharged (reacted) at the cathode, but OH- ions do not change at the anode. In other words, there is an excess amount of OH- ion left in the solution.


Consider CuSO4 solution. CuSO4 gives Cu2+ and SO42- ions, and H2O is slightly ionized to give H+ and OH- ions. In the solutions, there are two types of cation (Cu2+ and H+) and two types of anion (SO42- and OH-).

When the CuSO4 solution is electrolysed, Cu2+ and H+ ions are attracted by the cathode. Cu2+ ions are preferentially discharged because the Cu2+ ions are more concentrated than H+ ions.
Cu2+ + 2e- → Cu
At the same time, SO42- and OH- ions are attracted by the anode. OH- ions are preferentially discharged because OH- ions are oxidized more readily than SO42- ions.
4OH- → 2H2O + O2 + 4e-

Before the electrolysis, the concentration and H+ ions and the concentration of OH- ions are equal. During the electrolysis, OH- ions are discharged (reacted) at the anode, but H+ ions do not change at the cathode. In other words, there is an excess amount of H+ ion left in the solution.
2008-03-30 5:19 am
first of all, you should know that water is slightly ionized
into hydrogen ion and hydroxide ion.
around the cathode, the H+, which come from the dissociation of
water, will be discharged. so OH-will leave near the cathode.
at the anode, the reverse situation occurs.
OH- is discharged. so H+ is left near the anode.


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