Anode Cathode Chemistry Question?

Question

The question I am working on is as follows:

Write balanced half-reaction equations for the processes that occur at the anode and cathode when an aqueous solution of potassium chloride is electrolyzed. Identify the oxidizing agent and reducing agent. Calculate the number of kilowatt-hours of energy required to produce 2.00 kg of chlorine gas if the voltage applied is 4.8 V and the current is 3.5 x 105 amps. (1 kwh = 3.6 x 106 J)

I know that I am going to have the half reactions for Chlorine and water..but I am not sure by what it means when the question states that potassium chloride is "electrolyzed." does this mean that the Chlorine half reaction is for the cathode? And thus, the Chlornie will act as the oxidizing agent since it will be reduced? Argh I'm so confused! Can someone please help me? Thanks in advance :D

Answer 1

Cathode: 2H2O(l) + 2e- --> H2(g) + 2OH-(aq)

Anode: 2Cl-(aq) --> Cl2(g) + 2e-

Oxydation : anode
Reduction : cathode

oxydating agent: H2O
reducting agent : Cl-
Complete reactions:

Solution : 2KCl(aq) --> 2K+(aq) + 2Cl-(aq)
Cathode: 2H2O(l) + 2e- --> H2(g) + 2OH-(aq)
Anode: 2Cl-(aq) --> Cl2(g) + 2e-
--------------------- --------------------- ----------------- -------------------- --

2KCl + 2H2O --> 2KOH + H2 + Cl2

I can't do the rest because u didn't provide the amount of the moles.

Answer 2

electrolyis is forcing a non-spontaneous (negative E-cell) reaction to occur by applying electrical energy:

2 H2O(l) + 2e– → H2(g) + 2 OH–(aq), E0 =−0.83V

Cl2(g) + 2e− → 2Cl−(aq), E0 = +1.36V

since the reduction potential for Cl2 is higher, if the components were all mixed together it would be reduced - but you want to produce Cl2. the overall reaction for this is

2 H2O(l) + 2Cl−(aq) → Cl2(g) + H2(g) + 2 OH–(aq), E0 = -2.19V

here water is being reduced (so it's the oxidising agent), and chloride ion (NOT chlorine!) is being oxidised (so it's the reducing agent).

For the second part you need to find the number of moles of chlorine produced, and thus the number of moles of electrons that must be passed through the solution to produce that amount of chlorine. the current is in amps (coulombs / second), and you can find out Faraday's constant, the charge of a mole of electrons (coulombs / mole). From that and the number of moles of electrons calculated earlier you can find how long you need to run the current for. then you can use E = V*I*t to find the amount of electrical energy it corresponds to.

edit: steve_geo1 is correct i think, the terms 'oxidisng agent' and 'reducing agent' really only make proper sense when the reaction is spontaneous. when electrical energy is being provided, as here, that's the proper cause of the reaction, not the properties of the chemicals.

Answer 3

Anode: 2Cl- ===> Cl2(g) + 2e- Oxidation always occurs at the anode

Cathode: 2e- + 2H2O ===> H2(g) + 2OH- By a process of elimination, reduction occurs at the cathode.

The K+ is a spectator ion. It is unaffected by all these goings on. Cl- is oxidized. H+ is reduced. By "electrolysis" is meant that electrons are supplied from the outside to reduce. So electrons are the reducing agent. There is really no oxidizing agent.

For the rest of it, you have to work it out in moles of electrons. 1 Faraday = 1 mole e-'s = 96,500 coulombs.