# A hard electrochemistry question (in my opinion)

1. Mar 29, 2005

### chem_tr

Hello all,

As I am preparing for my PhD proficiency exam, I tried to solve some Analytical chemistry problems. and stuck in one.

Well, I have attempted to solve this, but have not found enough evidence to study on and solve the problem. What I've done is basicly the following:

I am in need of recommendations, thank you.

2. Mar 29, 2005

### GCT

Actually its quite simple...unfortunately I don't remember exactly how to do it off the top of my head. I'll get back to it later, I'm a bit busy at the moment. The method is usually introduced at the end of the electrochemistry chapter in general chemistry texts.

3. Mar 29, 2005

### Staff: Mentor

Or - 1A is a Coulomb/sec, you know the current, you know the charge per second. Every two electrons give one Cl2 particle. You know the charge, you know number of the Cl2 particles consumed per second. Just convert Coulombs to Avogadro number.

That's almost the same - Faraday constant is number of Coulombs in 1 mole of electrons :)

Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis

Last edited: Mar 29, 2005
4. Mar 30, 2005

### Gokul43201

Staff Emeritus
The number that comes to mind is about 96,500 somethings ( coulombs per mole ?)

5. Mar 30, 2005

### chem_tr

I know this. One faraday constant is equal to one mole of electrons - this is equal to $$6.023 \times 10^{23} \times 1.6\times 10^{-19} = 96368 \frac {Coulomb}{mol}$$

6. Mar 30, 2005

### chem_tr

Dear Borek, you said one Ampere is one coulomb per second. Then 1,000 amperes are equal to 1,000 coulombs per second. This in turn gives 6.25*1021 electrons per second, which means about 0.01 moles of electrons per second. In the redox, two electrons are used as you stated; so 0.005 moles of chlorine are consumed. This means 7.368*10-4 kilograms of chlorine gas are used.

Am I correct, pals? Thank you very much for your help, you all.

7. Mar 30, 2005

### Gokul43201

Staff Emeritus
1 mole Cl2 == 2 moles of e- = 2*6.023*10^23*1.6*10^(-19) C = 1.927*10^5 C

1 mole Cl2 == 71 g

So 71 g Cl2 == 1.927*10^5 C

Hence 10^3 C/s == 71/192.7 = 0.368 g of Cl2 consumed per second.

I think you made a typo in the very last step (everything up to that point looks good).

Note that one can only calculate the consumption RATE, since the time of use is not given.

8. Mar 30, 2005

### chem_tr

Thank you, I haven't remembered that one Ampere is equal to one Coulomb per second, so I was stuck. Again, thank you for your help.

9. Mar 30, 2005