August 09, 2020, 01:29:04 PM
Forum Rules: Read This Before Posting


Topic: Preparation of Cl2 (exam question)  (Read 7354 times)

0 Members and 1 Guest are viewing this topic.

Offline Aegiss

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Preparation of Cl2 (exam question)
« on: August 17, 2009, 07:08:42 AM »
I'm retaking my chemistry exam next friday. And I'm still having trouble solving a question from the previous exam.
I've translated it, so some things might not sound quite right. If anything is unclear, feel free to ask and I'll try to explain it a bit better.

On with the question:
A laboratory assitant gets the assignment to prepare Cl2.
Experiment1: To a 300ml solvent of potassium dichromate (0.15 mol) and chrome(III)nitrate (0.015 mol) with pH = 0 (acidulated by addition of nitric acid), 0.03 mol of salt(NaCl) is added.
Experiment2: To a 300ml solvent of potassium dichromate (0.15 mol) and chrome(III)nitrate (0.015 mol) with pH = 0 (acidulated by addition of nitric acid), 2.4 mol of salt(NaCl) is added.

E0=dichromate/chrome(III) = 1.33V
E0=Chloridegas(Cl2; 1atm)/salt = 1.36V

a) Work out the upper reaction completely. (oxidation state, oxidation, reduction, oxidizing agent, reducing agent, work out partial reactions)
b) Calculate the Equilibrium constant of the reaction.
c) Give the schematic representation of the matching galvanic cell and explain.
d) Explain if the goal is reached in both experiments using calculations.


Any help is welcome.. I'm mostly puzzled by the first question. But feel free to answer any of the others aswell.
What I've come up with so far is:
Oxidation - Cr2O72- + 14H+ + 12e- <--> 2Cr + 7H2O
Reduction - Cl2 + 2Na++2e- <--> 2NaCl
Total reaction - K2Cr2O7 + Cr(NO3)3 + HNO3 + 4NaCl + 3H2O <--> 2KCrO4 + 3H+ + Cr(OH)3 + 4NaNO3 + 2Cl2

I have no clue if any of this is right though. The oxidation/reduction aren't even included in the total reaction, so I'm assuming either of them, if not both, are wrong.

Also, sorry if this is in the wrong section. I didn't really know where to put it. Since the course is named general and inorganic chemistry I placed it here.

Thanks in advance for any replies!

Offline BluRay

  • Full Member
  • ****
  • Posts: 154
  • Mole Snacks: +9/-2
Re: Preparation of Cl2 (exam question)
« Reply #1 on: August 19, 2009, 04:23:16 PM »
CrVI in the chromate oxides Cl- to Cl2 becoming CrIII nitrate.

Offline Aegiss

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Preparation of Cl2 (exam question)
« Reply #2 on: August 20, 2009, 05:16:06 AM »
CrVI in the chromate oxides Cl- to Cl2 becoming CrIII nitrate.

Best I could get from that is: K2Cr2O7 + 3HNO3 <--> K2CrO4+3H++Cr(NO3)3

But that still doesn't involve the Cl2. I figure NaCl has to react with one of the other molecules, but I have no clue which.

Offline BluRay

  • Full Member
  • ****
  • Posts: 154
  • Mole Snacks: +9/-2
Re: Preparation of Cl2 (exam question)
« Reply #3 on: August 20, 2009, 06:30:28 AM »
Cl- of course come from NaCl...

Offline Aegiss

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Preparation of Cl2 (exam question)
« Reply #4 on: August 20, 2009, 07:13:32 AM »
Cl- of course come from NaCl...

I know that much... It just doesn't really add up. The reaction below leaves me with 2 positively charged ions, which I find... strange. And the fact that chrome(III)nitrate(the one present in the solvent before addition of NaCl) isn't used in the reaction.

K2Cr2O7 + 3HNO3 + 2NaCl <--> K2CrO4 + Cr(NO3)3 + Cl2 + 3H+ + 2Na+

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Preparation of Cl2 (exam question)
« Reply #5 on: August 20, 2009, 07:17:05 AM »
Oxidation - Cr2O72- + 14H+ + 12e- <--> 2Cr + 7H2O
This isn't oxidation, this is reduction as well. Cr3+ should form instead of Cr(s) I believe.

Offline Aegiss

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Preparation of Cl2 (exam question)
« Reply #6 on: August 20, 2009, 07:24:53 AM »
Oxidation - Cr2O72- + 14H+ + 12e- <--> 2Cr + 7H2O
This isn't oxidation, this is reduction as well. Cr3+ should form instead of Cr(s) I believe.

You're right. BluRay already pointed that out. The dichromate will form Cr(III)nitrate using nitric acid.
It does however still leave me wondering about all the positively charged ions..

Edit: Rebalanced the equation. Not too sure if it's right though...
K2Cr2O7 + 6HNO3 + 6NaCl <--> 2Cr(NO3)3 + 3Cl2 + K2O + 3H2O + 3Na2O

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Preparation of Cl2 (exam question)
« Reply #7 on: August 20, 2009, 07:32:43 AM »
K2Cr2O7 + 3HNO3 + 2NaCl <--> K2CrO4 + Cr(NO3)3 + Cl2 + 3H+ + 2Na+


This equation is not balanced in terms of the charges on each side. IMHO, you should leave out the spectator ions.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 25879
  • Mole Snacks: +1692/-401
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Preparation of Cl2 (exam question)
« Reply #8 on: August 20, 2009, 10:02:25 AM »
Edit: Rebalanced the equation. Not too sure if it's right though...
K2Cr2O7 + 6HNO3 + 6NaCl <--> 2Cr(NO3)3 + 3Cl2 + K2O + 3H2O + 3Na2O

K2O & Na2O in water?

Go net ionic, remove all spectators - the equation will be much easier to deal with.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Aegiss

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Preparation of Cl2 (exam question)
« Reply #9 on: August 21, 2009, 01:06:49 PM »
Alright, thanks for the help.

Exam's over, so you can just close/delete the thread.

Sponsored Links