[Aegiss]

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 - Cr₂O₇^2- + 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!

[BluRay]

Cr^VI in the chromate oxides Cl^- to Cl₂ becoming Cr^III nitrate.

[Aegiss]

Cr^VI in the chromate oxides Cl^- to Cl₂ becoming Cr^III 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.

[BluRay]

Cl^- of course come from NaCl...

[Aegiss]

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⁺

[UG]

Oxidation - Cr₂O₇^2- + 14H⁺ + 12e^- <--> 2Cr + 7H₂O

This isn't oxidation, this is reduction as well. Cr³⁺ should form instead of Cr(s) I believe.

[Aegiss]

Oxidation - Cr₂O₇^2- + 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

[UG]

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.

[Borek]

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.

[Aegiss]

Alright, thanks for the help.

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