[g.alonci]

Hi everybody! I'm an italian student of chemistry, and I need your help Razz
In the lab the teacher reduced a solution of tetrammine palladium with formic acid. He started with a yellow solution containing ammonia and the complex and began to drop HCl. Suddenly a yellow precipitate formed and disappeared. When the precipitate became hard to redissolve, he stopped dropping HCl and started with HCOOH. The reaction was very very slow at the beginning, but after about 30 minutes every drop of the acid caused a very huge amount of gas to be formed. At the end the resulting solution was very clear and incolor with palladium metallic precipitate. Now the problem is: how to work out how many heat was involved in the reduction?? I have not found any data on the web or in my books, so I have no idea!
Hel please Sad
I tought I may do something with the reduction potential, but I do not really know how to solve this problem Sad
I don't manage neither to find potential data for the complex, apart from the generic reduction of Pd(2+) or some chloro complexes.
Another question si about the stereochemistry of diclorodiamminopalladium. The prof said that the initial precipitate was CIS. Why?? I think it shoul be trans according to the trans effect!

[darko]

I am not so sure for Pd(II) complexess, but it should be same as with Pt(II), in example treatment of [PtCl₄]^2- with NH₃ gives cis, but [Pt(NH₃)₄]²⁺ with HCl gives trans, according to Cotton-Wilkinson

for heat of reduction

[g.alonci]

I found this data:
1)HCOOH + 2H+ +2e- --->C + 2H2O E=0,621
2)C02 + 4H+ + 4e- --------> C + 2H2O E=0,207
3)Pd(2+)-----Pd E=0,987
4)dfG(NH4+)=-79,31kJ
5)dfG(NH3) (aq) =-26,50 kJ
6)dfG(Hform) (aq)=-351
7)dfG(form-) (aq) =-351

Now I can write:
(1)+(3)-(2)= Pd(2+) +HCOOH ------>CO2 + Pd +2H+ E=0,621-0,207+0,987=1,407V
dG(1)=-nFE=-271.55
dG for formation of complex from ammonia and Pd(2+) is
dG(2)=(RTlnK)/1000 kJ (I consider K for formation);
dG for neutralization of ammonia with H+ and formic acid is
dG(3)=4*[(4)-(5)] (dG is equal for HCOOH and HCOO-) =-211,24 kJ

so the total dG for the reaction:
Pd(NH3)4 (2)+ + 3HCOOH -------> Pd + CO2 + 2HCOO- + 4NH4+
must be dG(tot)=-482,79 + (1/1000)RT ln K

I found that lnK for the (en) complex is 27,3 , so considering that complex with en are far more stable than that with ammonia, I can ipotize that k is about 20.
This way I obtain dG=-433,26
But this valour is not too high?? Where am I making a mistake?