# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Topic started by: sam on August 02, 2006, 06:21:37 PM

Title: electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions .masses of Pd and Ag ?
Post by: sam on August 02, 2006, 06:21:37 PM
The same amount of electricity (same number of moles of electrons) is used to
carry out the electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions in separate cells.
The masses of Pd and Ag produced were measured and compared. Which of the
following is true about the mass of Pd produced?

A. The mass of Pd produced is not related to the mass of Ag.
B. The mass of Pd produced is approximately half that of Ag.
C. The mass of Pd produced is approximately twice that of Ag.
D. The mass of Pd produced is approximately the same as that of Ag.

i have no idea from where i should start. could you help me figure it out?

Title: Re: electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions .masses of Pd and Ag ?
Post by: Borek on August 02, 2006, 06:32:28 PM
You have to compare number of electrons needed to reduce both ions and to compare both molar masses.

Title: Re: electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions .masses of Pd and Ag ?
Post by: BaO on August 02, 2006, 08:37:26 PM
because Pd and Ag will be produced ,there will be 2 half-reactions : Pd2++2e- ->Pd
Ag+ +e- -> Ag

according to Faraday's law,mPd=  #of moles of 2e , similarly to mass of Ag? (i'm not very sure)
so i think the answer is C. what you think?
Title: Re: electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions .masses of Pd and Ag ?
Post by: Will on August 02, 2006, 08:53:02 PM
so i think the answer is C. what you think?

I would say the answer is B.
Title: Re: electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions .masses of Pd and Ag ?
Post by: sam on August 03, 2006, 12:18:21 PM
Quote
I would say the answer is B

why? isnt mass of Pd = moles of 2e while mAg=e

Title: Re: electrolysis of PdCl2 (aq ) and AgNO3 (aq) solutions .masses of Pd and Ag ?
Post by: sam on August 03, 2006, 07:42:50 PM
oh, i get it .
mPd= moles of e-/2
mAg= moles of e- and because there are the same numbers of moles of e-, therefore the answer is B