A current is passed through three electrolytic cells containing silver trixonitrate(5),coppe(2)tetraoxonitrate(6) and brine respectively.if 12.7g of copper are deposited in the second electrolytic cell.calculate(a)the mass of silver deposited in d first cell.

Question

(b) d volume of liberated in the third cell at 17 degree and 800mmhg?

Answer 1

First you should know that silver nitrate is a correct IUPAC name for AgNO3 and copper(II) nitrate is a correct IUPAC name for Cu(NO3)2. Also, note that the oxidation number for N in your name for copper nitrate is (V) and not (6).
All of my numbers are close estimate. You should recalculate ALL of these numbers for all are my estimates only.
It will take 96,485 coulombs to deposit 63.54/2 = approx 32 g of Cu. So how many coulombs of electricity must have passed through the cell? That's
96,485 x (12.7/about 32) = approx 39,000.
It will take 96,485 coulombs to deposit 107.9 g Ag. You had approx 38,000 C so
107.9 g Ag x (approx 38,000/96,485) = approx ? g Ag deposited.

For the second part of the question note you don't specify the volume of what gas but it is done the same way as I've shown you for part 1 of the problem.

Answer 2

Ok