what weight of silver is deposited of silver nitrate at the same time as 2.40g of copper(ll) sulphate solution when the two solution are arrange in series connected to a battery
2 answers
pls help me out
Cu^2+(aq) + 2e --> Cu(s)
Ag^+(aq) + e ==> Ag(s)
First you must determine the coulombs needed to deposit 2.40 g Cu from CuSO4 solution.
63.54/2 = 31.77g Cu(s) can be deposited by 96,485 coulomb of electricity. So how many coulombs will be needed to deposit 2.40 g? That will be
96,485 x (2.40/31.77) = estimated 7,289 coulombs.
107.9 g/1 = 107.9 g Ag(s) will be deposited with 96,485 coulombs so how much Ag will be deposited with 7289 coulombs? That is
107.9 g Cu x (7,289/96,485) = ?
Check those numbers. Post your work if you get stuck.
Ag^+(aq) + e ==> Ag(s)
First you must determine the coulombs needed to deposit 2.40 g Cu from CuSO4 solution.
63.54/2 = 31.77g Cu(s) can be deposited by 96,485 coulomb of electricity. So how many coulombs will be needed to deposit 2.40 g? That will be
96,485 x (2.40/31.77) = estimated 7,289 coulombs.
107.9 g/1 = 107.9 g Ag(s) will be deposited with 96,485 coulombs so how much Ag will be deposited with 7289 coulombs? That is
107.9 g Cu x (7,289/96,485) = ?
Check those numbers. Post your work if you get stuck.
1 answer