How many coulombs were needed to deposit the 5.4 g Ag. That's 96,485 coulombs x (5.4/108) = approx 4824. How many grams Cu will be deposited with 96,485? That is 64/2 = 32 g. You didn't have 96,485; you had 4824. So
32 x (4,824/96,495) = ?
Redox reactions are
Ag^+ + e ==> Ag(s)
Cu^2+ + 2e ==> Cu(s)
The oxidation reaction(s) depend(s) upon the anions in the water solutions.
Two electrolysis cells are set up in series so that the same current flows through each. In one compartment 5.40 g of Ag (molar mass = 108 g/mol) is deposited from a Ag+ solution. What mass of Cu (molar mass = 64 g/mol) is deposited simultaneously if the other cell contains Cu2+?
I don't know how to do this. I am confused that both are positive? and it doesn't have a redox reaction? Help on how to do?
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