Given, initial mass of the Cu metal m 1 = 25g final mass of the Cu metal with th
ID: 699794 • Letter: G
Question
Given,
initial mass of the Cu metal m1 = 25g
final mass of the Cu metal with the silver deposit m2 = 28.2g
Now, the reactions occuring are:
Ag+ -----> Ag(s) + e-
Cu(s) + 2e- ------> Cu2+
The overall reaction is:
Cu(s) + 2Ag+ --------> Cu2+ + 2Ag(s)
From here we can observe that for every 2 moles of Ag deposited we lose 1 mol of Cu.
Let 2n be the number of moles of Ag deposited, then n will be the number of moles of Cu oxidised.
Molar mass of Ag: 108 g/gmol
Molar mass of Cu: 63.5 g/gmol
Increase in the mass of the copper piece = mass of silver deposited - mass of the Cu metal oxidised
=> m2 - m1 = 2n*MWAg - n*MWCu
=> 28.2 - 25 = 2n*108 - n*63.5 = 152.5n
=> n = 0.021 moles
Therefore the mass of silver deposited = 2n = 2*0.021*108 = 4.536 grams Ag.
Explanation / Answer
A 25g piece of copper (Cu) metal is placed in an aqueous silver nitrate (AgNO3) solution. The copper dissolves to form Cu2+ ions and the silver Ag+ ions plate out as solid silver on the piece of copper. After a short period of time, the weight of the piece of copper with silver deposit is 28.2 g. How many grams of silver are depostied on the piece of copper?
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