Question

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?

Answer 1

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*MW_Ag - n*MW_Cu

=> 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.