Four cells containing CuSO4(aq), AgNO3(aq), Pb(NO3)2(aq), and Au(NO3)2(aq) are connected power supplies that operate at 12.0A for 1 hour. At the end of this time , which cell will deposit the greatest mass of metal?
To determine which cell will deposit the greatest mass of metal, we need to compare the number of moles of metal deposited in each cell.
First, we need to find the charge transferred in each cell:
Charge (C) = Current (A) × Time (s)
Since the power supplies operate at 12.0A for 1 hour (3600 seconds), the charge transferred in each cell is:
Q = 12.0 A × 3600 s = 43200 C
Next, we need to find the moles of metal deposited in each cell using Faraday's law:
Moles of metal = Charge (C) / Faraday's constant
Faraday's constant = 96485 C/mol (charge of 1 mole of electrons)
Therefore, the moles of metal deposited in each cell are:
1. CuSO4(aq): Moles of Cu = 43200 C / 96485 C/mol = 0.447 mol of Cu
2. AgNO3(aq): Moles of Ag = 43200 C / 96485 C/mol = 0.447 mol of Ag
3. Pb(NO3)2(aq): Moles of Pb = 43200 C / 96485 C/mol = 0.447 mol of Pb
4. Au(NO3)2(aq): Moles of Au = 43200 C / 96485 C/mol = 0.447 mol of Au
Since all cells have the same number of moles of metal (0.447 mol), they will deposit the same mass of metal.