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Part c) is what I\'m stuck on, and it tells me that (in red) but I can\'t figure

ID: 822769 • Letter: P

Question

Part c) is what I'm stuck on, and it tells me that (in red) but I can't figure it out. Please someone help!! :)

Consider two cells, the first with Al and Ag electrodes, and the second with Zn and Ni electrodes, each in appropriate 1.00 M solutions of their ions. If connected as voltaic cells in series, which two metals are plated, and what is the total potential? If connected such that one cell acts as a battery to power the other as an electrolytic cell, which two metals are plated, and what is the total potential? If 3.50 g of metal is plated in the voltaic cell, how much metal is plated in the electrolytic cell? Incorrect. The Al/Ag cell is the voltaic cell and the Ni/Zn cell is the electrolytic cell. From the mass of Ag, you can find the number of moles of Ag, then the number of electrons transferred, then number of moles of Zn, then the mass of Zn.

Explanation / Answer

c) moles of Ag = 3.5 / 107.87 = 0.03245 moles

1 e- is involved in Ag(Ag+) and 2 e- involved in Zn(Zn2+)


no of moles of Zn formed = 0.03245 / 2 = 0.016223 moles

=> mass of Zn = 0.016223 x 65.38 = 1.061 gm