Two voltaic cells are to be joined so that one will run the other as an electrolytic cell.

In the first cell, one half-cell has Au foil in 1.00M Au(NO3)3, and the other half-cell has a Cr bar in 1.00M Cr(NO3)3.
In the second cell, one half-cell has a Co bar in 1.00M Co(NO3)2, and the other half-cell has a Zn bar in 1.00M Zn(NO3)2.

A) Calculate the E0cell for each cell.
[2.24V and 0.48V]
B) Calculate the total potential if the two cells are connected as voltaic cells in series.
[2.72V]
C) When the electrode wires are switched in one of the cells, which cell will run as the voltaic cell and which as the electrolytic cell.
[voltaic=cell 1, electrolytic=cell 2]
D) Which metal ion is being reduced in each cell?
[Au, Zn]
E) If 2.00 g of metal plates out in the voltaic cell, how much metal ion plates out in the eletrolytic cell.

I undertstand the answers until part C-E. For C, I have the right answer but my logic was based on the spontaneity of Au(s) & Cr reaction, and then with Co(s) with Zn. Is this right or do I have to take the series into account (which I don't know how).
D) this is based on C, but since Au(s) turns into its ionic form, shouldn't it be oxidized?
E) I tried to do a molar ratio (find out the number of moles Au (2.00g/197.0g) then multiply that with Zn molar mass (this is based on the answer for part D) but I still didn't get the answer.

Please explain this since final is approaching.
Thanks.

black people

Sure, let's go through each part of the question step by step:

A) To calculate the E0cell for each cell, you can use the Nernst equation and the standard reduction potentials given in a table. The Nernst equation is:

Ecell = E0cell - (0.0592/n) * log(Q)

For the first cell with Au and Cr, the half-reactions are:

Au3+ + 3e- -> Au (reduction at the cathode)
Cr -> Cr3+ + 3e- (oxidation at the anode)

Looking up the standard reduction potentials for these reactions, you'll find that E0cell for the first cell is 2.24V.

Similarly, for the second cell with Co and Zn, the half-reactions are:

Co2+ + 2e- -> Co (reduction at the cathode)
Zn -> Zn2+ + 2e- (oxidation at the anode)

Using the standard reduction potentials, you'll find that E0cell for the second cell is 0.48V.

B) To calculate the total potential if the two cells are connected in series, you simply add the E0cell values of the two cells. Therefore, the total potential is 2.24V + 0.48V = 2.72V.

C) When the electrode wires are switched in one of the cells, the direction of the current flow will be reversed. In a voltaic cell, the half-reaction with the more positive E0cell value will always occur at the cathode (positive electrode), and the half-reaction with the more negative E0cell value will occur at the anode (negative electrode). So, in this case, the first cell (Au and Cr) will run as the voltaic cell, and the second cell (Co and Zn) will run as the electrolytic cell.

D) In the first cell (Au and Cr), Au3+ ions are being reduced to Au metal, so Au3+ is the metal ion being reduced. In the second cell (Co and Zn), Zn2+ ions are being oxidized to Zn metal, so Zn2+ is the metal ion being reduced.

E) To determine how much metal ion plates out in the electrolytic cell (the second cell, Co and Zn), you need to consider the stoichiometry of the electrochemical reactions. In the given reaction, 2.00 g of metal plates out in the voltaic cell (the first cell, Au and Cr). To find the mass of the metal ion that plates out in the electrolytic cell, you'll need to set up a conversion.

First, calculate the number of moles of the metal that plates out in the voltaic cell:

moles of Au = mass of Au / molar mass of Au
= 2.00 g / 197.0 g/mol
= 0.0102 mol

Now, use the mole ratio between the metal ion and the metal in the electrolytic cell based on the balanced half-reaction equation (Zn2+ + 2e- -> Zn). The mole ratio is 1:2.

moles of Zn2+ = 2 * moles of Au
= 2 * 0.0102 mol
= 0.0204 mol

Finally, calculate the mass of the metal ion that plates out in the electrolytic cell:

mass of Zn2+ = moles of Zn2+ * molar mass of Zn2+
= 0.0204 mol * (65.38 g/mol)
= 1.33 g

Therefore, in the electrolytic cell, approximately 1.33 grams of Zn2+ will plate out.

I hope this explanation clarifies the process for you. Good luck with your final!