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. a) If connected as voltaic cells in series, which two metals are plated, and what is the total potential?
a) Ecell for Ag and Ni
b) Ecell for Ag and Zn
c) If 3.30 g of metal is plated in the voltaic cell, how much metal is plated in the electrolytic cell?
I will get you started on a and b.
a.
Ni ==> Ni^2+ Eo = ? as an oxiation
Ag^+ +e ==> Ag Eo = ? as a reduction
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Ni + 2Ag^+ ==> Ni^2+ + 2Ag
Ecell = sum E(oxdn) + E(redn) from above.
b is done the same way.
Zn + 2Ag^+ ==> Zn^2+ + 2Ag
a) To determine which two metals are plated in the voltaic cells, we need to compare their reduction potentials (or standard electrode potentials). The reduction potentials can be found in a table of standard electrode potentials.
For the first cell with Al and Ag electrodes, the reduction potential of Al is -1.66 V and the reduction potential of Ag is +0.80 V.
For the second cell with Zn and Ni electrodes, the reduction potential of Zn is -0.76 V and the reduction potential of Ni is -0.25 V.
The two metals that would be plated are the ones with the higher reduction potentials (more positive values). So, in this case, Ag and Ni would be plated.
b) The total potential of the voltaic cells connected in series is equal to the sum of the individual cell potentials. Therefore, we need to find the cell potentials for Ag and Ni, and for Ag and Zn.
The cell potential (Ecell) is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.
For Ag and Ni: Ecell = (reduction potential of Ni) - (reduction potential of Ag)
Ecell = -0.25 V - 0.80 V = -1.05 V
For Ag and Zn: Ecell = (reduction potential of Zn) - (reduction potential of Ag)
Ecell = -0.76 V - 0.80 V = -1.56 V
c) Since the two cells (voltaic and electrolytic) are not connected in any way, the metal plated in the electrolytic cell is not directly related to the amount of metal plated in the voltaic cell. Therefore, we cannot determine how much metal is plated in the electrolytic cell based on the information given.
To determine which metals are plated and the total potential in the given voltaic cells, we need to look at the reduction potentials for each half-reaction and understand the concept of cell potentials.
First, let's find the standard reduction potentials for each half-reaction:
Ag+ + e- → Ag (reduction potential = +0.80 V)
Zn2+ + 2e- → Zn (reduction potential = -0.76 V)
Ni2+ + 2e- → Ni (reduction potential = -0.23 V)
Al3+ + 3e- → Al (reduction potential = -1.66 V)
a) Ecell for Ag and Ni:
Since the half-reaction with the more positive reduction potential (Ag+ + e- → Ag) is assigned as the cathode, and the one with the more negative reduction potential (Ni2+ + 2e- → Ni) is assigned as the anode, the metals plated are silver (Ag) at the cathode and nickel (Ni) at the anode. To find the total potential, we subtract the anode potential from the cathode potential:
Ecell = Ecathode - Eanode
Ecell = (+0.80 V) - (-0.23 V)
Ecell = 1.03 V
Therefore, the plated metals are Ag and Ni, and the total potential is 1.03 V.
b) Ecell for Ag and Zn:
Following the same process, the metals plated would be silver (Ag) at the cathode and zinc (Zn) at the anode. The total potential is calculated as follows:
Ecell = Ecathode - Eanode
Ecell = (+0.80 V) - (-0.76 V)
Ecell = 1.56 V
Hence, the metals plated are Ag and Zn, and the total potential is 1.56 V.
c) To determine how much metal is plated in the electrolytic cell, we need to understand Faraday's laws of electrolysis. Faraday's laws state that the amount of substance produced or consumed in an electrolytic cell is directly proportional to the quantity of electricity passed through the cell.
To calculate the amount of metal plated, we can use the formula:
n = (mQ) / (MzF)
Where:
n is the number of moles of metal plated,
m is the mass of the plated metal,
Q is the total electrical charge passed through the cell in coulombs,
Mz is the molar mass of the metal, and
F is Faraday's constant (96,485 C/mol).
Since the charge passed through the voltaic cell can be determined from the total potential (Ecell) and time (t) as Q = It, where I is the current in amperes, we need to know the time the current flows through the cell.
Please provide the specific information about the time (t) the current flows so that we can calculate the amount of metal plated in the electrolytic cell.