1. According to the Standard Reduction Potential Table in your textbook, assign
ID: 1056361 • Letter: 1
Question
1. According to the Standard Reduction Potential Table in your textbook, assign the anode and cathode for thecell pairs, B1-B2, A1-B2, AND A2-B2 described in the Procedure.
2. Describe the Galvanic cell A4-B4 when the two cells are connected through a voltmeter to obtain a postive cell EMF. Which cell is anode? Which cell is cathode? What is the direction of electron flow?
3. Can you make a Galvanic cell with a positve EMF by connecting A3 and A4? If you can, which cell is cathode and which is anode? Why?
4. Explain the role of the filtering paper saturated with KNO3 solution in constructing each Galvanic cell in this experiment.
Explanation / Answer
1. a) B1-B2 - The Cu-Zn cell
According to standard reduction potentials for Cu and Zn, Zinc will undergo oxidation and copper will have less tendency to lose electrons, that is it will undergo reduction.
In the cell: Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Zn appears on the left side, indicating it is being oxidized. So, it acts as anode, while Cu acts as cathode.
b) A1-B2 - The Fe-Zn cell
Accoding to standard reduction potentials table, Fe(E0=-0.409V) is above the Zn(E0=-0.763V)
So, again Zn undergoes oxidation and Fe is reduced.
thus, Zn acts as anode and Fe acts as cathode.
The cell is represented as: Zn(s)|Zn2+(aq)||Fe2+(aq)|Fe(s)
c) A2-B2 - The Pb-Zn cell
Pb is above Zn in the standard reduction potentials table, so Zn will act as anode and Pb acts as cathode.
The cell is represented as: Zn(s)|Zn2+(aq)||Pb2+(aq)|Pb(s)
2. Galvanic cell A4-B4 (Cu(II)(1M) - Cu(II)(0.001M))
An electrochemical cell constructed with the same half reactions at the cathode and anode creates a non-zero voltage , when the concentration of reactants and products differ. This is called as concentration cell. the concentration cell acts to dilute the more concentrated solution and concentrate the more dilute solution, creating a voltage. This is achieved by transferring electrons from the cell with lower concentration to the cell with the higher concentration. So, electrons will flow from left to right.
So, the cell is represented as :
Cu(s)|Cu2+(0.001M)||Cu2+(1M)|Cu(s)
E0cell =0
Ecell = E0cell -RT/nF ln[Cu2+]left / [Cu2+]right
= 0 - 8.314*298/2*96500 ln 0.001M/1M
= 0.0887V
3. If a cell is constructed from two half cells with same chemistry as:
Cu(s)|Cu2+(1M)||Cu2+(1M)|Cu(s)
This is a concentration cell. The driving force in the concentration cell is to make the concentration in each half-cell equal. When the concentration of copper ion in each half cells are equal, the system is at equilibrium . At equilibrium, there is no driving force in the reaction, the voltage is zero and there is no net flow of electrons.
4. A filtering paper saturated with KNO3 solution, is used as a salt bridge to connect oxidation and reduction half cells. It maintains the electrical neutrality. Filter paper provides a solid medium for conduction. Conductivity depends on concentration of electrolyte solution, texture of filter paper and absorbing ability of filter paper. Generally, smoother texture and higher absorbency equates to higher conductivity.
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