Question

Part 3. Influence of Concentration on Cell Voltage   

Measured Temperature of Zn-Cu cell: 21.3C
Voltage of Zn-Cu cell : 0.977
(0.005 M Cu2+)
Measured Temperature of Cu-Cu cell: 21.1C
Voltage of Cu-Cu cell: -0.034
(0.500 M/0.005 M Cu2+)
Voltage of Zn-Cu cell : 1.057
(average voltage of 0.500 M Cu2+ from Part 2)

1) Describe what happened to the measured voltage (Zn-Cu) when the cathode solution was diluted. Do your observations match what you expected, based on your understanding of the Nernst equation?

2) What do you expect would have happened if the anode solution had been diluted instead of the cathode solution (Zn-Cu)?

3) Show how you calculate the theoretical voltage of the Zn-Cu cell using standard reduction potentials and the measured temperature.

4) Theoretical Voltage of Zn-Cu cell: ?
(0.005 M Cu2+)
To form a voltaic cell, should the voltage be positive or negative?

5) Which compartment in the Cu-Cu cell forms the anode to make a voltaic cell?

6) Show how you calculate the theoretical voltage of the voltaic Cu-Cu cell, using standard reduction potentials and the measured temperature.

7) What is the theoretical Voltage of the voltaic Cu-Cu cell ?
(0.005M/0.5M Cu2+)

Answer 1

Answer 1

On comparison of data, voltage of Zn-Cu cell was found to decrease with decrease in concentration of cathode.

As per Nernst equation

E = E^o - 0.0295Logan

where Q = [Zn²⁺]/[Cu²⁺]

This is because less is the concentration of cathode, more will be Q and hence the increase from standard potential will be more thus decreasing the cell voltage.

Answer 2

If anode would have been dilute, Q will increase which will decrease the value of 0.0295logQ and hence E would have been higher than that in 1.

Answer 3

Theoretical voltage of Zn-Cu cell

= E^o Cu - E^o Zn

= 0.34 -(-0.76) = 1.10 V

Answer 4

For cell to work, voltage must be positive.