- The cell that converts the chemical energy liberatedas a result of redox reaction to electrical energy is called a Daniel cell.
- It has anelectrical potential of 1.1 V.
- The setup for Daniel cell is as follows:
- In a beaker a plate of zinc is dipped in a solution of zinc sulfate (ZnSO4).
- In another beaker a plate of copper is dipped in a solution of copper (II) sulfate in another container. These plates of metal are called the electrodes of the cell.
- These electrodes behave as terminal to hold the electrons.
- The two electrodes are connected via wire.
- A salt bridge is placed between the two beakers. This provides a path for the movement of ions from one beaker to the other in order to maintain electrical neutrality.
- Zinc electrode gets oxidized and hence releases electrons that flow through the wire towards the copper electrode.
- The copper (II) sulfate solution releases copper ions Cu2+.
- At the anode:
Oxidation ---------------- loss of electrons.
Zn --> Zn2+ + 2e-
Reduction -------------gain of electrons.
Cu2+ + 2e- --> Cu
- Zinc atoms being more reactive have a greater tendency to lose electrons than that of copper.
The electrons in this cell moves from zinc anode to copper cathode through the wire connecting the two electrodes in the external circuit
- A bulb placed within this circuit will glow and a voltmeter connected within this circuit will show deflection.
- The net reaction of this cell is the sum of two half-cell reactions.
Zn(s) + Cu2+ (aq) --> Zn2+ (aq) + Cu(s)
In a Daniel cell a salt bridge is placed between the two beakers containing a solution of zinc sulfate (ZnSO4) and a solution of copper (II) sulfate respectively. This provides a path for the movement of ions from one beaker to the other in order to maintain electrical neutrality.