# construction working and uses of daniell cell

Daniell cell is the simplest voltaic or galvanic cell. It converts chemical energy into electrical energy spontaneously.

Construction:

Daniell cell consists of two half-cells. One half-cell is zinc rod dipped in 1M solutions. The other half-cell is copper rod dipped in a solution of 1M solutions. A porous partition or a salt bridge separates the two half-cells from each other. The two electrodes are connected together externally by a metal wire through a voltmeter are produced here so anode is the negative electrode. The copper rod acts as cathode where reduction takes place. Electrons are here so cathode is the positive electrode.

Working:

The following reactions take place at the two electrodes.

At the anode:

Oxidation ---------------- loss of electrons.

$Zn_{(s)}\rightarrow Zn^{2+}_{(aq)}+2e^-$

At cathode,

Reduction -------------gain of electrons.

$Cu^{2+}_{(aq)}+2e^{-} \rightarrow Cu _{(s)}$

The half reactions can be viewed as a competition between two kinds of metal atoms for electrons. In this case zinc atoms are more reactive and their tendency to lose electrons is greater than that of copper. Both the reactions take place simultaneously at both the half-cell s.
In this cell, electrons travel in the external circuit through the wire from the zinc anode to the copper cathode. If a bulb is in the circuit, it will light up. If there is a voltmeter, it will show the voltage. To complete the circuit, both the positive and negative ions move through the aqueous solutions via the salt bridge.

The total reaction is the sum of two half-cell reactions.

$Zn_{(s)}+Cu^{2+}\! _{(aq)}+Cu_{(s)}$

In this process, Zn electrode dissolves in the solution of ZnSO4 and reduces in size while Cu electrode grows in size due to the deposition of copper metal. Daniell cell generates an electric potential of 1.10 volt, when the solutions in the half cells are both 1M.

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Here Daniel cell uses are not given
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