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The electrochemical cell reaction of the Daniel cell is
 $ Zn(s) + C{u^{2 + }}(aq) \to Z{n^{2 + }}(aq) + Cu(s) $
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor $ 10 $ ?

Last updated date: 20th Jul 2024
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Hint : The best example of the galvanic cell is the Daniel cell. It converts chemical energy into electrical energy and it has two electrodes made up of dissimilar metals. The two metals are zinc and copper. Each of these electrodes is kept in contact with a solution of its own ion i.e. zinc sulphate solution and copper sulphate solution.

Complete Step By Step Answer:
In a Daniel cell, zinc acts as the anode and copper metal acts as cathode. $ Z{n^{2 + }} $ ions are in the anode compartment and $ C{u^{2 + }} $ ions are in the cathode compartment. Suppose the original concentration of the ions $ Z{n^{2 + }} $ and $ C{u^{2 + }} $ are $ 1M $ .
So, $ {E_{cell}} = E_{cell}^ \circ $
If the ion concentration in the anode compartment i.e. $ Z{n^{2 + }} $ is increased by a factor $ 10 $ , then $ {E_{cell}} = E_{cell}^ \circ - \left( {\dfrac{{0.0592v}}{2}} \right)\log \dfrac{{10}}{{12}} $
 $ {E_{cell}} = E_{cell}^ \circ - 0.03{\text{v}} $
The cell voltage is decreased by $ 0.03{\text{v}} $
Therefore, by increasing the $ Z{n^{2 + }} $ concentration by a factor $ 10 $ decreases the cell voltage by $ 0.03{\text{v}} $

Note :
In the Daniel cell, electrons flow from the zinc electrode towards the copper electrode through an external bridge whereas the metal ions from one half cell move to the other half cell through the salt bridge. So, the flow of current is from the copper electrode towards the zinc electrode.