Question

Calculate the EMF of a Daniel Cell when the concentration of \[ZnS{O_4}\] and \[CuS{O_4}\] are 0.001 M and 0.1M respectively. The standard potential of the cell is 1.1 V

Answer 1

Hint: Nernst equation is an equation is used in electrochemistry that establishes a relation between the reduction potential of an electrochemical cell with the other key characteristics like temperature, standard electrode potential, molal concentrations of constituent electrolytes, etc.

Formula used: E = \[{E^0} - \dfrac{{0.0592}}{n}{\log _{10}}Q\]

Complete Step-by-Step Answer:
Daniel Cell is a kind of electrochemical cell which uses electrodes made from copper and zinc. These electrodes are kept in their corresponding sulphate solutions. This means that zinc electrodes are immersed in zinc sulphate solution while copper electrodes are immersed in copper sulphate solution. The electrochemical action of the cell involves the following reactions:
At anode: zinc is oxidized
 \[Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }\]
At cathode: copper ion is reduced
 \[C{u^{2 + }}_{(aq)} + 2{e^ - } \to C{u_{(s)}}\]
The overall reaction for this entire process can be given by:
 \[Z{n_{(s)}} + C{u^{2 + }}_{(aq)} \to Z{n^{2 + }}_{(aq)} + C{u_{(s)}}\]
Applying Nernst equation for the above cell we get
E = \[{E^0} - \dfrac{{0.0592}}{n}{\log _{10}}\dfrac{{Z{n^{2 + }}}}{{C{u^{2 + }}}}\]
E = 1.1 - \[\dfrac{{0.0592}}{2}{\log _{10}}\dfrac{{0.001}}{{0.1}}\]
\[E{\text{ }} = {\text{ }}1.159{\text{ }}V\]

Hence, the EMF of the given Daniel Cell is calculated to be 1.159 V

Note: The Daniel cell is also the historical basis for the contemporary definition of the volt, which is the unit of electromotive force in the International System of Units. The definitions of electrical units that were proposed at the 1881 International Conference of Electricians were designed so that the electromotive force of the Daniel cell would be about 1.0 volts.