Question

In a Daniel cell constructed in the laboratory, the voltage observed was $0.9V$ instead of $1.1V$ of the standard cell. A possible explanation is:
A. $\left[ {Z{n^{2 + }}} \right] > \left[ {C{u^{2 + }}} \right]$
B. $\left[ {Z{n^{2 + }}} \right] < \left[ {C{u^{2 + }}} \right]$
C. $Zn$ electrode has twice the surface of $Cu$ electrode.
D. Mole ratio of $Z{n^{2 + }}:C{u^{2 + }}$ is $2:1$

Answer 1

Hint: The Daniel cell is a type of electrochemical cell invented in 1836 by John Frederic Daniell, a British chemist and meteorologist, and consists of a copper pot filled with a copper (II) sulfate solution, in which is immersed an unglazed earthenware container filled with sulfuric acid and a zinc electrode.

Complete answer:
In the Daniell cell, copper and zinc electrodes are immersed in a solution of copper(II) sulfate and zinc sulfate, respectively. At the anode (negative electrode), zinc is oxidized as per the following half reaction:
$Anode:Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }$
At the cathode (positive electrode), copper is reduced per the following reaction:
$Cathode:C{u^{2 + }}_{(aq)} + 2{e^ - } \to C{u_{(s)}}$
The overall reaction of the cell that takes place here is:
$Z{n_{(s)}} + C{u^{2 + }}_{(aq)} \to Z{n^{2 + }}_{(aq)} + C{u_{(s)}}$
The Nernst equation for the above reaction can be written as:
${E_{cell}} = E_{cell}^o - \dfrac{{0.0591}}{2}\log \dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}$
As per the question, the voltage observed was $0.9V$ instead of $1.1V$ of the standard cell. This means that when we substitute the value of $E_{cell}^o$ as $1.1V$, the value of the electrode potential of the cell comes out to be $0.9V$. A possible reason for this can be:
${E_{cell}} = 1.1 - \dfrac{{0.0591}}{2}\log \dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}$
Now, for the ${E_{cell}} = 0.9V$ or in other words, for ${E_{cell}} < 1$ , the value of concentration of zinc cation should be greater than the concentration of the copper ions. This can be shown mathematically as:
$ \Rightarrow {E_{cell}} = 1.1 - \log \left( {\dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}} \right)$
$ \Rightarrow {E_{cell}} < 1$ if $\log \left( {\dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}} \right) = + ve$ and this is positive only when: $\left( {\dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}} \right) > 1$.
Thus, $[Z{n^{2 + }}] > [C{u^{2 + }}]$.

Thus, the correct option is A.

Note:
In small scale demonstrations, a form of the Daniel cell known as two half cells is often used due to its simplicity. The two half cells each support one half of the reactions described above. A wire and light bulb may connect the two electrodes. Excess electrons produced by the oxidation of zinc metal are “pushed” out of the anode, which is therefore the negative electrode, travel through the wire and are "pulled" into the copper cathode where they are consumed by the reduction of copper ions. This provides an electrical current that illuminates the bulb.