Question

The emf of Daniel cell at $298K$ is ${E_1}{\text{ }}Zn|ZnS{O_4}(0.01M)||CuS{O_4}(1.0M)|Cu$. When the concentration of $ZnS{O_4}$ is $1.0M$ and that of $CuS{O_4}$ is $0.01M$, the emf changed to ${E_2}$. What is the relation between ${E_1}$ and ${E_2}$?
A.) ${E_1} = {E_2}$
B.) ${E_1} = 0 \ne {E_2}$
C.) ${E_1} < {E_2}$
D.) ${E_2} < {E_1}$

Answer 1

This type of question can be solved by using the concept of Nernst equation for two different cases in which zinc sulphate and copper sulphate is changed. The Nernst equation can be given as:
${E_{cell}} = E_{cell}^\circ - \dfrac{{0.059}}{n}\log \dfrac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}$

Complete step-by-step answer: In the Daniel cell, there are two half cells that are zinc half cell and other one is copper half cell. The zinc half cell consists of zinc sulphate solution and copper half cell consists of zinc sulphate solution. Also, there is a salt bridge between the two cells to complete the circuit.
The two half reactions that occurs in the Danie cell can be shown as:
Anode half reaction: $Zn(s) \to Z{n^{2 + }} + 2{e^ - }$
Cathode half reaction: $C{u^{2 + }} + 2{e^ - } \to Cu(s)$
Also, at anode the reduction reaction occurs. The reduction reaction is that reaction in which there is loss of electrons takes place.
Also, at cathode the oxidation reaction occurs. The oxidation reaction is that reaction in which there is gain of electrons takes place.
The overall reaction for the Daniel cell can be represented as:
$Zn(s) + C{u^{2 + }}(aq) \to Z{n^{2 + }}(aq) + Cu(s)$
Now, as we know that Nernst equation for the given cell is given as:
${E_{cell}} = E_{cell}^\circ - \dfrac{{0.059}}{n}\log \dfrac{{[anode]}}{{[Cathode]}}$
In the given cell $n = 2$ because both zinc and cathode lose or gain two electrons, here there is zinc at anode and copper at cathode. So, the above equation can be given as:
For the first case, $[Z{n^{2 + }}] = 0.01{\text{ and }}[C{u^{2 + }}] = 1.0$. Therefore, we can write:
${E_1} = E_{cell}^\circ - \dfrac{{0.059}}{2}\log \dfrac{{[0.01]}}{{[1.0]}}$
$\Rightarrow {E_1} = E_{cell}^\circ - \dfrac{{0.059}}{2}\log ({10^{ - 2}})$
$\Rightarrow {E_1} = E_{cell}^\circ - \dfrac{{0.059}}{2} \times ( - 2)$
$\Rightarrow {E_1} = E_{cell}^\circ + 0.059$ $- (1)$
For the second case, $[Z{n^{2 + }}] = 1.0{\text{ and }}[C{u^{2 + }}] = 0.01$. Therefore, we can write:
${E_2} = E_{cell}^\circ - \dfrac{{0.059}}{2}\log \dfrac{{[1.0]}}{{[0.01]}}$
$\Rightarrow {E_2} = E_{cell}^\circ - \dfrac{{0.059}}{2}\log ({10^2})$
$\Rightarrow {E_2} = E_{cell}^\circ - \dfrac{{0.059}}{2} \times (2)$
${E_2} = E_{cell}^\circ - 0.059$ $- (2)$
Now, by comparing equation $- (1){\text{ and - (2)}}$ we can conclude that:
${E_1} > {E_2}$

Hence, option D.) is the correct answer.

Note: Always remember that in Daniel cell representation like shown in the question $Zn|ZnS{O_4}(0.01M)||CuS{O_4}(1.0M)|Cu$ , here the left side of salt bridge shows the anodic half cell and the right side of the salt bridge( ||) represent cathodic half cell.