Question

In the electrochemical cell: \[Zn\mid ZnS{O_4}\left( {0.01M} \right)\mid \;\mid CuS{O_4}\left( {1.0M} \right)\mid Cu\], the emf of this Daniel cell is \[{E_1}\]. When the concentration of \[ZnS{O_4}\] is changed to \[1.0\;M\;\] and that of \[CuS{O_4}\] changed to \[0.01\;M\], the emf changes to \[{E_2}\]. From the following, which one is the relationship between \[{E_1}\] and \[{E_2}\]?
Given, \[\dfrac{{RT}}{F}\; = 0.059\]).
A. \[{E_2} = 0 \ne {E_1}\]
B. \[{E_1} = {E_2}\]
C. \[{E_1}\; < {E_2}\]
D. \[{E_1}\; > {E_2}\]

Answer 1

Hint: We know that,\[{E_{cell}} = E_{cell}^0\; - \dfrac{{0.059}}{n}\;log\dfrac{{[anode]}}{{[Cathode]}}\] is the equation of galvanic cell, in which the value of concentration of anode and cathode is given. Anode will oxidize by giving electrons and copper will gain electrons and reduce.
\[{E_{cell}} = E_{cell}^0\; - \dfrac{{0.059}}{n}\;log\dfrac{{[Z{n^{ + 2}}]}}{{[C{u^{ + 2}}]}}\] this is the equation of the cell we will be using to solve the question. In the question the cell is represented in which zinc is anode and copper is cathode. If we substitute the value of the concentration of the cell we can find the value of \[{E_1}\] and \[{E_2}\].

Complete step by step solution
Given data:
The value of the concentration of the cell is given.
\[{E_{cell}}\] is the electric cell potential of the cell.
\[E_{cell}^0\] is the standard electric potential of the cell.
n is the number of electron transfers here it is two.
The concentration of \[ZnS{O_4}\] is changed to \[0.01M\;\]and that of \[CuS{O_4}\] changed to \[1.0M\].
\[
\Rightarrow {E_{cell}} = E_{cell}^0\; - \dfrac{{0.059}}{n}\;log\dfrac{{[Z{n^{ + 2}}]}}{{[C{u^{ + 2}}]}}\\
\Rightarrow {E_1} = \,1.1\,\; - \dfrac{{0.059}}{n}\;log\dfrac{{0.01}}{{1.0}}\\
\Rightarrow {E_1} = \,1.1\,\; - \dfrac{{0.059}}{2} \times ( - 2)\;\\
\Rightarrow {E_1} = \,1.1\,\; - ( - 0.0592)\\
\Rightarrow {E_1} = \,1.16\;V\\
\;
\]
Now if the concentration of the cell is changed the concentration of \[ZnS{O_4}\] is changed to \[1.0\;M\;\]and that of \[CuS{O_4}\] changed to \[0.01\;M\] then the value is substituted, we get the other value.
\[
\Rightarrow {E_{cell}} = E_{cell}^0\; - \dfrac{{0.059}}{n}\;log\dfrac{{[Z{n^{ + 2}}]}}{{[C{u^{ + 2}}]}}\\
\Rightarrow {E_2} = \,1.1\,\; - \dfrac{{0.059}}{2}\;log\dfrac{{1.0}}{{0.01}}\\
\Rightarrow {E_2} = \,1.1\,\; - \dfrac{{0.059}}{2} \times 2\;\\
\Rightarrow {E_2} = \,1.1\,\; - (0.0592)\\
\Rightarrow {E_2} = \,1.041\;V\\
\;
\]

Hence the correct answer is (D) \[{E_1}\; > {E_2}\]

Note:
The value of concentration of the anode and cathode is always fixed and can be obtained from the call representation given in the question the anode is always on the left side and cathode always on right, that must be known in the Daniel cell or galvanic cell.