Question

Two students use the same stock solution $ZnS{{O}_{4}}$ and a solution of $CuS{{O}_{4}}$ . The emf of one cell is 0.03 V higher than that of the other. The concentration of $CuS{{O}_{4}}$ in the cell with higher emf value is 0.5 M. Find out the concentration of $CuS{{O}_{4}}$ in the other cell $\left( \dfrac{2.303RT}{F}=0.06 \right)$ .

Answer 1

The relation between the reduction potential of an electrochemical reaction to the standard electrode potential considering the temperature, and activities of the chemical species undergoing reduction and oxidation is known as the Nernst equation. Inserting the values in the given equation will give you the answer.
${{E}_{cell}}={{E}^{0}}-[RT/nF]\ln Q$

Complete answer:
-The Nernst equation is used for the calculation of the cell potential of an electrochemical cell at any given temperature, and reactant concentration. The Nernst reaction for single electrode potential is given as-
${{E}_{cell}}={{E}^{0}}-[RT/nF]\ln Q$
Where ${{E}_{cell}}$ is the cell potential of the cell
${{E}^{0}}$ is the cell potential under standard conditions
R is the universal gas constant
T is the temperature
n is the number of electrons transferred in the redox reaction
F is the Faraday constant
Q is the reaction quotient.
-Let there be two Daniel cell so, the cell will be given as-
$Zn(s)/ZnS{{O}_{4}}({{C}_{1}})//0.5\text{ M }CuS{{O}_{4}}/Cu(s)$
$Zn(s)/Z{{n}^{2+}}({{C}_{2}})//C{{u}^{2+}}({{C}_{1}})$
-Now calculating the cell potential for the first student,
${{E}_{1}}={{E}^{o}}-\dfrac{0.06}{2}\log \dfrac{{{C}_{1}}}{0.5}...(i)$
-The cell potential for the second student is,
{{E}_{2}}=E{}^\circ -\dfrac{0.06}{2}\log \dfrac{{{C}_{1}}}{{{C}_{2}}}...(ii)$ where ${{E}_{2}}>{{E}_{1}}
-According to question, ${{E}_{2}}-{{E}_{1}}$ = 0.03 and ${{C}_{1}}={{C}_{2}}$ .
-The cell reaction of the given cell is-
$Zn(s)+C{{u}^{2+}}(aq)Z{{n}^{2+}}(aq)+Cu(s)$
-Therefore ${{E}_{2}}-{{E}_{1}}=\dfrac{0.06}{2}\left( {{\log }_{10}}\dfrac{{{C}_{2}}}{C}\times \dfrac{0.5}{{{C}_{1}}} \right)$
Since, ${{C}_{1}}={{C}_{2}}$
$\Rightarrow 0.03=\dfrac{0.06}{2}{{\log }_{10}}\dfrac{0.5}{C}$
Hence, C = 0.05 M

Note:
There do are limitations to the Nernst equation. Nernst equations deviate at higher concentrations. Another limitation of this equation is that it cannot be used to measure the cell potential when there is no current flowing through the electrode. Factors such as resistive loss and overpotential must be taken care of while doing calculations in the presence through the electrode.