Question

During the electrolysis of $0.1M$$CuS{O_4}$ solution using copper electrodes, a depletion of $\left[ {C{u^{2 + }}} \right]$ occurs near the cathode with a corresponding excess near the anode, owing to insufficient stirring of the solution. If the concentration of $\left[ {C{u^{2 + }}} \right]$ near the anode and cathode are respectively$0.12M\& 0.08M$. Calculate the back emf developed. Temperature$ = 298K$.
A. $22mV$
B. $5.2mV$
C. $29mV$
D. $59mV$

Answer 1

Hint: We know the Hermann Nernst equation is commonly accustomed to calculate the cell potential of a chemical science cell at any given temperature, pressure, and chemical concentration.
\[{E_{cell}} = {E_0}-\left[ {RT/nF} \right]lnQ\]
Where,
\[{E_{cell}}\] = cell potential of the cell
The cell potential below customary conditions is \[{E_0}\]
R is the universal gas constant
T is temperature
n is amount of electrons transferred within the chemical reaction
Q is the reaction quotient.

Complete step by step answer:
First we see what back emf is.
The emf due to different concentrations of ions near cathode and anode is defined as back emf.
\[{E_{cell}} = {E_0}-\dfrac{{0.059}}{n}ln\dfrac{{\left[ {{\text{product}}} \right]}}{{\left[ {{\text{reactant}}} \right]}}\]
Let us see the reaction occur at cathode.
$C{u^{2 + }}\left( {aq} \right) + 2{e^ - } \to Cu\left( s \right)$
\[{E_{cathode}} = {E_0}-\dfrac{{0.059}}{2}ln\dfrac{{\left[ {Cu} \right]}}{{\left[ {C{u^{2 + }}} \right]}}\]
Let substitute the values in the above equation we get,
\[ \Rightarrow {E_{cathode}} = -\dfrac{{0.059}}{2}ln\dfrac{1}{{0.08}}\]
\[{E_{cathode}} = {\text{ - 0}}{\text{.032V}}\]
Let us see the reaction occur at anode.
$Cu\left( s \right) \to C{u^{2 + }}\left( {aq} \right) + 2{e^ - }$
\[{E_{anode}} = -\dfrac{{0.059}}{2}ln\dfrac{{\left[ {C{u^{2 + }}} \right]}}{{\left[ {Cu} \right]}}\]
Let substitute the values in the above equation we get,
\[ \Rightarrow {E_{anode}} = -\dfrac{{0.059}}{2}ln\dfrac{{\left[ {0.12} \right]}}{{\left[ 1 \right]}}\]
\[{E_{anode}} = 0.027V\]
We know that the potential at cathode is not equal to that at anode; the residual potential is the back emf.
${E_{back}} = \left| {{E_{cathode}}} \right|\, - \left| {{E_{anode}}} \right|$
${E_{back}} = 0.032\, - 0.027 = 5 \times {10^{ - 3}}V$
The obtained value is approximately equal to option B.
So, the correct answer is “Option B”.

Additional Information:
Now we discuss about the uses of Nernst equation as,
The Nernst equation may be accustomed to calculate Single conductor reduction or chemical reaction potential at any conditions.
Standard electrode potentials comparing the relative ability as a subtractive or aerobic agent. Finding the practicability of the mixture of such single electrodes to provide electrical potential. Emf of a chemistry cell Unknown ionic concentrations.
The \[pH\] scale of answers and solubility of meagerly soluble salts can be measured with the assistance of the Nernst equation.

Note:
Let us discuss about the limitations of Nernst Equation as,
The activity of an ion in an exceedingly} very dilute solution is on the point of eternity and can, therefore, be expressed in terms of the ion concentration. However, for solutions having very high concentrations, the ion concentration isn't up to the particle activity. So as to use the Nernst equation in such cases, experimental livements should be conducted to get actuality activity of the ion. Another defect of this equation is that it can't be accustomed to measure cell potential once there's a current flowing through the electrode. This can be as a result of the flow of current that affects the activity of the ions on the surface of the electrode. Also, further factors adore resistive loss and over potential must be thought-about when there is a current flowing through the electrode.