Question

Calculate the $E$ value for the cell reaction:
$Cu|C{{u}^{2+}}(0.001M)||C{{u}^{2+}}(0.1M)|Cu$

Answer 1

Hint: Nernst equation is used to calculate the cell potential during a reaction. It relates the reduction potential to the standard electrode potential, temperature and the concentration of product and reactant. This reaction is carried out in an electrochemical cell.

Formula used:
${{E}_{cell}}=E{}^\circ -2.303\dfrac{RT}{nF}\log \dfrac{[P]}{[R]}$
where, ${{E}_{cell}}$ is the cell potential, $E{}^\circ $ is the standard cell potential, $R$ is the gas constant, $T$ is the temperature in Kelvin, $F$ is the Faraday’s constant and $n$ is the exchange of electrons in a reactor.

Complete step by step answer:
Here, it is given that the concentration of $C{{u}^{2+}}$ at anode is $0.001M$ and concentration of $C{{u}^{2+}}$ at cathode is $0.1M$ . Here, the standard cell potential $(E{{{}^\circ }_{cell}})=0$ because it is a concentration cell that consists of two equivalent half cells of the same composition but differs only in concentration.
Now let us see the overall reaction of the cell.
$C{{u}_{(A)}}+Cu_{(C)}^{2+}\to C{{u}_{(A)}}+C{{u}_{(C)}}$
where, $A$ is the anode and $C$ is the cathode.
Now, let us write the Nernst equation.
${{E}_{cell}}=E{}^\circ -2.303\dfrac{RT}{nF}\log \dfrac{[P]}{[R]}$
where, ${{E}_{cell}}$ is the cell potential, $E{}^\circ $ is the standard cell potential, $R$ is the gas constant, $T$ is the temperature in Kelvin, $F$ is the Faraday’s constant and $n$ is the exchange of electrons in a reactor.
$R=0.0821L\,atm\,\,mo{{l}^{-1}}{{K}^{-1}}$
$T=398K$
$F=96500\,C\,mo{{l}^{-1}}$
\[{{E}_{cell}}=E{}^\circ -2.303\dfrac{RT}{nF}\log \dfrac{[Cu_{(A)}^{2+}][C{{u}_{(C)}}]}{[C{{u}_{(A)}}][Cu_{(C)}^{2+}]}\]
$C{{u}_{(C)}}=C{{u}_{(A)}}$
Now, substituting the values in the above formula, we get,
\[{{E}_{cell}}=0-\dfrac{2.303\times 0.0821\times 298}{2\times 96500}\log \dfrac{0.001}{0.1}\]
On further solving, we get,
$\Rightarrow {{E}_{cell}}=0-0.029\times \log 0.01$
As, $\log 0.01=-2$
\[\Rightarrow {{E}_{cell}}=-0.029\times -2\]
$\Rightarrow {{E}_{cell}}=0.058$

Therefore, the cell potential ($E$ value ) of this reaction is equal to $0.058$ .

Note: Cell potential is defined as the measure of potential difference between two cells in an electrochemical cell. It is used to measure voltage between the two half cells.
Nernst equation is defined as the equation which is used to calculate the cell potential.
In this equation, electrochemical cells are used, where oxidation takes place at one electrode and reduction takes place at another electrode. The electrode where oxidation takes place is anode and electrode where reduction takes place is cathode.
Faraday’s constant is used to represent the magnitude of electric charge per mole of electrons.
$1F=96500\,C\,mo{{l}^{-1}}$
Its SI unit is Coulombs per mol.

At $298K$ , the value of $\dfrac{RT}{F}$ is $0.0591$ .
To calculate the number of moles, you need to see the number of exchange of electrons. Here, $2{{e}^{-}}$ are exchanged.
Standard cell potential is a potential under $298K$, $1\,atm$ pressure and $1$ mole per litre concentration.