Question

At 25$^{o}C$ , the solubility product of CuCl is $2.0\times {{10}^{-7}}$ and $E_{C{{l}^{-}}/CuCl/Cu}^{o}$ is 0.128 V.
The value of $E_{C{{u}^{+}}/Cu}^{o}$ is: [(log 2 = 0.3), (2.303 RT/F = 0.06)]
A. -0.274 V
B. -0.402 V
C. +0.53 V
D. +0.402 V

Answer 1

Hint: There is a relationship between Nernst equation and solubility product and it is as follows.
\[E={{E}^{o}}-\dfrac{RT}{F}[\log {{K}_{sp}}]\]
Here, E = potential of the cell
${{E}^{o}}$ = Potential of the copper electrode
${{K}_{sp}}$ = solubility product.
R = ideal gas constant
T = Temperature
F = Faraday
We can simplify the Nernst equation and write the Nernst equation as follows.
\[E={{E}^{o}}-0.06[\log {{K}_{sp}}]\]

Complete step by step solution:
- In the question it is given to find the electrode potential of copper electrode.
- In the question it is given that the solubility product is copper chloride is $2.0\times {{10}^{-7}}$ and Potential of the standard electrode is 0.128 V.
- Substitute all the known values in the below formula to get the potential of the copper electrode.
\[E={{E}^{o}}-0.06[\log {{K}_{sp}}]\]
Here, E = Potential of the cell = 0.128 V
${{E}^{o}}$ = potential of the copper electrode
${{K}_{sp}}$ = solubility product of the copper chloride = $2.0\times {{10}^{-7}}$
\[\begin{align}
& E={{E}^{o}}-0.06[\log {{K}_{sp}}] \\
& {{E}^{o}}=E+0.06[\log {{K}_{sp}}] \\
& {{E}^{o}}=0.128+0.06[\log 2.0\times {{10}^{-7}}] \\
& {{E}^{o}}=0.128+0.06[6.7] \\
& {{E}^{o}}=0.53V \\
\end{align}\]
- From the above calculations we can say that the potential of the copper electrode is 0.53 V.

So, the correct option is C, 0.53 V.

Note: Always in any cell the potential of the total cell should be positive then only the cell exists. If the potential of the cell is negative we can say that the cell won’t exist. Generally simplified Nernst is very useful to calculate the potentials of the unknown electrodes in a cell.