Question

A concentration cell was constructed by placing two identical copper
electrodes in two Cu+2 ions solutions. If the concentrations of two Cu+2
solutions are 3.0 M and 0.038 M, calculate the potential of the cell at
29°C.​

Answer 1

Answer: The potential of the cell is 0.056 V

Explanation:

We are given:

$Cu(s)/Cu^{2+}(0.038M)||Cu^{2+}(3.0M)/Cu(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(0.038M)+2e^-$

Reduction half reaction: $Cu^{2+}(3.0M)+2e^-\rightarrow Cu(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}(diluted)]}{[Cu^{2+}(concentrated)]}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = 0 V

n = number of electrons exchanged = 2

$[Cu^{2+}(diluted)]=0.038M$

$[Cu^{2+}(concentrated)]=3.0M$

Putting values in above equation, we get:

$E_{cell}=0.0-\frac{0.059}{2}\times \log(\frac{0.038}{3.0})\\\\E_{cell}=0.056V$

Hence, the potential of the cell is 0.056 V