Question

A concentration cell consists of two Zn/Zn2+ half-cells. The concentration of Zn2+ in one of the half-cells is 2.0 M and the concentration in the other half-cell is 1.0*10^-3 .Indicate the half-reaction occurring at each electrode.

Answer 1

Answer: The EMF of the cell is coming when the cell having diluted concentration is getting oxidized.

Explanation:

We are given a cell which contains two $Zn/Zn^{2+}$ half cells. This means that the standard electrode potential of the cell will be 0.

For a reaction to be spontaneous, the EMF of the cell must be positive. If the EMF of the cell is negative, the reaction will be non-spontaneous and will not take place.

For a reaction to be spontaneous, the diluted cell must get oxidized.

The half reaction for the given cell follows:

Oxidation half reaction:  $Zn\rightarrow Zn^{2+}(1.0\times 10^{-3}M)+2e^-$

Reduction half reaction:  $Zn^{2+}(2.0M)+2e^-\rightarrow Zn$

Net reaction:  $Zn^{2+}(2.0M)\rightarrow Zn^{2+}(1.0\times 10^{-3}M)$

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

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

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Zn^{2+}_{\text{(diluted)}}]$ = $1.0\times 10^{-3}M$

$[Zn^{2+}_{\text{(concentrated)}}]$ = 2.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.0\times 10^{-3}M}{2.0M}$

$E_{cell}=0.097V$

Hence, the EMF of the cell is coming when the cell having diluted concentration is getting oxidized.