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.

Step-by-step 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* 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* 10^(-3)M)`

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

`E_(cell)=E^o_(cell)-(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* 10^(-3)M`

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

Putting values in above equation, we get:

`E_(cell)=0-(0.0592)/(2)\log (1.0* 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.