Represent the cell made up of the following cell reactions:

Mg>>>>Mg²⁺ (0.01 M)+2e E⁰ =+2.34 V

Sn⁺² (0.1 M) +2e>>>>Sn E⁰ =-0.136 V

Calculate the emf of the above cell at 25⁰ C

Dear Student,

Let us consider the reaction occurring at anode and cathode first;

Oxidation (loss of electron) occurs at anode, and reduction (gain of electrons) occurs at cathode. Therefore,

At anode:

Mg → Mg²⁺ + 2 e^-  E^o = +2.34 V

At cathode:

Sn²⁺ + 2 e^- → Sn  E^o = -0.136 V

Therefore cell can be represented as :

Mg|Mg²⁺(0.01 M) || Sn²⁺(0.1 M)|Sn

Now, the E.M.F of the cell can be calculated, with the help of Nernst equation:

We are given that,

[Mg²⁺] = 0.01 M

[Sn²⁺] = 0.1 M

Note: The sign of electrode potential given for the reaction Mg → Mg²⁺ + 2 e^- is reversed, during calculation, because we always consider the reduction electrode potential i.e., Mg²⁺ + 2 e^- → Mg which will be negative of oxidation potential.

Hope its clear now.

Enjoy learning!!