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-  Eo = +2.34 V

At cathode:

Sn2+ + 2 e- → Sn  Eo = -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!!

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