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
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,
Mg → Mg2+ + 2 e- Eo = +2.34 V
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.