Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.830 M and [Sn2 ] = 0.0140 M. Standard reduction potentials can be found here.

Mg(s) + Sn2+ (aq) <--> Mg2+ (aq) + Sn(s)

Standard... Mg2+(aq) + 2e– → Mg(s.....–2.38

Sn4+(aq) +2e– → Sn2+(aq)....+0.151

Answer 1

we know that

oxidation takes place at anode

and

reduction takes place at cathode

so

oxidation : anode

Mg ----> Mg+2

reduction : cathode

Sn+2 ----> Sn

now

Eo cell = Eo cathode - Eo anode

Eo cell = 0.151 + 2.38

Eo cell = 2.531

now

consider the cell equation

Mg (s) + Sn+2 ----> Mg+2 + Sn (s)

the reaction quotient is given by

Q = [Mg+]2 / [Sn+2]

according to nernst equation

E = Eo - ( 0.05916 / n) log Q

E = Eo - ( 0.05916 / n) log [Mg+2] / [Sn+2]

here

n = 2 as two electrons are transferred

so

E = 2.531 - ( 0.05916 / 2) log [ 0.83 / 0.014]

E = 2.47855

so

the cell potential is 2.47855 V