Question

Write the Nernst equation and the emf of the following cell at 298K:
Mg(s)/Mg2+(0.001M) ∕∕ Cu2+(0.0001M) ∕ Cu(s)
Given, E0Mg2+/Mg = -2.37 V; E0Cu2+/Cu = + 0.34 V​

Answer 1

Explanation:

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Answer 2

Answer: Emf of the cell = 2.68 V

Explanation:

Chemical equation involved for above cell is

Mg + Cu2+ ------> Mg2+ + Cu

from above

Cathode rxn : Cu2+ ----> Cu

Anode rxn : Mg -----> Mg2+

Nernst eqn at 298 K

Ecell = E°cell -.0591/n ×log[p]/[r]

( p= product , r= reactant )

Here , n = 2 ( loss and gain of 2 electrons in the above cell reaction )

Nernst equation for above cell

Ecell = E°cell -0.0591/2 × log [Mg2+][Cu]/ [Mg][Cu2+] .......1

E°cell = E°cathode - E°anode

 E° Cu2+/Cu - E° Mg2+/Mg

 +0.34 – (– 2.37) V

 = 2.71 V

Substituting the values in equations (1)

Ecell = 2.71 – 0.0591/2×log [0.001][ 1] / [1][0.0001]

after calculating, and taking log10=1

Ecell = 2.68 V (approx)