Question

E°_cell  for the given redox reaction is 2.71 V 

Mg(s) Cu²⁺ (0.01M)→ Mg²⁺ (0.001M) Cu(s) 

Calculate E_cell for the reaction. Write the direction of flow of current when an external opposite potential applied is 

(a) less than 2.71 V and 

(b) greater than 2.71 V

Answer 1

According to Nernst equation

E_cell=E^º_cell—\(\frac{2.303RT}{nF}\)log\(\frac{C_{2}}{C_{1}}\)

Substituting the values E^º_cell =2.71, n=2,F=96500 C, 

C₂=0.001M,C₁=0.01 M in Eq. (1),

we get

E_cell=2.71-\(\frac{2.303\times8.314\times298}{2\times96500}\)

log\(\frac{0.001}{0.01}\)=2.74 V

(a) When an external opposite potential is applied less than 2.71 V the direction of flow of current would remain same.

(b) When an external opposite potential is applied more than 2.71 V the direction of flow of current would be reversed.