Calculate the emf of the cell in which the following reactions takes place:

Ni(s) + 2Ag⁺(0.002M) ------- Ni²⁺(0.160M) +2Ag(s)

Given that E⁰ _(cell) = 1.05 V

Dear Student,

We have ,

[Ag⁺ ] = 0.002 M

[Ni⁺²] = 0.160 M

n =2

E_cell = E⁰_(cell)  - 0.059/ n log [Ni⁺² ]/ [Ag⁺ ]²

Her, no. of electrons involved , n =2 

So, E_cell = E⁰_(cell)  - (0.059/ 2) log [Ni⁺² ]/ [Ag⁺ ]²

E_cell = 1.05 - 0.059/ 2 log [0.160 ]/ [0.002 ]²  =1.05 -0.0295 x log 4 x 10⁴ = 1.05 -0.0295(4.6021)= 0.91 V