Question

A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.40 M and 0.130 M , respectively. The volume of half-cells is the same.

Part A: What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M ?  

Answer 1

Eo (Zn/Zn2+) = -0.76 V
E(Ni/Ni2+) = -0.25 V

reaction that will take place is:
Ni2+ + Zn ---> Zn2+ + Ni (n=2 since 2 electrons are transfered)

Eo cell = Eo cath - Eo ano
     = -0.25 - (-0.76)

use:
E cell = Eo cell - 0.059/2 log ([Zn2+]/[Ni2+] )

What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M ?

fall in Ni+2 = 1.4 - 0.5= 0.9 M

so Zn2+ = 0.13 + 0.9 = 1.03

E cell = Eo cell - 0.059/2 log ([Zn2+]/[Ni2+] )
   = 0.51 - 0.059/2 log {1.03/0.5}
   = 0.197V