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 Mand 0.130 M , respectively.

What is the initial cell potential?

What is the cell potential when the concentration of Ni2+ has fallen to 0.600 M ?

What is the concentrations of Ni2+when the cell potential falls to 0.46 V?

What is the concentration of Zn2+when the cell potential falls to 0.46 V?

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)
     = 0.51 V

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

a)
What is the initial cell potential?
E cell = Eo cell - 0.059/2 log ([Zn2+]/[Ni2+] )
   = 0.51 - 0.059/2 log {0.13/1.4}
   = 0.54 V

b)
What is the cell potential when the concentration of Ni2+ has fallen to 0.600 M ?
fall in Ni+2 = 1.4 - 0.6 = 0.8 M
so Zn2+ = 0.13 + 0.8 = 0.93
E cell = Eo cell - 0.059/2 log ([Zn2+]/[Ni2+] )
   = 0.51 - 0.059/2 log {0.93/0.6}
   = 0.49 V

c)
lets decrease in Ni2+ be x and increase in Zn2+ be x

E cell = Eo cell - 0.059/2 log ([Zn2+]/[Ni2+] )
0.46 = 0.51 - 0.059/2 log {0.13+x/1.4-x}
log {0.13+x/1.4-x} = 1.695
0.13+x/1.4-x = 49.5
0.13+x = 49.5 (1.4-x)
0.13+x = 69.3 - 49.5x
x= 1.36 M

Concetration of Ni2+ = 1.4-1.37 = 0.03 M
Concentration of Zn2+ = 0.13+1.37 = 1.5 M