Answer:
a) E = 0.477 V
b) E = 0.502 V
c) 0.02 M = [Ni+2]
d)[Zn+2] = 1.81 M
Explanation:
having the following reactions of each cell:
Zn =⇒  Zn+2  +  2e-   +0.76
Ni+2  +  2e- =⇒  Ni    -0.25
Zn  +  Ni+2  ==⇒  Ni   +   Zn+2   Eo = 0.51
a)
The number of electrons being transferred is 2, therefore n = 2 in the Nernst equation
E = Â Eo - 0.0592/2*log [Zn+2]/[Ni+2] = 0.51 - 0.0592/2*log[0.13/1.7] = 0.477 V
b)
using the formula above:
E = 0.51 Â - 0.0296*log [Zn+2]/[Ni+2] = 0.51 - 0.0296*log((0.13+0.5)/(1.7-0.5)) = 0.502 V
c)
using the formula above:
0.45 = 0.51 Â - 0.0296*log[Zn+2]/[Ni+2]
-0.06/-0.0296 = log[Zn+2]/[Ni+2]
2.02 = log [Zn+] / [Ni+2]
104.71 = [Zn+2] / [Ni+2]
x = change in [Ni+2]
[Ni+2] Â = 1.70 - Â x
[Zn+2] = Â 0.13 + Â x
0.13 + x/1.70 –x =  104.71
Resolving  x:
x = 1.68 M
[Ni+2] Â = 1.70 - Â 1.68 = 0.02 M
d)
[Zn+2] = Â 0.13 + Â x = 0.13 + 1.68 = 1.81 M