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 1.70M and 0.130M, respectively. The volume of half-cells is the same. Part AWhat is the initial cell potential? Express your answer using two significant figures. Part BWhat is the cell potential when the concentration of Ni2 + has fallen to 0.500M? Express your answer using two significant figures. Part CWhat is the concentrations of Ni2 + when the cell potential falls to 0.45V? Express your answer using one significant figure. Part DWhat is the concentration of Zn2 + when the cell potential falls to 0.45V? Express your answer using two significant figures.

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 = 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 - 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 - 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] = 1.70 - x

[Zn+2] = 0.13 + x

0.13 + x/1.70 - x = 104.71

Resolving x:

x = 1.68 M

[Ni+2] = 1.70 - 1.68 = 0.02 M

d)

[Zn+2] = 0.13 + x = 0.13 + 1.68 = 1.81 M