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 1.70M 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.

Answer 1

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 =  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