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

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

i believe it is a.) released into

explanation: