Zn + Ni^++ ==> Zn^++ + Ni
Ecell = Eocell - (0.05916/n)*Log Q where Q (I didn't have space to write it is)
(Zn^2+)(Ni)/(Ni^2+)(Zn)
You know of course that (Ni) and (Zn) = 1 since they are solids. Initially (Ni^2+) = 1.5*(Zn^2+) so at the end, since both are divalent, that ratio still will exist. Substitute the numbers into the equation and for (Ni^2+) put 1.5x and for (Zn^2+) put x and solve.
We will never agree on an answer if you don't show what you used for Eo values. I'm sure the tables in your book are not the same as in my books.
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 1.50 M and 0.10 M, respectively.
a. What is the initial cell potential?
My answer: 0.56 B
b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M?
My answer: 0.55 V
c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?
I am stuck on this one.
My answer
2 answers
Hope ya'll don't mind, but I have a little bit of a different approach to this kind of problem... Anyways, here's my take...
The answer for ‘a’ is correct, but for ‘b’, I compute a different results… Here’s my take on these problems… Hope it adds something that helps…
Zn⁰(s) + Ni⁺²(aq) <=> Zn⁺²(aq) + Ni⁰(s)
[Zn⁰(s)/ Zn⁺²(0.10M) Ni⁺²(1.5M)/Ni⁰(s)]
E = E⁰ - (0.0592/n)log([RA]/[OA])
->[RA] = Reducing Agent = [Zn⁺²]
->[OA] = Oxidizing Agent = [Ni⁺²]
From a Table of Standard Reduction Potentials
E⁰ = E⁰(OA) - E⁰(RA) = E⁰(Ni⁺²) - E⁰(Zn⁺²) = (-0.23v) – (-0.76v) = 0.53v (Standard Cell Potential)
a.Given:
E = E⁰ - (0.0592/n)log([RA]/[OA])
= E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v(0.0592/2)log[(0.10M)/(1.5M)]
= 0.56v
b.Before calculating E for [Ni⁺²], I determined the concentrations of reagents at would give the standard cell potential from the Nernst Equation… I found that 0.80M [Zn⁺²] and 0.80M[Ni⁺²] gave the exact E⁰-value using the Nernst Equation.
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²]) = 0.53v -(0.0592/2)log[(0.80M)/(0.80M)]
= 0.53v
The significance of this is that the non-standard cell potentials calculated using the Nernst Equation will be above the E⁰ if [RA] > [OA], and below E⁰ if [RA] < [OA]. Using [Ni⁺²] = 0.500M and [Zn⁺²] = 1.10M*
Here’s my calculation:
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v – (0.0593/2)log[(1.10M)/(0.50M)]
= 0.53v – 0.01v = 0.52v (slightly below the standard E^o value.
------------------------------
•[Zn⁺²] is determined by noting the change in concentration of Ni⁺² from 0.80M to 0.50M; i.e., a decrease of 0.30M = 0.80M – 0.50M change. Because the cell reagents react in a one to one reaction ratio, the amount of Zn⁺² delivered into the anodic cell solution also is 0.30M and thus brings the [Zn⁺²] = 1.10M. Since [Zn⁺²] = 1.10M is greater than [Ni⁺²] = 0.50M concentration then the non-standard cell potential should be below the standard cell potential of 0.53v; not above as your answer indicates in problem ‘b’.
Here’s my calculation:
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v – (0.0593/2)log[(1.10M)/(0.50M)]
= 0.53v – 0.01v = 0.52v
c.What are the [Zn⁺²] and [Ni⁺²] concentrations for a 0.45v Galvanic Cell Potential? I must admit, that my solution on this was by trial and error to determine the best ratio consistent with the rate change in cell ion concentrations. I found that if [Zn⁺²] = 1.598M and [Ni⁺²] = 0.0025M, the non-standard cell potential would be 0.45v, but this is again based upon trial and error.
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v – (0.0592/2)log(1.598M)/(0.0025M)]
= 0.53v – 0.08v = 0.45v
The answer for ‘a’ is correct, but for ‘b’, I compute a different results… Here’s my take on these problems… Hope it adds something that helps…
Zn⁰(s) + Ni⁺²(aq) <=> Zn⁺²(aq) + Ni⁰(s)
[Zn⁰(s)/ Zn⁺²(0.10M) Ni⁺²(1.5M)/Ni⁰(s)]
E = E⁰ - (0.0592/n)log([RA]/[OA])
->[RA] = Reducing Agent = [Zn⁺²]
->[OA] = Oxidizing Agent = [Ni⁺²]
From a Table of Standard Reduction Potentials
E⁰ = E⁰(OA) - E⁰(RA) = E⁰(Ni⁺²) - E⁰(Zn⁺²) = (-0.23v) – (-0.76v) = 0.53v (Standard Cell Potential)
a.Given:
E = E⁰ - (0.0592/n)log([RA]/[OA])
= E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v(0.0592/2)log[(0.10M)/(1.5M)]
= 0.56v
b.Before calculating E for [Ni⁺²], I determined the concentrations of reagents at would give the standard cell potential from the Nernst Equation… I found that 0.80M [Zn⁺²] and 0.80M[Ni⁺²] gave the exact E⁰-value using the Nernst Equation.
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²]) = 0.53v -(0.0592/2)log[(0.80M)/(0.80M)]
= 0.53v
The significance of this is that the non-standard cell potentials calculated using the Nernst Equation will be above the E⁰ if [RA] > [OA], and below E⁰ if [RA] < [OA]. Using [Ni⁺²] = 0.500M and [Zn⁺²] = 1.10M*
Here’s my calculation:
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v – (0.0593/2)log[(1.10M)/(0.50M)]
= 0.53v – 0.01v = 0.52v (slightly below the standard E^o value.
------------------------------
•[Zn⁺²] is determined by noting the change in concentration of Ni⁺² from 0.80M to 0.50M; i.e., a decrease of 0.30M = 0.80M – 0.50M change. Because the cell reagents react in a one to one reaction ratio, the amount of Zn⁺² delivered into the anodic cell solution also is 0.30M and thus brings the [Zn⁺²] = 1.10M. Since [Zn⁺²] = 1.10M is greater than [Ni⁺²] = 0.50M concentration then the non-standard cell potential should be below the standard cell potential of 0.53v; not above as your answer indicates in problem ‘b’.
Here’s my calculation:
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v – (0.0593/2)log[(1.10M)/(0.50M)]
= 0.53v – 0.01v = 0.52v
c.What are the [Zn⁺²] and [Ni⁺²] concentrations for a 0.45v Galvanic Cell Potential? I must admit, that my solution on this was by trial and error to determine the best ratio consistent with the rate change in cell ion concentrations. I found that if [Zn⁺²] = 1.598M and [Ni⁺²] = 0.0025M, the non-standard cell potential would be 0.45v, but this is again based upon trial and error.
E = E⁰ - (0.0592/2)log([Zn⁺²)/[Ni⁺²])
= 0.53v – (0.0592/2)log(1.598M)/(0.0025M)]
= 0.53v – 0.08v = 0.45v
3 answers