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

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.

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.
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•[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

To answer part c, we can use the Nernst equation to relate the cell potential to the concentrations of the ions involved.

The Nernst equation is given by:

E = E° - (RT/nF) * ln(Q)

where E is the cell potential, E° is the standard cell potential, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, n is the number of electrons transferred (the stoichiometric coefficients of the balanced equation), F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

In this case, since we are given that the reaction is happening at 25 °C, we can assume that the temperature is 298 K. The balanced equation for the cell reaction is:

Zn(s) + Ni2+(aq) -> Zn2+(aq) + Ni(s)

The half-cell reactions for the Zn and Ni electrodes are:

Zn(s) -> Zn2+(aq) + 2e-
Ni2+(aq) + 2e- -> Ni(s)

From the balanced equation, we can see that the number of electrons transferred (n) is 2.

To find the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V, we can rearrange the Nernst equation as follows:

ln(Q) = (E° - E)/(RT/nF)

ln(Q) = (0.45 V - 0.56 V)/[(8.314 J/mol·K)/(2 * 96485 C/mol) * 298 K]

ln(Q) = -0.1/(0.008314 J/mol·K) * 298 K

ln(Q) ≈ -36.8817

Therefore, Q ≈ e^(-36.8817) = 1.4563 x 10^(-16)

Now, we can write the expression for Q using the concentrations of Ni2+ and Zn2+:

Q = [Ni2+]/[Zn2+]

Substituting the initial concentrations, we have:

1.50 M / (0.10 M) ≈ 15

So, when the cell potential falls to 0.45 V, the ratio of [Ni2+] to [Zn2+] is approximately 15.

If you need more clarification or have further questions, feel free to ask!

To answer part c of the question, you can use the Nernst equation, which relates the cell potential to the concentrations of the species involved in the half-cell reactions. The Nernst equation is given by:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the gas constant (8.314 J/(mol⋅K))
T is the temperature in Kelvin
n is the number of electrons transferred in the balanced equation for the overall cell reaction
F is Faraday's constant (96485 C/mol)
Q is the reaction quotient, which is the ratio of the concentrations of the products over the concentrations of the reactants, raised to the power of their stoichiometric coefficients.

In this particular case, the initial cell potential (E°cell) is 0.56 V, and the concentrations of Ni2+ and Zn2+ are changing as the cell operates.

To find the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V, we need to rearrange the Nernst equation to solve for Q:

Q = e^((E°cell - Ecell) * nF/RT)

Substituting the given values into the equation, we have:

Q = e^((0.56 - 0.45) * 2 * 96485 / (8.314 * 298))

Using the above equation, calculate Q, and then set it equal to the concentrations of the products divided by the concentrations of the reactants, raised to the power of their stoichiometric coefficients. Then, solve for the unknown concentrations.