Chemistry/- Dr.Bob222

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?

b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M?

c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

I am stuck on this one.

1. 👍
2. 👎
3. 👁
1. 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.

1. 👍
2. 👎
2. 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

1. 👍
2. 👎

Similar Questions

1. chemistry

A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when E cell is 0.22 V.

2. Science

How many cell divisions would it take to produce at least 1,000 cells from one cell? (When one cell splits in half during cell division, the result is two new cells. Each of those cells can divide into two more and so on.) Answer

3. chemistry 2

A voltaic cell is constructed with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell. Measurement shows that thesilver electrode is positive. -Write balanced half-reactions and the overall spontaneous reaction.

4. AP CHEMISTRY

A voltaic cell is constructed that uses the following half-cell reactions. Cu+(aq) + e− -> Cu(s) I2(s) + 2 e− -> 2 I−(aq) The cell is operated at 298 K with [Cu+ ] = 2.7 M and [I− ] = 2.7 M. (a) Determine E for the cell at

1. Chemistry

The standard reduction potential for Cr3+(aq) is −0.74 V. The half-reaction for the reduction of Cr3+(aq) is the following. Cr3+(aq) + 3 e− → Cr(s) The standard reduction potential for Ni2+(aq) is −0.26 V. The

2. AP CHEMISTRY

A voltaic cell is constructed that is based on the following reaction. Sn2+(aq) + Pb(s) -> Sn(s) + Pb2+(aq) (a) If the concentration of Sn2+ in the cathode half-cell is 1.80 M and the cell generates an emf of +0.219 V, what is the

3. Chemistry

Little confusing!! I need help PLZ!!! Two half cells in a galvanic cell consist of one iron (Fe(s)) electrode in a solution of iron (II) sulphate (FeSO4(aq)) and a silver (Ag(s)) electrode in a silver nitrate solution. a. Assume

4. Chem 2

A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Al3+} when {Cd2+} = 0.401 M and Ecell = 1.299 V.Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40 V.

1. Chemistry

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.842 M and [Ni2 ] = 0.0100 M. Standard reduction potentials can be found here. reaction: Zn(s)+Ni^2+(aq)--->Zn^2+(aq)+Ni(s)

2. chem 2

Consider a chromium-silver voltaic cell that is constructed such that one half-cell consists of the chromium, Cr, electrode immersed in a Cr(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in

3. Chemistry

1) Assume that the reference half-cell is changed to a standard mercury-mercury (II) half-cell. a) What would be the reduction potential of a standard chlorine half-cell. c) What would be the cell potential of a standard

4. Chemistry

If the Fe2+(aq) + 2e---> Fe(s) was used as a reference for the electrode potential table instead of the standard hydrogen half cell what would the potential of the following half cell be? Ni2+(aq) + 2e---> Ni(s) A) -0.19 V B)