Using the Nernst equation, determine the diluted concentration for both cells. Show your work.

Question

Sn| Sn2+(diluted)||Cu2+|Cu

Zn| Zn2+ ||Sn2+(diluted)|Sn

Answer 1

I forgot to include that the above problem is accompanied with a virtual lab that would help determine E.
However, I don't know how to predict the value of E based on the virtual lab.

PS: I don't know how to post a link on this forum and so in order to visit the virtual lab website, follow these directions:

1. Type in "Chem collective" in the search bar

2. Click on the very first website

3. At the Resources by topic side bar, click on Oxidation/Reduction and Electrochemistry

4. Under the Galvanic Cell tab, click on Electrochemistry: Galvanic Cells and the Nernst Equation

5. At the left side bar under step 4, click on Practice with nonstandard cells

Answer 2

If you can figure out the non-standard cell potentials for your problem, then substitute in the Nernst Equations at the end of this dialog...

For [Zn⁰(s)│Zn⁺²(aq)ǁSn⁺²(aq)│Sn⁰(s)]
Half-Cell 1 => Zn⁰(s) => Zn⁺²(aq) +2e¯; E⁰ = - 0.0.76v (oxidation half-rxn)
Half-Cell 2 => Sn⁺²(aq)+ 2e¯ => Sn⁰(s); E⁰ = - 0.14v (reduction half-rxn)
Standard Cell Potential E⁰(Zn/Sn) = E⁰(Sn) - E⁰(Zn) = (-0.14v) – (-0.76v) = 0.62v

E = E⁰ - (0.0592/n)log([Oxid’n Cation]/[Red’n Cation])
E = Non-Standard Cell Potential
E⁰ = Standard Cell Potentials = E⁰(Reduction) - E⁰(Oxidation)
n = electron transfer in balanced oxidation-reduction reactions

Given: [Cu⁺²] and [Zn⁺²] are both 2M

Case I (Sn/Cu) Cell:
E(Sn/Cu) = E⁰(Sn/Cu) – (0.0592/n)logQ(c)
E(Sn/Cu) = E⁰(Sn/Cu) – (0.0592/n)logQ([Sn⁺²]/[Cu⁺²])
E(Sn/Cu) = (0.62v) – (0.0592/n)log([Sn⁺²]/[2.0M])
Case II (Zn/Sn) Cell:
E(Zn/Sn) = E⁰(Zn/Sn) – (0.0592/n)logQ(c)
E(Sn/Cu) = E⁰(Zn/Sn) – (0.0592/n)log([Zn⁺²]/[Sn⁺²])
E(Sn/Cu) = (0.62v) – (0.0592/n)log([2.0M]/[Zn⁺²])