Using the Nernst equation, determine the diluted concentration for both cells. Show your work.
Sn| Sn2+(diluted)||Cu2+|Cu
Zn| Zn2+ ||Sn2+(diluted)|Sn
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
To determine the diluted concentration for both cells using the Nernst equation, we need to follow these steps:
1. Identify the half-reactions for each cell:
- Cell 1: Sn -> Sn2+
- Cell 2: Zn -> Zn2+
2. Write the balanced equations and calculate the standard electrode potential (E°) for each half-reaction.
- Cell 1: Sn -> Sn2+ + 2e- (E°1)
- Cell 2: Zn -> Zn2+ + 2e- (E°2)
The values for E°1 and E°2 can be found in standard electrode potential tables.
3. Determine the overall reaction for each cell.
- Cell 1: Sn + 2Cu2+ -> Sn2+ + 2Cu
- Cell 2: Zn + Sn2+(diluted) -> Zn2+ + Sn
4. Calculate the cell potential (E) for each cell using the Nernst equation:
- Cell 1: E1 = E°1 - (RT/nF) ln([Sn2+]/[Cu2+])
- Cell 2: E2 = E°2 - (RT/nF) ln([Zn2+]/[Sn2+(diluted)])
In these equations, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred (2 in both cases), F is the Faraday constant, [Sn2+] is the diluted concentration of Sn2+ in cell 1, and [Zn2+] is the concentration of Zn2+ in cell 2.
5. Set up an equation comparing the cell potentials of both cells:
- E1 = E2
6. Solve the equation for [Sn2+(diluted)]:
- [Sn2+(diluted)] = [Cu2+] * e^((E°1 - E°2 + (RT/nF) ln([Zn2+]/[Sn2+(diluted)])) / (RT/nF))
Note: Remember to convert temperature to Kelvin and use the appropriate units for R and F.
Plug in the given concentrations or values and solve the equation to find the diluted concentration for both cells.
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])