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

Given that [Cu2+] and [Zn2+] are both 2M.

Note: I know that the nernst equation is
E= E*cell=(-0.0592 V/n)log Q

where Q in this case are [Sn2+]/[Cu2+] and [Zn2+]/[Sn2+].

I think that the E*cell = 0.30 for a)... but I am unsure of what the number for E is... I think the goal for part a) is to solve for [Sn2+] diluted and that in order to solve for it, the whole equation needs to be set equal to a number E that I am unsure of....

Are my thought processes correct?

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

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

I read your earlier post but fail to understand what you've written. Is E = E*cell = ---- or should that be

E = Eocell - (0.0592/n)*log Q?

Based upon the data given, this problem is indeterminate. One must have the non-standard cell potential for each cell in order to determine the cation concentrations of interest. Without the non-standard cell potentials, there will be two unknowns in each reaction and the reactions can't be couples through the non-standard potential because they are different values.

From standard reduction potential tables,
For [Sn⁰(s)│Sn⁺²(aq)ǁCu⁺²(aq)│Cu⁰(s)]
Half-Cell 1 => Sn⁰(s) <=> Sn⁺²(aq)+ 2e¯; E⁰ = - 0.14v (oxidation half-rxn)
Half-Cell 2 => Cu⁺²(aq) + 2e¯ <=> Cu; E⁰ = +0.34v (reduction half-rxn)
Standard Cell Potential E⁰(Sn/Cu) = E⁰(Cu) - E⁰(Sn) = (+0.34v) – (-0.14v) = 0.48v

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)logQ([Zn⁺²]/[Sn⁺²])
E(Sn/Cu) = (0.62v) – (0.0592/n)log([2.0M]/[Zn⁺²])

In order to calculate the Stannous (Tin) concentration in each cell one needs the Non-Standard Cell Potentials for each Reaction. That is E(Sn/Cu) @ ([Sn⁺²]/[2.0M]), and E(Zn/Sn) @ ([2.0M]/[Zn⁺²]).

Your thought processes are correct. The Nernst equation relates the cell potential (E*cell) to the concentrations of the redox species involved in the cell reaction.

Let's start with part a) first, involving the Sn| Sn2+(diluted)||Cu2+|Cu cell.

1. The balanced half-reactions for this cell are:

Sn2+(diluted) + 2e- -> Sn (reduction at the cathode)
Cu -> Cu2+ + 2e- (oxidation at the anode)

2. The standard cell potential (E*cell) for this cell is given as 0.30V.

3. The Nernst equation for this cell is written as:
E*cell = E - (0.0592 V/n)log(Q)

4. Since the cell reaction is occurring at 25 degrees Celsius, we can use n = 2 because 2 moles of electrons are involved in the balanced half-reactions.

5. For the given concentrations [Cu2+] and [Sn2+] as 2M, the Q value for this cell can be written as Q = [Sn2+]/[Cu2+].

6. To find the diluted concentration of Sn2+, we need to solve the Nernst equation for [Sn2+]. Rearranging the equation gives:
E = E*cell + (0.0592 V/n)log(Q)

7. Now we substitute the known values into the equation:
E = 0.30V + (0.0592 V/2)log([Sn2+]/[Cu2+])

8. Rearranging the equation to solve for [Sn2+]:
log([Sn2+]/[Cu2+]) = (E - E*cell) / (0.0592 V/2)

9. Taking the antilog of both sides:
[Sn2+]/[Cu2+] = 10^((E - E*cell) / (0.0592 V/2))

10. Finally, substitute the known values and calculate the diluted concentration [Sn2+].

Now you can follow the same process for part b) involving the Zn| Zn2+ ||Sn2+(diluted)|Sn cell, using the appropriate values and equations.

Yes, you are on the right track with your thought processes. Let's go through the steps to determine the diluted concentration for both cells using the Nernst equation.

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

To determine the diluted concentration of Sn2+, we can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the concentrations of the species involved. In this case, the cell potential (Ecell) is equal to the standard cell potential (E°cell) plus the product of the natural logarithm of the reaction quotient (Q) and the Faraday constant (F) divided by the number of moles of electrons transferred (n).

The Nernst equation for this cell is:

Ecell = E°cell - (0.0592 V/n) * log(Q)

Given that [Cu2+] and [Zn2+] are both 2M, we need to determine the value of the reaction quotient (Q), which is the ratio of the concentrations of Sn2+ to Cu2+. Since the diluted concentration of Sn2+ is what we are trying to find, let's set it as x. Therefore, the expression for Q becomes [x]/[2M].

Substituting the values into the Nernst equation:

Ecell = E°cell - (0.0592 V/n) * log([x]/[2M])

Given that E°cell = 0.30 V, we can further simplify the equation:

Ecell = 0.30 V - (0.0592 V/n) * log([x]/[2M])

Now we need to solve for [x], the diluted concentration of Sn2+. To do this, we need to know the value of Ecell. So, you're correct in saying that we need to set the whole equation equal to a specific value of E. This value could be given or it could be a specific condition you are considering.

Once you have the value of Ecell, you can rearrange the equation to solve for [x].

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

Similarly, for this cell, we can use the Nernst equation to determine the diluted concentration of Sn2+.

The Nernst equation for this cell is:

Ecell = E°cell - (0.0592 V/n) * log(Q)

The reaction quotient (Q) in this case is the ratio of the concentrations of Zn2+ to Sn2+. Since the diluted concentration of Sn2+ is what we are trying to find, let's set it as x. Therefore, the expression for Q becomes [2M]/[x].

Substituting the values into the Nernst equation:

Ecell = E°cell - (0.0592 V/n) * log([2M]/[x])

If you have a specific value of Ecell, you can similarly rearrange the equation to solve for [x].

Remember, in both cases, you will need specific values of Ecell to solve for the diluted concentrations of Sn2+.