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
see above
Yes, your thought processes are correct. To solve for the diluted concentration of Sn2+ in both cells using the Nernst equation, you will need to consider the cell potentials and the concentrations of the ions involved.
For part a), the cell reaction is Sn|Sn2+(diluted)||Cu2+|Cu. Assuming that the cell potential, E*cell, is 0.30 V, you can set up the Nernst equation as follows:
E = E*cell - (0.0592 V/n) * log(Q)
Here, Q represents the reaction quotient, which is the ratio of the product concentrations to the reactant concentrations. In this case, the reactant is Sn2+ and the product is Cu2+. The Nernst equation can be rewritten as:
E = 0.30 V - (0.0592 V/2) * log([Cu2+]/[Sn2+])
Since the concentrations of Cu2+ and Sn2+ are both given as 2M, the equation becomes:
E = 0.30 V - (0.0592 V/2) * log(2/2) = 0.30 V
Therefore, the diluted concentration of Sn2+ in cell a) remains the same at 2M.
For part b), the cell reaction is Zn|Zn2+||Sn2+(diluted)|Sn. Assuming the cell potential, E*cell, is also 0.30 V, you can set up the Nernst equation as follows:
E = E*cell - (0.0592 V/n) * log(Q)
In this case, the reactant is Sn2+ and the product is Zn2+. The Nernst equation can be rewritten as:
E = 0.30 V - (0.0592 V/2) * log([Zn2+]/[Sn2+])
Since the concentrations of Zn2+ and Sn2+ are both given as 2M, the equation becomes:
E = 0.30 V - (0.0592 V/2) * log(2/2) = 0.30 V
Therefore, the diluted concentration of Sn2+ in cell b) also remains the same at 2M.
In both cases, the diluted concentration of Sn2+ does not differ from the initial concentration of 2M.