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?
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
2 answers
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]).
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]).
2 answers