A voltaic cell is based on the following two half reactions:
Fe2+(aq) + 2 e¯ ---> Fe(s); Eo = -0.44 V
Cu2+(aq) + 2 e¯ ---> Cu(s); Eo = +0.34 V
When [Cu2+] = 0.0500 M, Ecell = + 0.80 V.
What is the [Fe2+] for these conditions?
Cu^2+(aq) + 2e ==> Cu(s) Eo = +0.34
Fe(s) ==> Fe^2+(aq) + 2e Eo = +0.44
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Cu^2+(aq) + Fe(s) =>Cu(s) + Fe^2+(aq)
Eo = 0.34 + 0.44 = 0.78v
Ecell = Eocell + (0.0592/n)*logQ
You know Eocell from above, Ecell from the problem, n = 2,
Q = (Fe^2+)(Cu)/(Fe)(Cu^2+) and all of this makes Q the only unknown.
In Q,[Fe^2+] is the unknown, (Cu) and (Fe)(solids) = 1 by definition, (Cu^2+) is given in the problem. Solve for Fe^2+.
To determine the concentration of Fe2+ ions ([Fe2+]) for these conditions, we can use the Nernst equation:
Ecell = E°cell - (0.0592 V/n) * log(Q)
where:
- Ecell is the cell potential under nonstandard conditions,
- E°cell is the standard cell potential,
- n is the number of electrons transferred in the balanced equation,
- Q is the reaction quotient.
In this case, n is 2 since two electrons are transferred in both the half-reactions. The reaction quotient (Q) can be calculated using the concentrations of the species involved in the half-reactions:
Q = [Fe2+]/[Cu2+]
Let's substitute the known values into the Nernst equation:
0.80 V = (0.34 V - (-0.44 V)) - (0.0592 V/2) * log([Fe2+]/0.0500 M)
Simplifying further:
0.80 V = 0.78 V - 0.0296 V * log([Fe2+]/0.0500 M)
Rearranging the equation:
0.0296 V * log([Fe2+]/0.0500 M) = 0.02 V
Taking the antilog of both sides:
([Fe2+]/0.0500 M) = 10^(0.02 V / 0.0296 V)
Now we can solve for [Fe2+] by multiplying both sides by 0.0500 M:
[Fe2+] = (10^(0.02 V / 0.0296 V)) * 0.0500 M
Using a calculator, evaluate the right-hand side of the equation, and you will get the concentration of [Fe2+] in M.