Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
If Ecell = 0.780 V and [Fe2+] = 1.0 M, what is [Cu2+]?
Ecell = Eocell - (0.05916/n)log[(Fe^2+/Cu2+)]
To find the concentration of Cu2+ in the given reaction, we can use the Nernst equation. The Nernst equation relates the standard cell potential (E°cell) with the actual cell potential (Ecell) under non-standard conditions.
The Nernst equation is given by:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the actual cell potential
- E°cell is the standard cell potential
- n is the number of electrons transferred in the balanced chemical equation
- Q is the reaction quotient
In this case, the equation is balanced as follows:
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
Since the reaction is written as a reduction half-reaction for Cu2+ and an oxidation half-reaction for Fe, the number of electrons transferred is 2.
Using the given Ecell value of 0.780 V and the known [Fe2+] concentration of 1.0 M, we can rearrange the Nernst equation to solve for log(Q):
log(Q) = (E°cell - Ecell) / (0.0592/n)
Plugging in the values:
log(Q) = (0.780 V - Ecell) / (0.0592/2)
Now, we need to rearrange the original balanced equation to match the format in the Nernst equation. The reaction becomes:
Cu2+(aq) + 2e- Cu(s)
So, Q = [Cu2+].
By substituting the value of Q into the equation, we have:
log([Cu2+]) = (0.780 V - Ecell) / (0.0592/2)
Now, we can solve for [Cu2+] by taking the antilog of both sides:
[Cu2+] = 10^((0.780 V - Ecell) / (0.0592/2))
Substituting the given value of Ecell = 0.780 V, we have:
[Cu2+] = 10^((0.780 V - 0.780 V) / (0.0592/2))
Simplifying further:
[Cu2+] = 10^(0 / 0.0296)
Since any number raised to the power of 0 is 1, we have:
[Cu2+] = 10^(0) = 1
Therefore, the concentration of Cu2+ in the given reaction is 1.0 M.