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?
See your(?) post above.
http://www.jiskha.com/display.cgi?id=1309009417
To find the concentration of Fe2+, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell = Cell potential
E°cell = Cell potential at standard conditions
n = Number of moles of electrons transferred in the balanced equation
Q = Reaction quotient
Let's start by calculating the value of Q:
Q = [Fe2+] / [Cu2+] = x / 0.0500
Now let's calculate the value of E°cell:
E°cell = E°reduction(cathode) - E°reduction(anode)
= E°red(Cu2+(aq) + 2e- -> Cu(s)) - E°red(Fe2+(aq) + 2e- -> Fe(s))
= +0.34 V - (-0.44 V)
= +0.78 V
Next, we can calculate the value of n. Looking at the balanced equations, it is clear that 2 moles of electrons are transferred for each reaction.
Now we can substitute the values into the Nernst equation and solve for [Fe2+]:
0.80 V = 0.78 V - (0.0592/2) * log(x / 0.0500)
0.80 V - 0.78 V = -0.0296 * log(x / 0.0500)
0.02 V = -0.0296 * log(x / 0.0500)
log(x / 0.0500) = -0.02 V / -0.0296
log(x / 0.0500) = 0.6757
x / 0.0500 = 10^0.6757
x = 0.0500 * 10^0.6757
x ≈ 1.882 M
Thus, the concentration of Fe2+ is approximately 1.882 M for these conditions.
To find the concentration of Fe2+ ([Fe2+]) for the given conditions, we can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the standard cell potential (Eo), the gas constant (R), the temperature (T), and the reaction quotient (Q).
The Nernst equation is given as follows:
Ecell = Eo - (RT / nF) * ln(Q)
Where:
- Ecell is the cell potential
- Eo is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (K)
- n is the number of moles of electrons transferred in the balanced half-reaction
- F is the Faraday constant (96485 C/mol)
- Q is the reaction quotient, which is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
In this case, since the cell potential is given (Ecell = +0.80 V) and we want to find the concentration of Fe2+ ([Fe2+]), we rearrange the Nernst equation as follows:
Ecell = Eo - (RT / nF) * ln(Q)
Ecell - Eo = (RT / nF) * ln(Q)
Now, we substitute the given values:
Ecell = +0.80 V
Eo = -0.44 V
T = (assumed to be 298 K at standard conditions)
n = 2 (since 2 moles of electrons are transferred in the balanced half-reaction)
F = 96485 C/mol
Next, we need to determine the value of Q. The reaction quotient (Q) is calculated using the concentrations of the species involved in the half-reactions. In this case, we know that Cu2+ concentration ([Cu2+]) is 0.0500 M. However, we don't have the [Fe2+] concentration. So, let's assume it as x M, as we are trying to find its value.
The balanced equation for the cell reaction is:
Fe2+(aq) + Cu(s) → Fe(s) + Cu2+(aq)
The concentration of Fe2+ is x M, and the concentration of Cu2+ is 0.0500 M. Cu(s) and Fe(s) are solids, so their concentrations are considered to be 1 (since they are in their standard states).
Using the Nernst equation, we substitute the values and solve for [Fe2+]:
Ecell - Eo = (RT / nF) * ln(Q)
0.80 V - (-0.44 V) = (8.314 J/(mol·K)) * (298 K) / (2 * 96485 C/mol) * ln((1) / ((0.0500)(x^2)))
1.24 V = 2.7326 * 10^-2 ln(1 / (0.0500x^2))
To solve for x, we rearrange the equation and solve for x:
ln(1 / (0.0500x^2)) = (1.24 V) / (2.7326 * 10^-2)
ln(1 / (0.0500x^2)) = 45.388
1 / (0.0500x^2) = e^45.388
(0.0500x^2) = 1 / e^45.388
x^2 = (1 / e^45.388) / 0.0500
x^2 = 6.46433 * 10^-20
x = √(6.46433 * 10^-20)
x ≈ 8.0429 * 10^-11
Therefore, the concentration of Fe2+ ([Fe2+]) is approximately 8.0429 * 10^-11 M for the given conditions.