a cell is based on the following reaction
Fe(s)+2Fe+3(aq) -----> 3Fe+2 (aq) E cell= 1.18 V.
Calculate the concentration of iron (II), in M, if the cell emf is 1.28 V. when [Fe^(3+)]= 0.5 M
the answer is 0.047 M
Ecell = Eocell - (0.05916/n)*log Q
You have Ecell and Eocell. n is 2 and
Q = (Fe^2+)^3/(Fe^3+)^2
Substitute and solve for (Fe^2+).
0.047M is the correct answer.
thank you
To find the concentration of iron (II) in the cell, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n)log(Q)
where:
Ecell is the cell potential (1.28 V)
E°cell is the standard cell potential (1.18 V)
n is the number of electrons transferred in the reaction (3 in this case)
Q is the reaction quotient
First, let's calculate the reaction quotient Q using the concentrations provided. From the balanced equation:
Fe(s) + 2Fe+3(aq) -> 3Fe+2(aq)
we can see that the concentrations of Fe+3 and Fe+2 are directly proportional, and that the ratio is 2:3. Therefore, if [Fe+3] = 0.5 M, then [Fe+2] = (2/3) * [Fe+3] = (2/3) * 0.5 M = 0.333 M.
Now, substitute the values into the Nernst equation:
1.28 V = 1.18 V - (0.0592/3) log(Q)
Subtract 1.18 V from both sides:
0.1 V = (-0.0592/3) log(Q)
Rearrange the equation:
log(Q) = (0.1 V * 3) / -0.0592
log(Q) = -5.07
Now, solve for Q:
Q = 10^(-5.07)
Q ≈ 1.08 * 10^(-6)
Finally, substitute this value of Q back into the original equation and solve for [Fe+2]:
1.28 V = 1.18 V - (0.0592/3)log(1.08 * 10^(-6))
log(1.08 * 10^(-6)) ≈ -5.97
1.28 V = 1.18 V - (0.0592/3)(-5.97)
Multiply -5.97 by (0.0592/3):
1.28 V = 1.18 V + 0.1192
Now, subtract 1.18 V from both sides:
0.1 V = 0.1192
Divide both sides by 0.1 V:
1 = 1.192 M
Finally, multiply by the ratio of Fe+2 to Fe+3 concentrations:
[Fe+2] = (2/3) * 1.192 M = 0.795 M
Therefore, the concentration of Fe+2 is approximately 0.795 M. However, this does not match the given answer of 0.047 M. Double-check your calculations or provide more information if necessary.
To calculate the concentration of iron (II) (Fe^2+) in this cell reaction, we can use the Nernst equation, which relates the cell potential (E_cell) to the concentrations of the reactants and products involved in the cell reaction.
The Nernst equation is given by:
E_cell = E°_cell - (RT / (nF)) * ln(Q)
Where:
E_cell = Cell potential
E°_cell = Standard cell potential (given as 1.18 V in this case)
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of electrons transferred in the cell reaction
F = Faraday's constant (96485 C/mol)
Q = Reaction quotient
In this case, the number of electrons transferred (n) is 3, as indicated by the balanced equation. The reaction quotient (Q) can be calculated using the concentrations of the reactants and products involved.
The balanced equation is:
Fe(s) + 2 Fe^3+(aq) → 3 Fe^2+(aq)
Given:
E_cell = 1.28 V
[Fe^3+] = 0.5 M
Now let's calculate Q, the reaction quotient:
Q = ([Fe^2+]^3) / ([Fe^3+]^2)
We need to determine the value of [Fe^2+]. Let's assume it to be x M.
Substituting the values into the Nernst equation:
1.28 V = 1.18 V - (8.314 J/(mol·K) * T / (3 * 96485 C/mol)) * ln(x^3 / (0.5^2))
Simplifying the equation:
0.1 V = (- (8.314 J/(mol·K) * T / (3 * 96485 C/mol)) * ln(x^3 / 0.25)
Now, to isolate the x term, we can rearrange the equation:
ln(x^3 / 0.25) = -0.1 V * (3 * 96485 C)/(8.314 J/K)
Taking the exponential form of both sides:
x^3 / 0.25 = exp(-0.1 V * (3 * 96485 C)/(8.314 J/K))
Simplifying further:
x^3 = 0.25 * exp(-0.1 V * (3 * 96485 C)/(8.314 J/K))
x = (0.25 * exp(-0.1 V * (3 * 96485 C) / (8.314 J/K)))^(1/3)
Now you can plug in the given values for T and calculate the value of x.
Once you have the value of x, it will represent the concentration of iron (II) in Molarity (M).