Calculate [Fe^2]+ when the cell reaction reaches equilibrium?
A voltaic cell is constructed based on the following reaction and initial concentrations:
Fe^2+,( 0.0050 M ) + Ag+,( 2.5 M ) ----> Fe^3+,( 0.0050 M ) + Ag(s)
See answer below:
http://www.jiskha.com/display.cgi?id=1238125870
To calculate the concentration of Fe^2+ at equilibrium in the given cell reaction, we can use the Nernst equation. The Nernst equation relates the concentration of reactants and products to the cell potential at equilibrium.
The Nernst equation is given as:
E = E° - (RT/nF) * ln(Q)
where:
E is the cell potential at equilibrium
E° is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of electrons transferred in the balanced cell reaction
F is the Faraday constant (96,485 C/mol)
ln(Q) is the natural logarithm of the reaction quotient
In the given cell reaction, Fe^2+ is oxidized to Fe^3+ while Ag+ is reduced to Ag(s). The number of electrons transferred (n) in this reaction is 1.
The reaction quotient (Q) can be calculated by dividing the product of the concentrations of the products by the product of the concentrations of the reactants, each raised to the power of their coefficient in the balanced equation.
Q = ([Fe^3+]/[Ag+])
Using the initial concentrations given in the question, we can substitute these values into the reaction quotient (Q) equation.
Q = (0.0050 M / 2.5 M) = 0.002
Now, we need to determine the standard cell potential (E°) for this reaction. We can refer to standard reduction potentials to find the values for the individual half-cell reactions and then combine them to find the overall cell potential.
The standard reduction potential for the Fe^3+/Fe^2+ half-cell reaction can be found in tables as +0.77 V. The reduction potential for Ag+/Ag(s) is +0.80 V.
The standard cell potential (E°) can be calculated by subtracting the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction).
E° = E°(cathode) - E°(anode)
E° = 0.80 V - 0.77 V
E° = 0.03 V
Now, we have all the values needed to calculate the cell potential at equilibrium (E) using the Nernst equation.
E = E° - (RT/nF) * ln(Q)
E = 0.03 V - [(8.314 J/(mol·K)) * (298 K) / (1 * (96,485 C/mol))] * ln(0.002)
Calculating this equation will give us the cell potential (E) at equilibrium.
Once we have the cell potential (E) at equilibrium, we can use it to calculate the concentration of Fe^2+ at equilibrium.
To do this, we will reverse the Nernst equation and solve for Q.
E = E° - (RT/nF) * ln(Q)
ln(Q) = (E° - E) * (nF/RT)
Q = exp((E° - E) * (nF/RT))
Substituting the known values, we can calculate Q and then solve for [Fe^2+].
Q = exp((0.03 V - E) * (1 * (96,485 C/mol)) / ((8.314 J/(mol·K)) * (298 K)))
Finally, we can substitute the value of Q into the equation for [Fe^2+] at equilibrium.
[Fe^2+] = ([Fe^3+]/[Ag+]) = Q
Calculating this equation will give us the concentration of Fe^2+ at equilibrium in the given cell reaction.