A voltaic cell consists of Ag/Ag+2 electrode E= 0.80 and a Fe+2/Fe+3 electrode E=0.77 with the following initial molar concentrations: [Fe+2}= 0.30 M [Fe+3]= 0.10 M [Ag+]=0.30 M. What is the equilibrium concentration of Fe+3? (Assume the anode and the cathode solutions are of equal volume, and at the temperature 25 degreees Celcius)
Did you intend Ag/Ag^+ or Ag/Ag^2+. I will assume you intended Ag/Ag^+
Ag^+ + e ==> Ag Eo = 0.80
Fe^2+ ==> Fe^3+ Eo = -0.77
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Ag^+ + Fe^2+ ==> Ag + Fe^3+ Eocell = 0.03
Then Ecell = EoCell - (0.0592/n)log Q
where Q = (Fe^3+)/(Fe^2+)
To determine the equilibrium concentration of Fe+3 in the voltaic cell, we can calculate the cell potential using the Nernst equation.
The Nernst equation is given as:
Ecell = Eºcell - (RT/nF) * ln(Q)
Where:
Ecell = cell potential
Eºcell = standard cell potential
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin (25 degrees Celsius = 298 K)
n = number of moles of electrons transferred in the balanced redox equation
F = Faraday's constant (96,485 C/mol)
Q = reaction quotient
Let's start by determining the balanced redox equation for the cell reaction:
Ag+ + Fe+2 --> Ag + Fe+3
The balanced equation shows that one mole of electrons is transferred.
Now we can calculate the cell potential:
Ecell = 0.80 V - (8.314 J/mol·K * 298 K / (1 * 96,485 C/mol)) * ln((0.30 M)/(0.10 M))
Ecell = 0.80 V - (2481.9642 J/C) * ln(3)
Ecell ≈ 0.80 V - (2481.9642 J/C) * 1.0986
Ecell ≈ 0.80 V - 2726.203 J/C
Ecell ≈ -1926.203 J/C
Now, we can rearrange the Nernst equation to solve for Q:
Q = e^((Eºcell - Ecell) * nF / (RT))
Q = e^((0.80 V - (-1926.203 J/C)) * 1 * 96,485 C/mol / (8.314 J/mol·K * 298 K))
Q = e^(1989.203 J/C * 96,485 C/mol / (8.314 J/mol·K * 298 K))
Q ≈ e^(231542.755 J/mol / 6216.868 J/mol·K)
Q ≈ e^37.259
Q ≈ 2.54 x 10^16
Now, we can substitute the Q value back into the Nernst equation to solve for the concentration of Fe+3:
Ecell = Eºcell - (RT/nF) * ln(Q)
0.77 V = 0.80 V - (8.314 J/mol·K * 298 K / (1 * 96,485 C/mol)) * ln((0.30 M)/x)
0.77 ≈ 0.80 - 2481.9642 * ln(0.30/x)
0.03 ≈ -2481.9642 * ln(0.30/x)
ln(0.30/x) ≈ -0.000012072
0.30/x ≈ e^(-0.000012072)
0.30/x ≈ 0.999988926
x ≈ 0.30 / 0.999988926
x ≈ 0.300011897 M
Therefore, the equilibrium concentration of Fe+3 is approximately 0.300 M.
To determine the equilibrium concentration of Fe+3 in the voltaic cell, we can use the Nernst equation. The Nernst equation relates the cell potential (E) of an electrochemical cell to the concentrations of the species involved.
The Nernst equation is given by:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential (in volts)
- E°cell is the standard cell potential (in volts)
- (0.0592/n) is the Nernst constant at 25 degrees Celsius (where n is the number of electrons involved in the cell reaction)
- log(Q) is the logarithm of the reaction quotient (Q)
In our case, we have an Ag/Ag+2 electrode with E= 0.80 V and a Fe+2/Fe+3 electrode with E=0.77 V. The reaction involved in the cell is:
Ag+ + e- -> Ag
Fe+2 -> Fe+3 + e-
Since the anode and cathode solutions are of equal volumes, we can assume the number of electrons involved in the cell reaction is one (n = 1).
To calculate Q, we need to substitute the initial concentrations into the reaction quotient:
Q = ([Ag+]/[Ag]) * ([Fe+3]/[Fe+2])
Q = (0.30 M / 1 M) * (x / 0.30 M)
Where x is the concentration of Fe+3 at equilibrium.
Using the Nernst equation, we can write:
Ecell = E°cell - (0.0592/1) * log(Q)
0.80 V = 0.80 V - 0.0592 * log((0.30 M / 1 M) * (x / 0.30 M))
Simplifying the equation:
0 = -0.0592 * log(x / 0.30)
Solving for x:
log(x / 0.30) = 0
x / 0.30 = 10^0
x = 0.30 M
Therefore, the equilibrium concentration of Fe+3 in the voltaic cell is 0.30 M.