3 Ce4+(aq) + Cr(s) �¨ 3 Ce3+(aq) + Cr3+(aq)
What is the emf of this cell when [Ce4+] = 2.9 M, [Ce3+] = 0.11 M, and [Cr3+] = 0.015 M?
for the E0cell i got 2.35 and for Q i got 8.186 then when i plugged that into the equation i get 2.7101V is that correct? I plugged 1 in for n.
To calculate the EMF (electromotive force) of the cell, we need to use the Nernst equation. The Nernst equation relates the EMF of an electrochemical cell to the concentration of the species involved. The equation is as follows:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell is the EMF of the cell
E°cell is the standard EMF of the cell at standard conditions
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of moles of electrons transferred in the balanced equation
F is Faraday's constant (96,485 C/mol)
ln is the natural logarithm
Q is the reaction quotient, which is the ratio of concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients.
In this case, the balanced equation is:
3 Ce4+(aq) + Cr(s) → 3 Ce3+(aq) + Cr3+(aq)
The number of electrons transferred is 3, so n = 3.
First, we need to find the value of Q using the given concentrations:
Q = ([Ce3+]^3 * [Cr3+]) / ([Ce4+]^3)
Substituting the given values, we have:
Q = (0.11^3 * 0.015) / (2.9^3)
Next, we can calculate the EMF of the cell using the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
However, the standard EMF (E°cell) of this cell at standard conditions is not given. Without that information, we cannot calculate the exact EMF of the cell.
To determine the emf of this cell, we can use the Nernst equation. The Nernst equation relates the emf of an electrochemical cell to the concentrations of the species involved.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E = emf of the cell
E° = standard emf of the cell (under standard conditions)
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin
n = number of electrons transferred in the balanced equation
F = Faraday's constant (96,485 C/mol)
ln = natural logarithm
Q = reaction quotient
In this case, we need to find two things: the standard emf (E°) and the reaction quotient (Q).
To find E°, we need to look up the standard reduction potentials for the half-reactions involved in this cell:
Ce4+ + e- → Ce3+ (E° = +1.61 V)
Cr3+ + 3e- → Cr (E° = -0.74 V)
The overall reaction is the sum of these two half-reactions, so the balanced equation is:
3 Ce4+ + Cr → 3 Ce3+ + Cr3+
Next, we can find the reaction quotient (Q) by plugging in the given concentrations into the formula for Q.
Q = ([Ce3+]^3 * [Cr3+]) / ([Ce4+]^3)
Plugging in the given concentrations:
Q = (0.11^3 * 0.015) / (2.9^3)
Now we have all the necessary values to calculate the emf (E) using the Nernst equation:
E = E° - (RT/nF) * ln(Q)
Substituting the values we have:
E = (1.61 V) - [(8.314 J/mol·K)(298 K)/(3 mol)(96,485 C/mol)] * ln([(0.11^3 * 0.015) / (2.9^3)])
Now, we can solve this equation to find the emf (E) of the cell.
Do it this way.
Ce^4+ + e ==> Ce^3+ Eo = Eo1
Cr ==> Cr^3+ + 3e Eo = Eo2
------------------------
Eocell = Eo1 + Eo2
Then Ecell = Eocell - (0.05916/n)log Q where
Q = (Ce^3+)^3x(Cr^3+)/(Ce^4+)^3
Substitute the values and solve for Ecell.