Calculate the emf of the cell
CR|Cr3+(0.1M)||Fe2+(0.1M)Fe
Given, E°Cr3+|CR = -0.75v
E°Fe2+|Fe = -0.45v
Cell reactn 2Cr(s)+3Fe2+(aq) -> 2Cr3+(aq)+3Fe(s)
To calculate the cell's electromotive force (EMF), we need to use the Nernst equation, which takes into account the concentration of the species involved in the half-cell reactions.
The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell is the cell potential (EMF)
E°cell 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 equation
F is the Faraday constant (96,485 C/mol)
Q is the reaction quotient (the ratio of product concentrations over reactant concentrations raised to their stoichiometric coefficients)
In this case, we have the following information:
E°Cr3+|CR = -0.75 V
E°Fe2+|Fe = -0.45 V
The given cell reaction is: 2Cr(s) + 3Fe2+(aq) -> 2Cr3+(aq) + 3Fe(s)
We can see that there are 2 moles of electrons transferred in this balanced equation, so n = 2.
And since the concentrations of Cr3+, Fe2+, and Cr are all equal to 0.1 M, the reaction quotient (Q) can be calculated as follows:
Q = ([Cr3+]^2 * [Fe]^3) / ([Fe2+]^3 * [Cr]^2)
Now, we can substitute all the known values into the Nernst equation to calculate the cell potential (EMF):
Ecell = E°cell - (RT/nF) * ln(Q)
Ecell = (-0.75 V) - [(8.314 J/(mol·K)) * (298 K)/(2 * 96,485 C/mol)] * ln(Q)
Evaluate the value of ln(Q) by using the concentration values given:
Q = (0.1^2 * 0.1^3) / (0.1^3 * 0.1^2)
Q = 1
So, ln(Q) = ln(1) = 0
Substituting and solving the equation:
Ecell = (-0.75 V) - [(8.314 J/(mol·K)) * (298 K)/(2 * 96,485 C/mol)] * 0
Ecell = (-0.75 V)
Therefore, the EMF of the cell is -0.75 V. Note that the negative sign indicates that the reaction is non-spontaneous under standard conditions.
To calculate the emf (Ecell) of the cell, you can use the equation:
Ecell = Eocathode - Eoanode
Where Eocathode is the standard reduction potential of the cathode and Eoanode is the standard reduction potential of the anode.
Step 1: Identify the cathode and anode in the cell:
Cathode: Fe2+(aq) + 2e- -> Fe(s)
Anode: Cr(s) -> Cr3+(aq) + 3e-
Step 2: Determine the standard reduction potentials for the cathode and anode:
Eocathode = E°Fe2+|Fe = -0.45 V
Eoanode = E°Cr3+|Cr = -0.75 V
Step 3: Calculate the emf (Ecell) of the cell:
Ecell = Eocathode - Eoanode
Ecell = (-0.45 V) - (-0.75 V)
Ecell = (-0.45 V) + (0.75 V)
Ecell = 0.30 V
Therefore, the emf (Ecell) of the cell is 0.30 V.
Ecell=Ecathode-Eanode
=-0.44+0.74
=0.30V