calculate cell potential at 25C for:
3Fe3+(aq) + AL(s)= 3Fe2+(aq)+ Al3+ (aq)
if Fe3+ = 0.300 M, Fe2+ = 0.150M
Al3+ = 0.300M. The standard reduction potentials are as follows:
Fe3+ (aq) + e- = Fe2+ (aq)
Al3+ (aq) + 3e-=Al (s)
Eo = +0.771 V
Eo = -1.66 V
To calculate the cell potential at 25°C, you can use the Nernst equation:
E = E° - (RT / nF) * ln(Q)
where:
E = cell potential at 25°C
E° = standard cell potential
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin (25°C = 298 K)
n = number of moles of electrons transferred in the balanced chemical equation
F = Faraday's constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient
First, let's write the balanced equation for the reaction:
3Fe3+(aq) + Al(s) -> 3Fe2+(aq) + Al3+(aq)
From the balanced equation, we can see that the number of moles of electrons transferred (n) is equal to 3.
Now we can calculate the value of Q, the reaction quotient, using the concentrations given:
Q = ([Fe2+]^3 * [Al3+]) / ([Fe3+]^3 * [Al])
Substituting the given concentrations into the equation:
Q = (0.150^3 * 0.300) / (0.300^3 * 1)
Simplifying:
Q = 0.003375 / 0.027 = 0.125
Now we can calculate the cell potential using the Nernst equation:
E = E° - (RT / nF) * ln(Q)
Substituting the given values:
E = 0.771 V - ((8.314 J/mol·K * 298 K) / (3 * 96485 C/mol)) * ln(0.125)
Calculating the natural logarithm:
E = 0.771 V - ((8.314 J/mol·K * 298 K) / (3 * 96485 C/mol)) * (-2.079)
Calculating the expression inside the parentheses:
E = 0.771 V - (0.08189 / 0.28975) * (-2.079)
Simplifying:
E = 0.771 V - (-0.7148)
Calculating the final result:
E = 0.771 V + 0.7148 V = 1.485 V
Therefore, the cell potential at 25°C for the given reaction is 1.485 V.
To calculate the cell potential at 25°C for the given half-reactions, we need to use the Nernst equation, which relates the cell potential to the standard reduction potentials and the concentrations of the species in the cell.
The Nernst equation is represented as follows:
E = E° - (0.0592/n) * log(Q)
Where:
E = Cell potential at a given temperature
E° = Standard cell potential at 25°C
n = Number of electrons involved in the reaction
Q = Reaction quotient
Let's start by determining the number of electrons involved in the reaction. Looking at the balanced equation:
3Fe3+(aq) + AL(s) = 3Fe2+(aq) + Al3+(aq)
We can see that 3 electrons are involved in the reaction.
Next, we calculate the reaction quotient (Q) by taking the product of the concentration of the products (raised to their respective stoichiometric coefficients) divided by the product of the concentration of the reactants (raised to their respective stoichiometric coefficients).
Q = [Fe2+]^3 * [Al3+] / [Fe3+]^3
Now, substitute the values given:
[Fe2+] = 0.150 M
[Al3+] = 0.300 M
[Fe3+] = 0.300 M
Q = (0.150)^3 * (0.300) / (0.300)^3
Q = 0.0207
Now, we can substitute the values of E°, n, and Q into the Nernst equation to calculate the cell potential (E) at 25°C.
E = E° - (0.0592/n) * log(Q)
E = 0.771 V - (0.0592/3) * log(0.0207)
After performing the necessary calculations, the final answer will give you the cell potential at 25°C for the given reaction.
Fe^+3 + e ==> Fe^+2 Eo = +0.771 v
Al==> Al^+3 + 3e Eo = +1.66 v
Ecell = .771 + 1.66 = ??