Al+(Cr2O7)2- = Al3+ + Cr3+
[al3+]=0.01M
[(cr2o7)2-]= 0.1M
[cr3+]=0.001M
find the E cell if we add Naoh to each electrode until al(oh)3 precipitates pH=8
KPT of al(oh)3=5 * 10 -33
Eal=-1.66 V
Ecr=1.33V
To find the E cell, we can use the Nernst equation, which relates the concentration of reactants and products to the cell potential (E cell). The Nernst equation is given by:
E cell = E° cell - (0.0592/n) * log(Q)
Where:
E cell is the cell potential
E° cell is the standard cell potential
n is the number of electrons transferred in the balanced equation
Q is the reaction quotient, which is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.
Given the balanced chemical equation:
Al + (Cr2O7)2- -> Al3+ + Cr3+
We can see that n = 3 (as three electrons are transferred).
To determine the value of Q, we need to calculate the concentrations of the products and reactants at equilibrium after the addition of NaOH.
Since Al(OH)3 precipitates, we can assume that the reaction has reached equilibrium, and we can determine the equilibrium concentrations.
The balanced net ionic equation for the reaction between Al and OH- is:
Al(OH)3(s) + 3H2O(l) -> Al(OH)4-(aq) + 3H+(aq)
From this equation, we can see that [Al3+] = [Al(OH)4-] and [H+] = 3[OH-].
Given that the pH is 8, we can calculate the concentration of OH- ions:
pOH = 14 - pH
pOH = 14 - 8 = 6
[OH-] = 10^-pOH
[OH-] = 10^-6 = 1 * 10^-6 M
[Al3+] = [Al(OH)4-] = [OH-] = 1 * 10^-6 M
[H+] = 3 * [OH-] = 3 * 1 * 10^-6 M = 3 * 10^-6 M
Now, we can calculate Q:
Q = ([Al3+] * [Cr3+]) / ([Al(OH)4-] * [Cr2O7]2-)
Q = (0.01 * 0.001) / (1 * 10^-6 * 0.1)
Q = 0.01 / 10^-7
Q = 10^5
Now we have all the necessary values to calculate the E cell using the Nernst equation:
E cell = E° cell - (0.0592/3) * log(Q)
E cell = (E° Al - E° Cr) - (0.0592/3) * log(10^5)
Given:
E° Al = -1.66 V
E° Cr = 1.33 V
E cell = (-1.66 - 1.33) - (0.0592/3) * log(10^5)
Calculating further using the above values, we can find the value of E cell.
To find the cell potential (Ecell), we need to calculate the potential difference between the Al and Cr electrodes.
Step 1: Write the balanced net ionic equation for the reaction:
2Al(s) + 3Cr2O7^2-(aq) + 14H+(aq) -> 2Al^3+(aq) + 3Cr^3+(aq) + 7H2O(l)
Step 2: Identify the half-reactions:
Half-reaction at the anode (Al electrode): 2Al(s) -> 2Al^3+(aq) + 6e-
Half-reaction at the cathode (Cr electrode): 3Cr2O7^2-(aq) + 14H+(aq) + 6e- -> 3Cr^3+(aq) + 7H2O(l)
Step 3: Determine the standard reduction potentials for each half-reaction:
E°Al = -1.66 V (already given)
E°Cr = 1.33 V (already given)
Step 4: Calculate the cell potential:
Ecell = E°cathode - E°anode
Ecell = 1.33 V - (-1.66 V)
Ecell = 2.99 V
Step 5: Adjust the cell potential for non-standard conditions:
The cell potential in non-standard conditions can be calculated using the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Step 6: Calculate the concentrations of reactants and products:
[Al^3+] = 0.01 M (already given)
[Cr2O7^2-] = 0.1 M (already given)
[Cr^3+] = 0.001 M (already given)
Step 7: Calculate the reaction quotient (Q):
Q = ([Cr^3+]/[Cr2O7^2-]^3) * ([Al^3+]^2)
= (0.001/0.1^3) * (0.01^2)
= 0.001 * 0.01^2 / 0.1^3
= 0.001 * 0.0001 / 0.001
= 0.0000001 / 0.001
= 0.0001
Step 8: Calculate the pH at which Al(OH)3 precipitates:
Since the Ksp for Al(OH)3 is given as 5 * 10^-33, we can use the relation:
[Al^3+][OH^-]^3 = Ksp
Let [OH^-] be x (molar concentration)
Then, [Al^3+] = x
Since we know [Al^3+] = 0.01 M, we can substitute:
0.01 * x^3 = 5 * 10^-33
Solving for x, we get:
x = (5 * 10^-33 / 0.01)^(1/3)
≈ 7.28 * 10^-12 M
Taking the negative logarithm of [OH^-], we get:
pOH = -log10 ([OH^-])
= -log10 (7.28 * 10^-12)
≈ 11.14
Using the relation: pH = 14 - pOH, we can calculate the pH:
pH = 14 - 11.14
≈ 2.86
Step 9: Calculate the adjusted cell potential:
R = 8.314 J/(mol·K) (gas constant)
T = temperature in Kelvin (usually 298 K at room temperature)
n = number of electrons transferred (6 in this case)
F = Faraday's constant (approximately 96485 C/mol)
Assuming room temperature (298 K):
Ecell = 2.99 - (8.314 * 298 / (6 * 96485)) * ln(0.0001)
Ecell = 2.99 - (0.10335) * ln(0.0001)
Ecell ≈ 2.99 - (-9.195)
Ecell ≈ 11.185 V
Therefore, the estimated Ecell value, when NaOH is added to each electrode until Al(OH)3 precipitates at pH 8, is approximately 11.185 V.