2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)
For the above Al-H2 cell, what is the voltage if [Al3+] = 6.0461 M, [H+] = 6.1 M, and the hydrogen pressure is 0.0557 atm. Eo = 1.66 V.
.00004-(.05916/6)*log(7.1e-4)
=.03 WRONG ANSWER I GOT
.00004=(6.0461*6.0461*.0557)/(6.1^6)
Can you please help?
I have Eo Al = +1.66 written as an oxidation.
Eo for H^+ to H2 = 0
Eocell = 1.66 + 0 = 1.66.
Ecell = 1.66 - (0.05916/6)*log Q where
Q = (Al^3+)^2* (pH2)^3/(H^)^6
You didn't show where the 0.00004 came from but shouldn't that be 1.66 v? Second, the 0.0557 is cubed. I went through the calculation and obtained 1.7 v but I have a new calculator and I may have punched the wrong buttons.
I have it set up as
Ecell = 1.66 + (0.05916/6)*log Q and
Q = (6.0461)^2(0.0557)^3/(6.1)^6
Well, it seems like you're in quite a pickle there! Unfortunately, as a clown bot, I'm not really equipped to solve complex chemistry equations. But fear not, my friend! I can try to lighten the mood with a chemistry joke while you work on your calculations.
Why do chemists like nitrates so much?
Because they're cheaper than day rates!
Keep up the good work, and I hope you find the correct answer soon!
To find the voltage (E) of the Al-H2 cell, we can use the Nernst equation:
E = Eo - (0.05916 / n) * log(Q)
Where:
Eo = Standard electrode potential (given as 1.66 V)
n = Number of electrons transferred in the balanced equation (2 in this case)
Q = Reaction quotient
First, we need to calculate the reaction quotient Q:
Q = ([Al3+] / [Al]°) * ([H2] / [H+]°)^3 * (PH2 / P°)
Where:
[Al3+] = Concentration of Al3+ ions (given as 6.0461 M)
[H2] = Concentration of H2 gas (which is not given)
[H+] = Concentration of H+ ions (given as 6.1 M)
PH2 = Partial pressure of H2 gas (given as 0.0557 atm)
P° = Standard pressure (1 atm)
We are missing the concentration of H2 gas, which needs to be determined using the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (given as 0.0557 atm)
V = Volume (unknown)
n = Number of moles (unknown)
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (which is not given)
Since the volume and temperature are not provided, we will assume they remain constant throughout the calculations.
Solving for n in the ideal gas law equation, we get:
n = PV / RT
Plugging in the given values, we find:
n = (0.0557 atm) * V / (0.0821 L·atm/(mol·K) * T)
Now we can substitute the value of n into the reaction quotient Q:
Q = ([Al3+] / [Al]°) * ([H2] / [H+]°)^3 * (PH2 / P°)
= (6.0461 M / 1 M) * ([H2] / 6.1 M)^3 * (0.0557 atm / 1 atm)
= 6.0461 * ([H2] / 6.1)^3 * 0.0557
Substitute the calculated Q value into the Nernst equation:
E = Eo - (0.05916 / n) * log(Q)
= 1.66 V - (0.05916 / 2) * log(6.0461 * ([H2] / 6.1)^3 * 0.0557)
Remember, we need the concentration of H2 gas to proceed. If you provide the concentration or any missing information (such as the temperature or volume), I can help you further.
To calculate the voltage for the given Al-H2 cell, we can use the Nernst equation. The Nernst equation relates the cell voltage to the concentration of the species involved in the reaction and the gas pressure of hydrogen (H2). The equation is as follows:
E = Eo - (0.05916/n) * log(Q)
Where:
E is the cell voltage
Eo is the standard cell potential
n is the number of electrons transferred in the balanced redox 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.
In our case, the balanced redox equation is:
2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)
Let's calculate the reaction quotient (Q) first:
Q = ([Al3+]^2 * [H2]^3) / [H+]^6
Given:
[Al3+] = 6.0461 M
[H2] = the hydrogen pressure = 0.0557 atm
[H+] = 6.1 M
Plugging in these values into the equation, we get:
Q = (6.0461^2 * 0.0557^3) / 6.1^6
Now, let's substitute the values into the Nernst equation:
E = 1.66 V - (0.05916/6) * log(Q)
Calculating:
E = 1.66 V - (0.05916/6) * log((6.0461^2 * 0.0557^3) / 6.1^6)
This will give you the correct voltage for the Al-H2 cell.