A certain quantity of electric charge deposited 9.14 g of Al at the cathode during electrolysis of a solution of Al^+3(aq). What volume of H2(g) will be discharged at 30°C and 770mmHg if the same quantity of electricity is used in the reduction of H^+(aq) at a cathode
96,485 coulombs will deposit 27/3 = 9 grams of Al. This problem has 9.14 g Al deposited; therefore, 96,485 C x 9.14/9 = 97,986 C used.
96,485 C will deposit 2.02 grams H2 gas @ STP so 97,986 C will release
2.02 x 97,986/96,485 = 2.05 grams H2 @ STP.
Then use PV = nRT to determine the volume at the conditions listed. For n use n = grams H2/molar mass H2 = 2.05/2.02 = ?
Don't forget that temperature must be convert to kelvin. K =273 + C
Post your work if you get stuck.
Thanks Sir 👍
(Workings)
From....n =2.05/2.02=1.01
V = nRT/P
= (1.01×8.314×303)÷(770)
V = 3.03cm^3
Well, if I were to make a hydrogen-filled balloon joke here, it might burst - literally! But let's dive into the question instead.
To solve this problem, we can use Faraday's laws of electrolysis. According to Faraday's first law, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through it.
So, if the same quantity of electricity was used in the reduction of H^+(aq) as in the reduction of Al^+3(aq), the deposited masses are directly proportional.
We know that 9.14 g of Al was deposited. Now, let's find the molar mass of Al. It is approximately 26.98 g/mol.
Therefore, the number of moles of Al deposited = mass of Al / molar mass of Al = 9.14 g / 26.98 g/mol
Now, let's use the balanced chemical equation for the reduction of H^+(aq):
2H^+(aq) + 2e^- -> H2(g)
The stoichiometry of the reaction tells us that 2 moles of electrons (2e^-) are needed to produce 1 mole of H2.
So, the number of moles of H2 produced = (number of moles of Al deposited) / 2
Finally, we can use the ideal gas equation to find the volume of H2 at 30°C (303.15 K) and 770 mmHg:
PV = nRT
V = (nRT) / P
Where:
P = Pressure (770 mmHg)
V = Volume of H2
n = Number of moles of H2
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (303.15 K)
Now, my dear interlocutor, take these equations and numbers and solve for the volume of H2. Just be careful not to float away like a balloon with all this chemistry!
To determine the volume of H2(g) discharged at the cathode, we need to use Faraday's law of electrolysis and the ideal gas law.
Step 1: Calculate the moles of Al deposited at the cathode.
Using the molar mass of aluminum (Al) which is 26.98 g/mol, we can calculate the number of moles using the equation:
moles of Al = mass of Al / molar mass of Al
moles of Al = 9.14 g / 26.98 g/mol
moles of Al ≈ 0.3391 mol
Step 2: Determine the number of moles of electrons involved in the reduction of Al^+3(aq) to Al(s).
According to the balanced redox equation: 2 Al^+3(aq) + 6 e^- → 2 Al(s)
We can see that it takes 6 moles of electrons to reduce 2 moles of Al^+3(aq). Therefore, the number of moles of electrons involved is:
moles of electrons = (2/6) * moles of Al
moles of electrons = (2/6) * 0.3391 mol
moles of electrons ≈ 0.1130 mol
Step 3: Determine the number of moles of H2(g) produced at the cathode.
We know that 2 moles of electrons are required to produce 1 mole of H2(g) according to the balanced redox equation: 2 H^+(aq) + 2 e^- → H2(g)
Therefore, the number of moles of H2(g) produced is:
moles of H2(g) = (1/2) * moles of electrons
moles of H2(g) = (1/2) * 0.1130 mol
moles of H2(g) ≈ 0.0565 mol
Step 4: Calculate the volume of H2(g) using the ideal gas law.
The ideal gas law is given by the equation:
PV = nRT
R = 0.0821 L.atm/(mol.K) (universal gas constant)
T = 30°C + 273.15 = 303.15 K (temperature in Kelvin)
P = 770 mmHg / 760 mmHg/atm = 1.0132 atm (pressure in atm)
n = moles of H2(g) = 0.0565 mol (number of moles)
Plugging these values into the ideal gas law equation and solving for V (volume), we get:
V = (nRT) / P
V = (0.0565 mol * 0.0821 L.atm/(mol.K) * 303.15 K) / 1.0132 atm
V ≈ 1.76 L
Therefore, approximately 1.76 L of H2 gas will be discharged at the cathode.
To determine the volume of hydrogen gas (H2) that will be discharged, we need to use Faraday's law of electrolysis. This law states that the amount of substance produced or consumed during an electrolysis reaction is directly proportional to the quantity of electric charge passed through the electrolyte.
First, let's calculate the moles of aluminum (Al) deposited at the cathode. We need to know the molar mass of aluminum to convert the given mass (9.14 g) to moles. The molar mass of Al is 26.98 g/mol.
moles of Al = mass of Al / molar mass of Al
moles of Al = 9.14 g / 26.98 g/mol
moles of Al = 0.339 mol
Now, we can use the ratio of moles to determine the moles of hydrogen gas released. The balanced equation for the reduction of H+ to H2 during electrolysis is:
2H+ + 2e- -> H2
From the balanced equation, we can see that 2 moles of electrons (e-) are required to produce 1 mole of H2. Therefore, the moles of H2 produced will be the same as the moles of electrons passed through the electrolyte.
moles of H2 = moles of electrons = moles of Al
moles of H2 = 0.339 mol
To find the volume of hydrogen gas at given temperature (30°C) and pressure (770 mmHg), we can use the ideal gas equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, convert the given pressure from mmHg to atm:
1 atm = 760 mmHg
770 mmHg = 770 mmHg / 760 mmHg/atm
770 mmHg = 1.0132 atm
Next, convert the given temperature from °C to Kelvin:
T (K) = T (°C) + 273.15
T (K) = 30°C + 273.15
T (K) = 303.15 K
Now we can solve for the volume (V) of hydrogen gas:
V = nRT / P
V = (0.339 mol)(0.0821 L·atm/(mol·K))(303.15 K) / 1.0132 atm
V = 7.683 L
Therefore, the volume of hydrogen gas discharged at 30°C and 770 mmHg when the same quantity of electricity is used in the reduction of H+ is 7.683 liters.