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
Oh, I see we're taking a dip in the world of chemistry! Well, let me just put on my lab coat and clown nose and I'll do my best to answer your question with a sprinkle of humor.
Now, to calculate the volume of H2(g) that will be discharged, we need to consider some molar relationships and use a little bit of stoichiometry. So let's electrify our brains and dive into it!
First, we need to find the number of moles of Al that were deposited. Since we know the molar mass of aluminum (Al) is approximately 27 g/mol, we can simply divide the mass (9.14 g) by the molar mass to get the number of moles.
9.14 g Al / 27 g/mol ≈ 0.339 moles of Al
Now, here comes the fun part! The balanced equation for the reduction of H+(aq) to H2(g) is:
2H+(aq) + 2e- → H2(g)
Notice that for every 2 moles of electrons (2e-), we produce 1 mole of hydrogen gas (H2). So, since the same quantity of electricity is being used, we can say that the number of moles of electrons that flowed through the circuit is equal to the number of moles of aluminum deposited (0.339 moles). This means that we will produce the same number of moles of hydrogen gas.
Now, before calculating the volume of H2(g), we need to convert the temperature to Kelvin because we're dealing with gases. So, 30°C + 273.15 = 303.15 K. And we also need to convert the pressure to atm, so 770 mmHg ÷ 760 mmHg/atm ≈ 1.01 atm.
Finally, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we know the values for P, n, R (0.0821 L·atm/mol·K), and T, we can rearrange the equation to solve for V, the volume of H2(g):
V = (nRT) / P
= (0.339 moles) * (0.0821 L·atm/mol·K) * (303.15 K) / (1.01 atm)
≈ 8.19 liters
So, my electrifying friend, the volume of H2(g) that will be discharged at 30°C and 770 mmHg when the same quantity of electricity is used in the reduction of H+(aq) will be approximately 8.19 liters.
I hope my amusing explanation brought a little spark of laughter to your chemistry journey! If you have any more questions, feel free to zap them my way.
To determine the volume of H2(g) discharged, we need to use Faraday's laws of electrolysis.
First, let's calculate the moles of Al deposited at the cathode:
1. Determine the molar mass of aluminum (Al):
The molar mass of Al is 26.98 g/mol.
2. Calculate the moles of Al deposited:
Moles of Al = Mass of Al deposited / Molar mass of Al
= 9.14 g / 26.98 g/mol
≈ 0.3386 mol
According to Faraday's laws of electrolysis, the moles of H2(g) released at the cathode will be equivalent to the moles of Al deposited.
3. Calculate the volume of H2(g) discharged:
We'll use the Ideal Gas Law, which states:
PV = nRT
Given:
T = 30°C = 273.15 + 30 = 303.15 K (convert to Kelvin)
P = 770 mmHg = 770/760 atm (convert to atm)
R = 0.0821 (gas constant in L.atm/mol.K)
Rearranging the equation, we get:
V = nRT / P
Substitute the known values:
V = (0.3386 mol) * (0.0821 L.atm/mol.K) * (303.15 K) / (770/760 atm)
≈ 0.360 L or 360 mL
Therefore, approximately 360 mL of H2(g) will be discharged at 30°C and 770 mmHg when the same quantity of electricity is used in the reduction of H+(aq) at the cathode.
To answer this question, we need to use Faraday's law of electrolysis, which states that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
First, let's determine the number of moles of aluminum (Al) that were deposited at the cathode. We can use the molar mass of aluminum (26.98 g/mol) to convert the mass of aluminum deposited (9.14 g) into moles:
Number of moles of Al = Mass of Al / Molar mass of Al
= 9.14 g / 26.98 g/mol
≈ 0.3391 mol (rounded to four decimal places)
Since the same quantity of electricity is used in the reduction of H+(aq) at the cathode, we can use this number of moles of aluminum as a reference to determine the volume of hydrogen gas (H2) that will be discharged.
To do this, we will equate the moles of Al deposited to the moles of H2 discharged. The balanced chemical equation for the reduction of H+ to H2 is:
2H+ + 2e- → H2
From the balanced equation, we can see that 2 moles of H+ ions are required to produce 1 mole of H2 gas. Therefore, the number of moles of H2 gas generated will be half of the number of moles of Al deposited:
Number of moles of H2 = Number of moles of Al / 2
= 0.3391 mol / 2
≈ 0.1696 mol (rounded to four decimal places)
Now, to determine the volume of H2 gas at the given conditions of temperature (30°C or 303 K) and pressure (770 mmHg), we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure of the gas (in atm)
V = Volume of the gas (in liters)
n = Number of moles of the gas
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature of the gas (in Kelvin)
First, we need to convert the given pressure from mmHg to atm:
Pressure (P) = 770 mmHg / 760 mmHg/atm (since 1 atm = 760 mmHg)
≈ 1.0132 atm (rounded to four decimal places)
Now, we can rearrange the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
Substituting the known values into the equation:
V = (0.1696 mol * 0.0821 L·atm/(mol·K) * 303 K) / 1.0132 atm
≈ 4.992 L (rounded to three decimal places)
Therefore, approximately 4.992 liters of H2 gas will be discharged at 30°C and 770 mmHg when the same quantity of electricity is used in the reduction of H+(aq) at the cathode.