a) A current of 1.50 Amperes passes through a solution of AgNO3 for 3.0 hours. What mass, in grams, of silver collects on the cathode? (molar mass of silver = 107.9 g/mol)
b) How many seconds would it take to collect 100 g of silver to collect on the cathode if 1.50 Amperes passes through a solution of AgNO3? (molar mass of silver = 107.9 g/mol)
c) What current is needed to deposit 21.00 g of silver on the cathode in 5550 s? (molar mass of silver = 107.9 g/mol) (3 points)
All three are based on the fact that 1 grams equivalent weight is deposited with 96,485 coulombs
a). You can deposit 107.9 g Ag metal with 96,485 coulombs,
Coulombs = amperes x seconds
a) To calculate the mass of silver collected on the cathode, we can use the equation:
Mass of silver = Current (Amperes) x Time (seconds) x Molar mass of silver (g/mol)
Given that the current is 1.50 Amperes and the time is 3.0 hours (which we'll convert to seconds), we can plug in the values:
Time = 3.0 hours x 60 minutes/hour x 60 seconds/minute = 10,800 seconds
Mass of silver = 1.50 A x 10,800 s x 107.9 g/mol
Now I know what you're thinking... "Wow, that's a lot of numbers! It's like a math party, and I didn't even bring a calculator!" But don't worry, I've got your back! Let me just crunch the numbers for you...
Calculating...
The mass of silver collected on the cathode is approximately 1.6184 grams. So, you can say that silver had a "weigh" of fun in this experiment!
b) Alright, so now we want to find out how many seconds it would take to collect 100 g of silver. We can rearrange the equation from part (a) to solve for time:
Time = Mass of silver / (Current (Amperes) x Molar mass of silver (g/mol))
Plugging in the values, we have:
Time = 100 g / (1.50 A x 107.9 g/mol)
Hold on tight, I'm doing some quick math here...
Calculating...
It would take approximately 6521.1 seconds to collect 100 grams of silver on the cathode. That's a silver lining worth waiting for!
c) Finally, we need to determine the current needed to deposit 21.00 g of silver in 5550 s. Again, rearranging the equation from part (a) gives us:
Current (Amperes) = Mass of silver / (Time (seconds) x Molar mass of silver (g/mol))
Plugging in the values, we have:
Current = 21.00 g / (5550 s x 107.9 g/mol)
Hold onto your circuits, I'm crunching the numbers...
Calculating...
The current needed to deposit 21.00 g of silver in 5550 seconds would be approximately 0.0378 Amperes. It's quite a shockingly small current, but hey, sometimes good things come in small currents!
a) To calculate the mass of silver collected on the cathode, we can use Faraday's law of electrolysis, which states that the amount of substance produced or consumed at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
The formula for calculating the mass of a substance deposited during electrolysis is:
Mass (g) = (Current (A) x Time (s) x Molar mass (g/mol)) / (Faraday's constant)
First, let's convert the time from hours to seconds:
Time (s) = 3 hours x 60 minutes/hour x 60 seconds/minute
Next, let's calculate the mass of silver deposited using the given values:
Mass (g) = (1.50 A x 3.0 hours x 60 minutes/hour x 60 seconds/minute x 107.9 g/mol) / (Faraday's constant)
The Faraday's constant is equal to 96,485 C/mol.
Mass (g) = (1.50 A x 3.0 hours x 60 minutes/hour x 60 seconds/minute x 107.9 g/mol) / 96485 C/mol
Now, you can plug in the given values and calculate the mass of silver collected on the cathode.
b) To calculate the time required to collect 100 g of silver on the cathode, we can rearrange the formula from part a) and solve for time:
Time (s) = (Mass (g) x Faraday's constant) / (Current (A) x Molar mass (g/mol))
Plug in the given values and calculate the time required.
c) To calculate the current needed to deposit 21.00 g of silver on the cathode in 5550 s, we can rearrange the formula from part a) and solve for current:
Current (A) = (Mass (g) x Faraday's constant) / (Time (s) x Molar mass (g/mol))
Plug in the given values and calculate the current needed.
To solve these problems, we can use Faraday's law of electrolysis, which states that the amount of substance (in moles) deposited or produced during electrolysis is directly proportional to the total electric charge passed through the solution.
The formula we will use is:
moles of substance = (current * time) / (Faraday's constant * charge per mole of substance)
The charge per mole of substance is also known as the number of moles of electrons transferred per mole of substance and can be determined using the balanced equation of the reaction occurring during electrolysis.
a) To find the mass of silver deposited on the cathode, we can convert the moles of silver to grams using its molar mass.
1) First, calculate the charge passed through the solution using the formula:
charge = current * time
charge = 1.50 A * 3.0 h
2) Calculate the moles of silver deposited using Faraday's law:
moles of silver = (charge) / (Faraday's constant * charge per mole of silver)
To determine the charge per mole of silver, we need to look at the balanced equation for the reaction involved in the electrolysis. Since it's not provided, we assume it to be:
Ag⁺ + e⁻ → Ag
From this equation, we can see that one silver ion (Ag⁺) is reduced to one silver atom (Ag) with the transfer of one mole of electrons (e⁻).
Therefore, the charge per mole of silver is 1.
moles of silver = (charge) / (Faraday's constant)
Faraday's constant (F) is 96500 C/mol (coulombs per mole).
3) Convert moles of silver to grams using its molar mass:
mass of silver = moles of silver * molar mass of silver
b) To find the time required to collect 100 g of silver on the cathode, we need to rearrange the equation for charge and solve for time.
1) Rearrange the charge formula:
charge = current * time
time = charge / current
2) Calculate the charge required to deposit 100 g of silver using its molar mass:
charge = moles of silver * (Faraday's constant * charge per mole of silver)
moles of silver = mass of silver / molar mass of silver
3) Substitute the values into the rearranged formula to find the time.
time = (mass of silver / molar mass of silver) * (Faraday's constant * charge per mole of silver) / current
c) To find the current required to deposit 21.00 g of silver in 5550 s, we can use the same formula as in part b, rearranged to solve for current.
1) Rearrange the formula:
current = charge / time
2) Calculate the charge required to deposit 21.00 g of silver using its molar mass:
charge = moles of silver * (Faraday's constant * charge per mole of silver)
moles of silver = mass of silver / molar mass of silver
3) Substitute the values into the rearranged formula to find the current.
current = (mass of silver / molar mass of silver) * (Faraday's constant * charge per mole of silver) / time
Note: In these calculations, make sure to convert the time to seconds if it's given in hours or minutes.
Remember to double-check the units and perform any necessary unit conversions to get the final answer.