A current is passed through three electrolytic cells containing silver trixonitrate(5),coppe(2)tetraoxonitrate(6) and brine respectively.if 12.7g of copper are deposited in the second electrolytic cell.calculate(a)the mass of silver deposited in d first cell.
(b) d volume of liberated in the third cell at 17 degree and 800mmhg?
First you should know that silver nitrate is a correct IUPAC name for AgNO3 and copper(II) nitrate is a correct IUPAC name for Cu(NO3)2. Also, note that the oxidation number for N in your name for copper nitrate is (V) and not (6).
All of my numbers are close estimate. You should recalculate ALL of these numbers for all are my estimates only.
It will take 96,485 coulombs to deposit 63.54/2 = approx 32 g of Cu. So how many coulombs of electricity must have passed through the cell? That's
96,485 x (12.7/about 32) = approx 39,000.
It will take 96,485 coulombs to deposit 107.9 g Ag. You had approx 38,000 C so
107.9 g Ag x (approx 38,000/96,485) = approx ? g Ag deposited.
For the second part of the question note you don't specify the volume of what gas but it is done the same way as I've shown you for part 1 of the problem.
Ok
To calculate the mass of silver deposited in the first cell, we need to use the concept of Faraday's Law of Electrolysis. According to Faraday's Law, the amount of substance deposited or liberated at an electrode is directly proportional to the amount of electric charge passed through the electrolyte.
The formula to calculate the mass of substance deposited is as follows:
Mass (g) = (Current (A) × Time (s) × Molar mass (g/mol)) / (Faraday's constant (C/mol))
For the first cell, we'll assume that the current is passed for the same time as in the second cell.
Let's calculate the mass of copper deposited in the second cell first.
Given:
Mass of copper deposited (m) = 12.7 g
Charge (Q) = ?
Time (t) = ?
Molar mass of copper (M) = 63.56 g/mol
Faraday's constant (F) = 96,485 C/mol
Using the formula, we can rearrange it to solve for charge:
Q = (m × F) / M
Q = (12.7 g × 96,485 C/mol) / 63.56 g/mol
Q = 19,356.06 C
Now, we can use the charge value (Q) to calculate the mass of silver deposited in the first cell.
Given:
Charge (Q) = 19,356.06 C
Time (t) = ?
Molar mass of silver trixonitrate (AgNO3) (Ms) = 169.87 g/mol
Faraday's constant (F) = 96,485 C/mol
Using the formula and rearranging it to solve for time:
t = (Q × Ms) / (F × n)
The n value is the number of moles of AgNO3 required for one mole of Ag deposited. From the balanced equation of AgNO3 decomposition during electrolysis, it is 3.
Substituting the values:
t = (19,356.06 C × 169.87 g/mol) / (96,485 C/mol × 3)
t = 33.47 seconds (approx.)
(a) The mass of silver deposited in the first cell is approximately 33.47g.
Now, let's move on to part (b) and calculate the volume of chlorine gas liberated in the third cell.
Given:
Temperature (T) = 17 degrees Celsius
Pressure (P) = 800 mmHg (which can be converted to atm by dividing by 760 mmHg/atm)
Gas constant (R) = 0.0821 L.atm/(mol.K)
Number of moles of Cl2 (n) = ?
Volume (V) = ?
We can use the Ideal Gas Law to calculate the volume:
PV = nRT
Rearranging the formula to solve for volume:
V = (nRT) / P
To calculate the number of moles of Cl2, we need to use Faraday's Law again.
Given:
Charge (Q) = 19,356.06 C
Faraday's constant (F) = 96,485 C/mol
The Faraday's constant represents the charge required to liberate one mole of an element during electrolysis.
n = Q / F
n = (19,356.06 C) / (96,485 C/mol)
n = 0.201 mol (approx.)
Now, let's calculate the volume of chlorine gas liberated:
V = (0.201 mol × 0.0821 L.atm/(mol.K) × (17 + 273) K) / (800 mmHg / 760 mmHg/atm)
V = 0.058 L (approx.)
(b) The volume of chlorine gas liberated in the third cell at 17 degrees Celsius and 800 mmHg (assuming ideal gas behavior) is approximately 0.058 L.
To calculate the mass of silver deposited in the first cell, we need to know the amount of charge passed through the cell and the Faraday's constant.
(a) Calculate the amount of charge passed through the second cell:
Since 12.7g of copper is deposited in the second cell, we can use the equation Q = n F, where Q represents the charge passed through the cell, n represents the number of moles of electrons transferred, and F represents Faraday's constant (F = 96,485 C/mol).
To find n, we need to use the stoichiometry of the reaction. From the balanced equation of the electrolysis of copper(II) tetraoxonitrate(6), we know that for every mole of copper deposited, 2 moles of electrons are transferred:
Cu²⁺ + 2e⁻ → Cu
Therefore, n = (12.7 g Cu) / (63.55 g/mol Cu) × (2 mol e⁻ / 1 mol Cu) = 0.4 mol e⁻
Now, we can calculate the amount of charge passed through the second cell:
Q = (0.4 mol e⁻) × (96,485 C/mol) = 38,594 C
(b) To calculate the volume of chlorine gas liberated in the third cell, we need to know the number of moles of chlorine gas produced.
First, convert the given temperature from Celsius to Kelvin:
T = 17°C + 273.15 = 290.15 K
Then, use the Ideal Gas Law equation PV = nRT, where P represents the pressure, V represents the volume, n represents the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
Rearranging the equation, we have:
n = PV / RT
Plug in the given values:
P = 800 mmHg (convert to atm by dividing by 760 mmHg/atm)
V = unknown
R = 0.0821 L·atm/(mol·K)
T = 290.15 K
n = (800 mmHg / 760 mmHg/atm) × V / (0.0821 L·atm/(mol·K)) × 290.15 K
Now, you can calculate the volume of chlorine gas liberated in the third cell.