This questions has 5 parts but I already have the answers for a and b. I need help on the last three parts.
In an electrolytic cell, a current of 0.250 ampere is passed through a solution of a chloride of iron producing Fe(s) and Cl2(g).
a) write the equation fot he half-reaction that occurs at the anode
b) when the cell operates for 2.00 hrs, 0.521 g iron is deposited at one electrode. Determine the formula of the chloride of iron in the original solution.
c) write the balanced equation for the overall reaction that occurs in the cell.
d) how many liters of Cl2 (g), measured at 25 degrees celsius and 750 mm Hg are produced when the cell operates as described in part (b) ?
e. Calculate the current that would produce chlorine gas from the solution at a rate of 3.00 grams per hour.
ANSWERS:
a) 2 Cl- - 2e- --> Cl2
b) 0.0187 mol e-
0.00933 mol Fe-
about a 1:2 ratio therefore = Fe3+
I need help on c,d, and e please
a)Your equation is not balanced by charge. The charge is -4 on the left and zero on the right. The equation should be
2Cl^- ==> Cl2(g) + 2e
b)0.521=55.85= 0.00933 mol Fe(s)
0.0187 Faradays.
divide 0.0187/0.0187 = 1.00
0.00933/0.0187 = 0.499 = 0.500
A Faraday will deposit 1 mol of a univalent metal, 0.5 mol of a divalent metal, 0.333 mol of a trivalent metal, etc. SO, the change in electrons must have been 2 and not 3 as your answer suggests, and the formula is FeCl2. Another way is ox state = mol e/mol Fe = 0.01866/0.00933 = 2.000.
c). So anode is 2Cl^- ==> Cl2(g) + 2e
cathode is Fe^+2 + 2e ==> Fe(s)
Add the two to obtain the cell reaction.
d). 0.01866 C x 70.91/2 = ??grams
Convert to L at STP, then use PV = nRT to convert to the non-standard conditions.
e. I would put all of this into a formula and solve for the unknown.
A x hrs x 3600 s/hr x molar mass = 96,485 x grams x delta e.
A=??
hrs=1.00
molar mass = 70.906 g Cl2/mol Cl2.
grams = 3.00
delta e = 2
Solve for A.
Check my thinking. Check my arithmetic.
b)0.521=55.85= 0.00933 mol Fe(s)
I made a typo on (b). It should be
0.521 g/55.85 = 0.00933 mol Fe(s).
c) The balanced equation for the overall reaction that occurs in the cell is:
2FeCl2 + 2Cl- --> 2Fe + 4Cl2
d) 0.01866 C x 70.91/2 = 1.30 g Cl2
1.30 g Cl2 x 1 L/22.4 L = 0.058 L Cl2 at STP
0.058 L Cl2 x (750 mmHg/760 mmHg) x (273.15 K/298.15 K) = 0.053 L Cl2 at 25°C and 750 mmHg
e. A x 1.00 hrs x 3600 s/hr x 70.906 g Cl2/mol Cl2 = 96,485 x 3.00 g x 2
A = 0.039 A
c) To write the balanced equation for the overall reaction that occurs in the cell, we need to combine the half-reactions from part a) and part b).
The half-reactions are:
Anode: 2Cl^- -> Cl2(g) + 2e^-
Cathode: Fe^3+ + 3e^- -> Fe(s)
To balance the charges, we multiply the anode half-reaction by 3 and the cathode half-reaction by 2:
3 x (2Cl^- -> Cl2(g) + 2e^-) gives us 6Cl^- -> 3Cl2(g) + 6e^-
2 x (Fe^3+ + 3e^- -> Fe(s)) gives us 2Fe^3+ + 6e^- -> 2Fe(s)
Combining the two half-reactions, we get:
6Cl^- + 2Fe^3+ -> 3Cl2(g) + 2Fe(s)
So, the balanced equation for the overall reaction that occurs in the cell is: 6Cl^- + 2Fe^3+ -> 3Cl2(g) + 2Fe(s).
d) To determine the volume of Cl2(g) produced at the given conditions, we first need to calculate the moles of Cl2 produced. From part b), we know that 0.521 g of Fe corresponds to 0.00933 mol of Fe(s).
Since the stoichiometry of the reaction from part c) is 6:3 (6 moles of Cl2 produced per 3 moles of Fe), we can calculate the moles of Cl2 produced as:
(0.00933 mol Fe) x (6 mol Cl2 / 3 mol Fe) = 0.01866 mol Cl2
Next, we can use the ideal gas law to calculate the volume of Cl2(g):
PV = nRT
Assuming the temperature (T) is 25 degrees Celsius (which is 298 Kelvin) and the pressure (P) is 750 mm Hg, we have:
V = (0.01866 mol Cl2) x (0.0821 L atm / mol K) x (298 K) / (750 mm Hg)
Convert the pressure to atm: 750 mm Hg x (1 atm / 760 mm Hg) = 0.987 atm
Substituting the values, we get:
V = (0.01866 mol Cl2) x (0.0821 L atm / mol K) x (298 K) / (0.987 atm)
Simplifying, we find the volume of Cl2(g) produced is approximately V = 0.1 L.
e) To calculate the current needed to produce 3.00 grams of Cl2 per hour, we can use Faraday's law:
I = (m x M) / (n x F x t)
Where:
I is the current (in amperes)
m is the mass of Cl2 produced (in grams)
M is the molar mass of Cl2 (in g/mol)
n is the number of electrons transferred (from the balanced equation in part c), which is 6 for Cl2
F is Faraday's constant (approximately 96,485 C/mol)
t is the time in seconds (given as 1 hour, which is 3600 seconds)
Substituting the values, we have:
I = (3.00 g x 70.906 g/mol) / (6 x 96,485 C/mol x 3600 s)
Simplifying, we find the current needed to produce chlorine gas at this rate is approximately I = 0.027 A.
c) To write the balanced equation for the overall reaction that occurs in the cell, we need to combine the half-reactions at the anode and cathode. The anode half-reaction is:
2Cl^- --> Cl2 + 2e^-
And the cathode half-reaction is:
Fe^+2 + 2e^- --> Fe(s)
We need to balance the number of electrons on both sides to combine the half-reactions. So, we multiply the anode half-reaction by 2:
4Cl^- --> 2Cl2 + 4e^-
Now, the number of electrons matches, and we can add the two half-reactions together:
4Cl^- + Fe^+2 --> 2Cl2 + Fe(s)
d) To find the number of liters of Cl2 gas produced, we need to convert the moles of Cl2 gas produced to liters. We already know from part (b) that 0.521 g of iron is deposited, so we can calculate the moles of Cl2 produced using the balanced equation from part (c):
4 moles Cl^- = 2 moles Cl2
0.00933 mol Fe = x moles Cl2 (where x is what we want to find)
Using the ratio from the balanced equation, we can calculate x:
(0.00933 mol Fe)(2 mol Cl2 / 4 mol Fe) = 0.00466 mol Cl2
Now, we can use the ideal gas law to convert the moles of Cl2 to liters at the given temperature and pressure:
PV = nRT
P = 750 mmHg (convert to atm by dividing by 760 mmHg)
V = ?
n = 0.00466 mol
R = 0.0821 L*atm/(mol*K)
T = 25 degrees Celsius = 273 + 25 = 298 K
Substituting the values into the ideal gas law equation:
(750 mmHg / 760 mmHg) * V = (0.00466 mol)(0.0821 L*atm/mol*K)(298 K)
V = (0.00466 mol)(0.0821 L*atm/mol*K)(298 K) / (750 mmHg / 760 mmHg)
V = 0.0109 L
So, approximately 0.0109 liters of Cl2 gas is produced when the cell operates as described in part (b).
e) To calculate the current that would produce chlorine gas at a rate of 3.00 grams per hour, we need to use Faraday's Law of Electrolysis. Faraday's Law states that the amount of substance produced or consumed during electrolysis is directly proportional to the quantity of electricity passed through the cell.
The equation to calculate the amount of substance produced is:
Amount of substance (in moles) = (Current (in amperes) * Time (in seconds)) / (Faraday's constant)
We can rearrange the equation to solve for the current:
Current = (Amount of substance * Faraday's constant) / Time
Given:
Amount of substance = 3.00 grams
Time = 1 hour = 3600 seconds
Faraday's constant = 96,485 C/mol
Substituting the values into the equation:
Current = (3.00 grams * (1 mol / molar mass of Cl2) * 96,485 C/mol) / 3600 seconds
Note: We need to convert grams to moles by dividing by the molar mass of Cl2. The molar mass of Cl2 is approximately 70.906 g/mol.
Using the molar mass of Cl2:
Current = (3.00 grams * (1 mol / 70.906 g) * 96,485 C/mol) / 3600 seconds
Calculating the current:
Current = 386.97 C/s
Therefore, a current of approximately 386.97 amperes would produce chlorine gas from the solution at a rate of 3.00 grams per hour.