AP Chemistry

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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).

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