A student followed the procedure of this experiment, but instead of using lead s
ID: 972203 • Letter: A
Question
A student followed the procedure of this experiment, but instead of using lead strips, the student placed iron strips in a solution of iron(II) nitrate. The electrolysis apparatus was then set up according to Figure 1. The following data were obtained:
Average milliamps 122
Time of electrolysis, sec 3600
Initial mass of iron strip serving as cathode, g 56.4501
Final mass of iron strip serving as cathode, g 56.5811
Initial mass of iron strip serving as anode, g 57.6810
(a) How many grams of iron were deposited?
(b) Calculate the number of coulombs passed.
(c) Calculate the number of faradays. (answer in scientific notation)
(d) Calculate the number of electrons transferred. (answer in scientific notation)
(e) Calculate the number of Fe2+ ions deposited. (answer in scientific notation)
(f) How many moles of iron were deposited? (answer in scientific notation)
(g) Calculate the molecular mass of iron.
(h) How many moles of electrons were transferred? (answer in scientific notation)
(i) Find the ratio of the number of moles of electrons transferred to the number of moles of iron deposited.
(j) What is the final mass of the iron strip serving as the anode?
Explanation / Answer
a) grams of iron deposited 56.5811 - 56.4501 = 0.1310 g
b) coulombs of current passed = 122 X 10-3 X 3600 = 439 C
c) No. of Faradays = 439/96500 = 0.00455 F
d) 1 F = 1 mole of electron . therefore 0.00455 F = 0.00455 mols of electrons = 0.00455 * 6.022 * 1023 = 2.7 * 1021 electrons
e) No. of Fe2+ deposited = (2.7 * 1021)/2 = 1.35 * 1021
f) moles of iron deposited = (1.35 * 1021)/(6.02 * 1023) = 0.00224 moles
g) molecular mass of iron = mass/ moles = 0.1310 / 0.00224 = 58.4 g/mol
h) moles of electrons transferred = 0.00455 moles
i) 2
j) 57.6810g - 0.1310 g = 57.5500 g
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