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?

Answer 1

For the given electrolysis experiment

(a) mass of Fe deposited = 56.5811 - 56.4501 = 0.131 g

(b) Number of coulombs passed = 0.122 amp x 3600 s = 439.2 C

(c) Number of faraday = 2 x 96500 = 193000 F

(d) Number of electrons transferred = 2

(f) moles of Fe deposited = 439.2/193000 = 2.27 x 10^-3 mol

(g) molecular mass of Fe = 0.131 g/2.27 x 10^-3 = 57.71 g/mol

(h) moles of electron transferred = 2.27 x 10^-3 x 2 = 4.54 x 10^-3 mol

(i) ratio (number of moles of e-/number of moles of Fe deposited) = 2

(j) final mass of Fe serving as anode = 57.6810 - 0.131 = 57.550 g