Question

C2 T14HW Question 17 Homework . Unanswered Fill in the Blanks How long would it take to plate 5.00 grams of iron if a current of 5.18 A is passed through an Fe(NO3)2 solution? time = hrs

Answer 1

Given:

current (I) = 5.18 A = 5.18 C/s

Time (t) =?

F = 96485 C/ mol

Mass of Fe deposited = 5.00 g

We have relation, No. of moles = Mass / Molar Mass

$\therefore$ No. of moles of Fe = 5.00 g / ( 55.85 g / mol ) = 0.08952 mol

Consider reaction, Fe ²⁺ (aq) + 2 e ^-$\rightarrow$ Fe (s)

From reaction it is clear that, for one Fe deposited 2 electrons are required.

$\therefore$ Moles of Fe deposited= 1/2 (moles of electrons)

$\therefore$ Moles of electrons used in the reaction = 2 ( Moles of Fe deposited ) = 2 ( 0.08952 mol ) = 0.1790 mol

We have relation, Moles of electrons = I x t / F

$\therefore$ time (t) = F $\times$ Moles of electrons / I

$\therefore$ time (t) required to deposit 5.00 g Fe = 96485 C / mol $\times$ 0.1790 mol / 5.18 C /s

$\therefore$ time (t) required to deposit 5.00 g Fe = 3334 s

We have relation, 1 min = 60 s & 1 hour = 60 min.

$\therefore$ time (t) required to deposit 5.00 g Fe = 3334 s ( 1 min / 60 s ) ( 1 hr / 60 min ) = 0.926 hr

ANSWER : 0.926 hr