Question

(a) \(\mathrm{A} \mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of \(7.60 \mathrm{~A}\). What mass of \(\mathrm{Cr}(s)\) is plated out after \(2.00\) days? (b) What amperage is required to plate out \(0.250 \mathrm{~mol} \mathrm{Cr}\) from a \(\mathrm{Cr}^{3+}\) solution in a period of \(8.00 \mathrm{~h}\) ?

Short answer

(a) After electrolyzing a Cr³⁺ solution with a current of 7.60 A for 2.00 days, the mass of Cr plated out is 235 g. (b) The amperage required to plate out 0.250 mol Cr from a Cr³⁺ solution in a period of 8.00 hours is 2.514 A.

Step-by-step solution

Understand Faraday's Law of Electrolysis

Faraday's Law of Electrolysis states: \(q = n \cdot F\) where: - \(q\) is the charge in coulombs (C) - \(n\) is the amount of substance in moles - \(F\) is the Faraday constant (\(96485 \; C \cdot mol^{-1} \)) The charge can be calculated as the product of current (I) and time (t), in the form: \(q = I \cdot t\) In this exercise, Chromium has a charge of +3 when it forms Cr³⁺ ions. That means, \(n = \frac{q}{3F}\)

Calculate the charge for part (a)

First, convert the time of 2.00 days to seconds: \(t = 2.00 \; days \cdot \frac{86400}{day} = 172800 \; s\) Then, calculate the charge: \(q = I \cdot t = 7.60 \; A \cdot 172800 \; s = 1313280 \; C\)

Calculate the number of moles of Cr plated out in part (a)

Using Faraday's Law and the charge calculated earlier, we can find the number of moles of Cr plated out: \(n = \frac{q}{3F} = \frac{1313280 \; C}{3 \cdot 96485 \; C/mol} = 4.53 \; mol\)

Calculate the mass of Cr plated out in part (a)

Convert moles of Cr into mass using the molar mass of Cr, which is 52 g/mol: \(mass = n \cdot molar \; mass = 4.53 \; mol \cdot 52 \; g/mol = 235 \; g\)

Calculate the charge needed for part (b)

We have moles of Cr in part (b), so we can find the required charge using Faraday's Law: \(q = n \cdot 3F = 0.250 \; mol \cdot 3 \cdot 96485 \; C/mol = 72488 \; C\)

Calculate the necessary current (amperage) in part (b)

First, convert the time of 8.00 h to seconds: \(t = 8.00 \; h \cdot \frac{3600}{h} = 28800 \; s\) Then calculate the current (amperage) using the charge and time: \(I = \frac{q}{t} = \frac{72488 \; C}{28800 \; s} = 2.514 \; A\) The amperage required to plate out 0.250 mol Cr from a Cr³⁺ solution in a period of 8.00 h is 2.514 A.