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Chapter 20: Problem 88

Metallic magnesium can be made by the electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) What mass of \(\mathrm{Mg}\) is formed by passing a current of \(4.55\) A through molten \(\mathrm{MgCl}_{2}\), for \(3.50\) days? (b) How many minutes are needed to plate out \(10.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\), using \(3.50 \mathrm{~A}\) of current?

Short Answer

Expert verified
(a) Approximately 692.00 g of Mg is formed by passing a current of 4.55 A through molten MgCl2 for 3.50 days. (b) It takes approximately 94.63 minutes to plate out 10.00 g of Mg from molten MgCl2 using 3.50 A of current.

Step by step solution

01

(a) Calculate the charge passed through molten MgCl2

First, we need to find the total charge passed through the molten MgCl2. We are given the current (4.55 A) and time (3.50 days). Convert the time to seconds and calculate the charge. Time = 3.50 days 脳 24 hours/day 脳 3600 seconds/hour = 302400 seconds Charge (Q) = Current (I) 脳 Time (t) Q = 4.55 A 脳 302400 s Q = 1375920 Coulombs
02

(a) Calculate the moles of Mg formed

Now, we will calculate the number of moles of Mg formed during electrolysis. Magnesium has a valency of +2, so z = 2. n = (Q 脳 z) / F n = (1375920 C 脳 2) / 96485 C/mol n = 28.490 moles
03

(a) Calculate the mass of Mg formed

Now, we will calculate the mass of Mg formed during electrolysis. The molar mass of Mg is 24.305 g/mol. Mass of Mg = Moles of Mg 脳 Molar mass of Mg Mass of Mg = 28.490 moles 脳 24.305 g/mol Mass of Mg 鈮� 692.00 g So, approximately 692.00 g of Mg is formed by passing a current of 4.55 A through molten MgCl2 for 3.50 days.
04

(b) Calculate the moles of Mg to plate out

We are given that 10.00 g of Mg needs to be plated out. We will first calculate the number of moles of Mg. Moles of Mg = Mass of Mg / Molar mass of Mg Moles of Mg = 10.00 g / 24.305 g/mol Moles of Mg 鈮� 0.4116 moles
05

(b) Calculate the required charge for plating out Mg

Now, we will find the required charge to plate out the given moles of Mg. We will use the formula n = (Q 脳 z) / F and solve for Q. Q = n 脳 F / z Q = 0.4116 moles 脳 96485 C/mol / 2 Q 鈮� 19874.42 Coulombs
06

(b) Calculate the time in minutes to plate out Mg

Now, we will find the time needed to plate out Mg using 3.50 A of current. We will use the formula Q = It and solve for t. Time (t) = Charge (Q) / Current (I) t = 19874.42 C / 3.50 A t 鈮� 5678.12 seconds Now, convert the time to minutes: Time = 5678.12 seconds 脳 (1 minute / 60 seconds) 鈮� 94.63 minutes So, it takes approximately 94.63 minutes to plate out 10.00 g of Mg from molten MgCl2 using 3.50 A of current.

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Most popular questions from this chapter

Two important characteristics of voltaic cells are their cell potential and the total charge that they can deliver. Which of these characteristics depends on theamount of reactants in the cell, and which one depends on their concentration?

If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(8.7 \times 10^{4}\), calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {cell }}^{0}\)

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g)-\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow\) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)-\cdots\) \(3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q)-\mathrm{-} \rightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

A common shorthand way to represent a voltaic cell is to list its components as follows: anode|anode solution || cathode solution|cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}\); sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2} ;\) sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: \(\mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q)-\rightarrow\) $$ \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ Pt is used as an inert electrode in contactwith the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Sketch the cell.

Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(a q)\), (b) \(\mathrm{Ca}(\mathrm{s})\), (c) \(\mathrm{ClO}_{3}^{-}(a q)\), (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ?
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