Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 89

How long will it take to plate out each of the following with a current of \(100.0 \mathrm{~A}\) ? a. \(1.0 \mathrm{~kg} \mathrm{Al}\) from aqueous \(\mathrm{Al}^{3+}\) b. \(1.0 \mathrm{~g} \mathrm{Ni}\) from aqueous \(\mathrm{Ni}^{2+}\) c. \(5.0 \mathrm{~mol} \mathrm{Ag}\) from aqueous \(\mathrm{Ag}^{+}\)

Short Answer

Expert verified
The time it will take to plate out the given substances from their respective aqueous ions using a current of 100.0 A are as follows: a. 1.0 kg of Al will take approximately \(107,524.2~s\). b. 1.0 g of Ni will take approximately \(32.7~s\). c. 5.0 mol of Ag will take approximately \(4,824.25~s\).

Step by step solution

01

Calculate the charges per mole of the given ions

First, we need to determine the charges per mole for each ion. a. Al³⁺ → Al requires 3 moles of electrons for each mole of aluminum (3+ charge). b. Ni²⁺ → Ni requires 2 moles of electrons for each mole of nickel (2+ charge). c. Ag⁺ → Ag requires 1 mole of electron for each mole of silver (1+ charge).
02

Calculate the moles of the desired substances

We'll now calculate the number of moles of the given substances using their molecular weights. The atomic weights of Al, Ni, and Ag are 26.98 g/mol, 58.69 g/mol, and 107.87 g/mol, respectively. a. \(1.0~kg \mathrm{Al} = 1000~g = \frac{1000}{26.98} = 37.06~mol \mathrm{Al}\) b. \(1.0~g \mathrm{Ni} = \frac{1.0}{58.69} = 0.0170~mol \mathrm{Ni}\) c. \(5.0~mol \mathrm{Ag}\) (already given)
03

Calculate the total charge required for each deposition

Now, we'll multiply the moles of each substance by their respective charges per mole and Faraday's constant to get the total amount of charge required. a. \(Q_{Al} = 37.06~mol \times 3 ~mol \frac{e^-}{mol}(Al) \times 96,485~C/mol(e^-) = 10,752,421.3 ~C\) b. \(Q_{Ni} = 0.0170~mol \times 2 ~mol \frac{e^-}{mol}(Ni) \times 96,485~C/mol(e^-) = 3,273.77 ~C\) c. \(Q_{Ag} = 5.0~mol \times 1 ~mol \frac{e^-}{mol}(Ag) \times 96,485~C/mol(e^-) = 482,425 ~C\)
04

Calculate the time required for the deposition using the provided current

We'll now use the given current of 100.0 A to find the time it takes to deposit each substance using the calculated total charge by dividing the charge by the current. a. \(t_{Al} = \frac{Q_{Al}}{I} = \frac{10,752,421.3 ~C}{100.0 ~A} = 107,524.2~s\) b. \(t_{Ni} = \frac{Q_{Ni}}{I} = \frac{3,273.77 ~C}{100.0 ~A} = 32.7~s\) c. \(t_{Ag} = \frac{Q_{Ag}}{I} = \frac{482,425 ~C}{100.0 ~A} = 4,824.25~s\) Thus, it will take approximately 107,524.2 seconds for (a), 32.7 seconds for (b), and 4,824.25 seconds for (c) with a current of 100.0 A to plate out the given substances from their respective aqueous ions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the cell described below: $$\mathrm{Al}\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb}$$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Al}^{3+}\right]\) to have changed by \(0.60 \mathrm{~mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$

Although aluminum is one of the most abundant elements on earth, production of pure Al proved very difficult until the late 1800 s. At this time, the Hall- Heroult process made it relatively easy to produce pure Al. Why was pure Al so difficult to produce and what was the key discovery behind the Hall-Heroult process?

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} M\) and the measured cell potential is \(1.82 \mathrm{~V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=?$$

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\) c. \(\begin{aligned} \mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Acid}}{\longrightarrow} \\ & \mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q) \\ \text { d. } \mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow} \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \\ \text { e. } \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow} \\\ \mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q) \end{aligned}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free