/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 You want to "plate out" nickel m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 3

You want to "plate out" nickel metal from a nickel nitrate solution onto a piece of metal inserted into the solution. Should you use copper or zinc? Explain.

Short Answer

Expert verified
You should use copper to plate out nickel metal from the nickel nitrate solution. This is because copper has a higher standard reduction potential (+0.34 V) than nickel (-0.25 V), indicating a greater tendency for reduction to occur at the electrode surface. Zinc would not be effective as its reduction potential is lower than that of nickel, and it would more likely dissolve into the solution instead of plating the nickel.

Step by step solution

01

Recall the reduction half-reactions and potentials

To solve this problem, we need to recall the reduction half-reactions and their standard reduction potentials (E掳) for the involved metals. We have: - Copper (Cu): \( Cu^{2+} + 2e^- \rightarrow Cu(s), \) \(E掳 = +0.34 V\) - Zinc (Zn): \( Zn^{2+} + 2e^- \rightarrow Zn(s), \) \(E掳 = -0.76 V\) - Nickel (Ni): \( Ni^{2+} + 2e^- \rightarrow Ni(s), \) \(E掳 = -0.25 V\)
02

Compare reduction potentials

Now, we compare the standard reduction potentials of copper, zinc, and nickel. A positive potential indicates a greater tendency for reduction to occur at the electrode surface, while a negative potential indicates a lower tendency. Looking at the values, we can see that copper has the highest potential, nickel has the intermediate potential, and zinc has the lowest potential. When two metals are placed in contact with each other in a solution, the metal with the higher potential will have a better chance of reducing the ions in solution.
03

Determine the best metal for plating out nickel

Since the goal is to plate out nickel metal from a nickel nitrate solution, we should choose the metal with a higher reduction potential than nickel. In this case, copper has a higher potential (+0.34 V) than nickel (-0.25 V). Therefore, you should use copper to plate out nickel metal from the nickel nitrate solution. Using zinc would not be effective since its reduction potential is lower than that of nickel, and zinc would more likely dissolve into the solution instead of plating the nickel.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

A galvanic cell is based on the following half-reactions: $$ \begin{array}{ll} \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{V} \end{array} $$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} \mathrm{M}\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00\) atm, and a weak acid, HA, at an initial concentration of \(1.00 \mathrm{M}\). If the observed cell potential is \(0.333 \mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between \(\Delta G\) and cell potential to do this.)

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at \(\mathrm{STP}\) are produced from the electrolysis of water by a current of \(2.50 \mathrm{A}\) in \(15.0 \mathrm{min} ?\)

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu} $$ The mass of each electrode is \(200 .\) g. a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after 10.0 A of current has flowed for \(10.0 \mathrm{h}\). (Assume each half-cell contains \(1.00 \mathrm{L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{h}\). d. How long can this battery deliver a current of 10.0 A before it goes dead?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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.