Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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°) for the involved metals. We have: - Copper (Cu): \( Cu^{2+} + 2e^- \rightarrow Cu(s), \) \(E° = +0.34 V\) - Zinc (Zn): \( Zn^{2+} + 2e^- \rightarrow Zn(s), \) \(E° = -0.76 V\) - Nickel (Ni): \( Ni^{2+} + 2e^- \rightarrow Ni(s), \) \(E° = -0.25 V\)
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 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

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. $$\begin{array}{c} \mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \quad \quad \mathscr{E}^{\circ}=0.90 \mathrm{V} \\ \mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{V} \end{array}$$

Consider the following half-reactions: \(\begin{aligned} \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & & \mathscr{E}^{\circ}=1.188 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{E}^{\circ}=0.755 \mathrm{V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} & \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=0.96 \mathrm{V} \end{aligned}\) Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)

The overall reaction in the lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow \\ 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\) that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} .\) At \(25^{\circ} \mathrm{C}, 8^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \quad \mathrm{Br}_{2}, \quad \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s) ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

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