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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°) 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.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

Which of the following statements concerning corrosion is(are) true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

Consider the cell described below: $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M)\right|\left|\mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{mol} / \mathrm{L}\). (Assume \(T=25^{\circ} \mathrm{C} .\) )

It took 2.30 min using a current of 2.00 A to plate out all the silver from 0.250 L of a solution containing Ag \(^{+} .\) What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
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