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Chapter 18: 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
Zinc should be used instead of copper for the plating process because it has a more negative reduction potential (\(E°_{Zn} = -0.76 V\)), making it more reactive than copper (\(E°_{Cu} = +0.34 V\)). This higher reactivity will allow zinc to more easily lose electrons, enabling the deposition of nickel metal from the nickel nitrate solution.

Step by step solution

01

Identify the Metals' Standard Reduction Potentials

The first step is to identify the standard reduction potentials of the metals in question, which will allow us to determine which metal is more reactive. The standard reduction potentials of copper, zinc, and nickel are as follows: Copper (Cu): \[E°_{Cu} = +0.34 V\] Zinc (Zn): \[E°_{Zn} = -0.76 V\] Nickel (Ni): \[E°_{Ni} = -0.25 V\]
02

Compare the Reactivities

Next, compare the reactivity of the metals by comparing their standard reduction potentials. Remember that the more negative the reduction potential, the more likely a metal is to lose electrons, which is indicative of its reactivity. Since \(E°_{Zn} = -0.76 V\) and \(E°_{Cu} = +0.34 V\), it is clear that zinc (\(Zn\)) has a more negative reduction potential than copper (\(Cu\)). Consequently, zinc is more reactive than copper.
03

Choose the Metal for the Plating Process

Given that zinc is more reactive than copper, when placed in a nickel nitrate solution, zinc would be more likely to lose electrons. This would result in the zinc dissolving into the solution and the nickel ions in the solution gaining electrons and depositing as metallic nickel onto the zinc metal surface. Therefore, to plate out nickel metal from a nickel nitrate solution, zinc should be used instead of copper.

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Most popular questions from this chapter

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{KF}\) b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)

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 \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) 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.

Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$\mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q)$$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

Electrolysis of an alkaline earth metal chloride using a current of \(5.00 \mathrm{~A}\) for \(748 \mathrm{~s}\) deposits \(0.471 \mathrm{~g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?
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