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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
Copper should be used to plate out nickel from a nickel nitrate solution because of its higher standard reduction potential (+0.34 V) compared to nickel (-0.26 V) and zinc (-0.76 V). This makes it more favorable for copper to be oxidized, allowing nickel ions to be reduced and deposited onto the copper surface.

Step by step solution

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1. Identify the relevant half-reactions

First, we'll need to consider the half-reactions for the reduction of the different metal ions to their metallic form: Ni^(2+) + 2 e^(-) -> Ni (Reduction half-reaction for nickel) Cu^(2+) + 2 e^(-) -> Cu (Reduction half-reaction for copper) Zn^(2+) + 2 e^(-) -> Zn (Reduction half-reaction for zinc)
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2. Check standard reduction potentials

Standard reduction potentials (E°) are used to compare the tendency of a species to get reduced in a chemical reaction. We will use these values to determine which metal is more easily oxidized. Keep in mind that we want to find the metal that has a higher standard reduction potential than nickel, as it will be more favorable for the reverse reaction, which is oxidation. According to standard reduction potential tables for 25 °C: E°(Ni^(2+)/Ni) = -0.26 V E°(Cu^(2+)/Cu) = +0.34 V E°(Zn^(2+)/Zn) = -0.76 V
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3. Compare the standard reduction potentials

Comparing the standard reduction potentials, we can see that copper has a higher reduction potential than nickel, while zinc has a lower reduction potential. E°(Cu^(2+)/Cu) > E°(Ni^(2+)/Ni) > E°(Zn^(2+)/Zn) This means that it is more favorable for copper to be reduced and for nickel to be oxidized, causing copper to lose electrons (oxidation) and nickel ions to gain electrons (reduction).
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4. Conclusion and Recommendation

Based on the standard reduction potentials, copper is the better choice to plate out nickel metal from a nickel nitrate solution because copper has a higher reduction potential than nickel; thus, it will be more easily oxidized, allowing the nickel ions to be reduced and deposited onto the copper surface.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Reduction Potential
Standard Reduction Potential (SRP) is a crucial concept in electrochemistry which helps us predict the direction of electron flow in a reaction. It is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-reaction has an associated standard reduction potential, measured in volts (V), which is reported under standard conditions: a temperature of 25°C, a 1 molar concentration, and a pressure of 1 atmosphere.

The higher the SRP value, the greater the species' affinity for electrons and the more likely it is to be reduced. When comparing different metals, such as in electroplating, we can use these SRP values to determine which metal will act as the cathode (where reduction occurs) and which will act as the anode (where oxidation occurs).

For instance, in the provided exercise, we compared the SRP of nickel, copper, and zinc to determine the best metal to use for plating. Copper, with its positive SRP, indicates it is more likely to be reduced than nickel and zinc, making it the preferred choice for the cathode where nickel will be plated out of solution.
Half-Reactions
Half-reactions are a way of breaking down the overall chemical reaction into its oxidation and reduction components. Every electrochemical reaction can be divided into two half-reactions: one that involves the loss of electrons (oxidation) and one that involves the gain of electrons (reduction).

In the context of electroplating, like plating nickel from a solution onto a metal, it is important to identify the oxidation and reduction half-reactions for each metal involved. This allows us to determine which metal will give up electrons ('be oxidized') and which metal will gain electrons ('be reduced').

Understanding Half-Reactions in Electroplating

For electroplating nickel onto a surface, we look at the half-reaction of nickel reduction. By comparing this half-reaction to those of other metals, like copper and zinc, in the provided exercise, we determined the suitability of each metal to serve as an electrode for the plating process.
Oxidation and Reduction
Oxidation and reduction are chemical processes involving the transfer of electrons between species. Oxidation is the loss of electrons, while reduction is the gain of electrons. These two processes always occur together; when one species is oxidized, another is reduced, which is why we refer to them as redox reactions.

In an electroplating scenario, the metal that oxidizes, or loses electrons, corrodes and dissolves into the solution, whereas the metal that reduces, or gains electrons, is deposited onto a surface. The choice of which metal to use as the anode for electroplating is guided by its tendency to oxidize.

To choose the correct metal for 'plating out' nickel, we must consider the ease with which the metals can be oxidized. From the exercise solution, it is clear that the positive standard reduction potential of copper means it can be oxidized more readily than nickel, whereas zinc, with a lower potential, is less likely to oxidize. Hence, copper is the suitable metal for nickel electroplating.

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Most popular questions from this chapter

Combine the equations $$ \Delta G^{\circ}=-n F \mathscr{C}^{\circ} \quad \text { and } \quad \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} $$ to derive an expression for \(\mathscr{E}^{\circ}\) as a function of temperature. Describe how one can graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) from measurements of \(\mathscr{E}^{\circ}\) at different temperatures, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{~V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}\) \(\mathscr{C}^{\circ}=0.68 \mathrm{~V}\) \(\begin{array}{ll}\text { b. } \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} & \mathscr{E}^{\circ}=-1.18 \mathrm{~V} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.036 \mathrm{~V}\end{array}\)

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3} . \mathrm{Ox}-\) ide ions can move through this solid at high temperatures (about \(\left.800^{\circ} \mathrm{C}\right) . \Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is \(-380 \mathrm{~kJ}\). Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

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.

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