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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
In conclusion, to plate out nickel metal from a nickel nitrate solution, zinc should be used instead of copper. This is because zinc is more reactive than nickel and is lower in the electrochemical series (Zn²⁺/Zn = -0.76 V) compared to nickel (Ni²⁺/Ni = -0.25 V). Therefore, zinc will displace nickel ions in the solution, allowing the plating process to occur. On the other hand, copper is less reactive than nickel and will not be able to displace the nickel ions, rendering it unsuitable for plating nickel.

Step by step solution

01

Understand the Electrochemical Series

The electrochemical series is a list of elements arranged in the order of their standard electrode potentials. Elements higher up in the series are more likely to accept electrons and undergo reduction, while elements lower in the series are more likely to lose electrons and undergo oxidation. In relatively simple words, elements higher in the series are more reactive, and elements lower in the series are less reactive.
02

Find the Positions of Nickel, Copper, and Zinc in the Electrochemical Series

We need to locate the positions of nickel, copper, and zinc in the electrochemical series. Their standard electrode potentials are as follows: Nickel (Ni²⁺/Ni) = -0.25 V Copper (Cu²⁺/Cu) = +0.34 V Zinc (Zn²⁺/Zn) = -0.76 V
03

Compare the Reactivity of Copper and Zinc Relative to Nickel

From the standard electrode potentials, we can see that: - Nickel is higher in the electrochemical series compared to copper (Ni²⁺/Ni = -0.25 V and Cu²⁺/Cu = +0.34 V). This means that copper will not displace nickel from the nickel nitrate solution, as it is less reactive than nickel. - Zinc is lower in the electrochemical series compared to nickel (Ni²⁺/Ni = -0.25 V and Zn²⁺/Zn = -0.76 V). This means that zinc will displace nickel from the nickel nitrate solution, as it is more reactive than nickel.
04

Determine the Suitable Metal for Plating Nickel

Based on the comparison of reactivity, using zinc (the more reactive metal) will allow us to plate out nickel metal from the nickel nitrate solution successfully, as it will displace the nickel ions in the solution. On the other hand, using copper (the less reactive metal) will not result in plating nickel, as it won't be able to displace the nickel ions. In conclusion, we should use zinc to plate out nickel metal from the nickel nitrate solution.

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Key Concepts

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

Reactivity
Reactivity refers to how easily an element or compound can undergo a chemical reaction with another substance. In the context of electrochemistry, particularly when discussing plating, reactivity is influenced by standard electrode potentials. Elements higher up in the electrochemical series are considered more reactive because they are more prone to reduction — they tend to gain electrons easily. Conversely, elements lower in the series are more prone to oxidation, meaning they lose electrons readily.

When deciding whether copper or zinc is more suitable for plating nickel, their relative reactivity compared to nickel is key. Copper, with a higher standard electrode potential than nickel, is less reactive and won’t lose electrons to allow nickel ions to convert into nickel metal. Zinc, on the other hand, has a lower standard electrode potential, making it more reactive than nickel. This reactivity will enable zinc to displace nickel ions in the solution, effectively facilitating the plating process.
Standard Electrode Potentials
Standard electrode potentials (often denoted as E° values) are a measure of the tendency of a chemical species to be reduced, which is to accept electrons. It is measured in volts under standard conditions, with all solutions having a concentration of 1 mol/L at a temperature of 25°C (298 K).

For nickel, copper, and zinc, their E° values are as follows:
  • Nickel (Ni²⁺/Ni) = -0.25 V
  • Copper (Cu²⁺/Cu) = +0.34 V
  • Zinc (Zn²⁺/Zn) = -0.76 V
These values indicate that copper is most likely to undergo reduction (positive E°), whereas zinc is least likely (most negative E°). Nickel lies between the two in terms of reactivity. The more negative the potential, the greater the tendency to lose electrons and undergo oxidation. This characteristic is crucial for determining which metal displaces another in plating processes.
Nickel Plating
Nickel plating is a process of coating a surface with nickel using an electrochemical method. This process involves the transfer of nickel ions from a solution onto a metallic surface, creating a lustrous and corrosion-resistant coating.

In nickel plating, choosing the correct metal to facilitate the transfer efficiently from solution to the target metal is essential. Ideally, the chosen metal should have a more negative standard electrode potential than nickel. Zinc fits this criterion well, given its higher reactivity.

The steps in nickel plating typically involve:
  • Cleaning the base metal to remove impurities.
  • Preparing a nickel nitrate solution for the ion transfer process.
  • Submerging the base metal in the solution.
  • Allowing the more reactive metal (zinc) to displace nickel ions, leading to nickel deposition on the base metal.
This methodology ensures effective plating, leveraging the principles of reactivity and standard electrode potentials to achieve a durable nickel coating.

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Most popular questions from this chapter

One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is $$2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}$$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to \(\mathrm{H}_{2} \mathrm{N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2},\) which is used to produce production of nylon. What current must be used to produce \(150 . \mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?

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, 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.

A galvanic cell is based on the following half-reactions: $$\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \quad \mathscr{E}^{\circ}=0.34 \mathrm{V}$$ $$\mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) \quad \mathscr{E}^{\circ}=-1.20 \mathrm{V}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M,\) and the vanadium compartment contains a vanadium electrode and \(\mathrm{V}^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with 0.0800\(M \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) , resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=?$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 \(\mathrm{mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) 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.

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Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

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