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Chapter 20: Problem 9

\(\mathrm{Ni}^{2+}\) has a more positive reduction potential than \(\mathrm{Cd}^{2+}\) (a) Which ion is more easily reduced to the metal? (b) Which metal, Ni or Cd, is more easily oxidized?

Short Answer

Expert verified
a) \(\mathrm{Ni}^{2+}\) is more easily reduced to a metal. b) \(\mathrm{Cd}\) is more easily oxidized.

Step by step solution

01

Interpret Reduction Potential

Reduction means a species gaining electrons. The reduction potential of a species measures how likely it is to gain electrons, in other words, be reduced. When comparing two species, the one with a higher (more positive) reduction potential is more likely to be reduced. In our case, \(\mathrm{Ni}^{2+}\) has a more positive reduction potential than \(\mathrm{Cd}^{2+}\).
02

Identify Ion Easier to Reduce

According to the first step, the ion with a higher reduction potential is more likely to be reduced. Therefore, since \(\mathrm{Ni}^{2+}\) has a higher reduction potential than \(\mathrm{Cd}^{2+}\), \(\mathrm{Ni}^{2+}\) is more easily reduced to the metal.
03

Identify Metal Easier to Oxidize

Oxidation is the process of a species losing electrons. A metal with a lower reduction potential is more likely to lose electrons, i.e., it is more easily oxidized. Thus, since \(\mathrm{Cd}\) has a lower reduction potential than \(\mathrm{Ni}\), \(\mathrm{Cd}\) is more readily oxidized

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Key Concepts

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

Oxidation and Reduction
In chemistry, oxidation and reduction refer to processes involving the transfer of electrons between substances. They are two sides of what is known as a redox reaction, with oxidation being the loss of electrons by a molecule, atom, or ion, and reduction being the gain of electrons.

Easy ways to remember these concepts are:
  • Oxidation: Loss of electrons (OIL = Oxidation Is Loss)
  • Reduction: Gain of electrons (RIG = Reduction Is Gain)
When a substance is reduced, it gains electrons, making it more negative or less positive. This makes reduction comparable to filling a bucket with electrons. Conversely, oxidation is like emptying that bucket.

When considering two or more substances, the one that holds its electrons more tightly, or attracts additional electrons more effectively, is said to reduce more easily. In this context, we compare the reduction potential of different ions.
Electrode Potential
Electrode potential is a measure of how much an ion or element "wants" to gain electrons and be reduced. Think of it as an indicator of how greedy an element is for electrons.

Higher reduction potential indicates a greater tendency for the species to be reduced, meaning it can attract electrons effectively. Thus, in chemical reactions, comparing electrode potentials helps us predict which ions are reduced or oxidized.

Reduction potential is measured in volts (V), and substances are compared using a reference electrode. A positive reduction potential means an element is more likely to accept electrons. This is why \(\mathrm{Ni}^{2+}\) with a more positive reduction potential than \(\mathrm{Cd}^{2+}\) is more easily reduced to metallic nickel.
Electrochemistry
Electrochemistry is the branch of chemistry concerned with the interrelation of electrical and chemical processes. It plays a crucial role in real-world applications such as batteries, electroplating, and electrolysis.

In electrochemical cells, redox reactions occur, where oxidation happens at the anode and reduction at the cathode. These cells can be galvanic, like batteries, where chemical energy is converted into electrical energy, or electrolytic, where electrical energy drives non-spontaneous reactions.

  • Galvanic Cells: Spontaneously generate electricity through redox reactions.
  • Electrolytic Cells: Require external electricity to prompt a chemical change.
Understanding electrochemistry is essential for harnessing these processes, whether storing energy in a battery or refining metals. It relies heavily on concepts such as electrode potentials, where the difference in potential between electrodes determines the overall voltage of the cell.

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Most popular questions from this chapter

Consider the following electrochemical cell: $$ \operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm})\right| \mathrm{H}^{+}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(x \mathrm{M}) | \mathrm{Ag}(\mathrm{s}) $$ (a) What is \(E_{\text {cell }}^{\circ}-\) that is, the cell potential when \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M} ?\) (b) Use the Nernst equation to write an equation for \(E_{\text {cell }}\) when \(\left[\mathrm{Ag}^{+}\right]=x\) (c) Now imagine titrating \(50.0 \mathrm{mL}\) of \(0.0100 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) in the cathode half-cell compartment with 0.0100 M KI. The titration reaction is $$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{AgI}(\mathrm{s})$$ Calculate \(\left[\mathrm{Ag}^{+}\right]\) and then \(E_{\text {cell }}\) after addition of the following volumes of \(0.0100 \mathrm{M} \mathrm{KI}:(\mathrm{i}) 0.0 \mathrm{mL} ;(\mathrm{ii}) 20.0 \mathrm{mL}\) (iii) \(49.0 \mathrm{mL} ;(\text { iv }) 50.0 \mathrm{mL} ;(\mathrm{v}) 51.0 \mathrm{mL} ;(\mathrm{vi}) 60.0 \mathrm{mL}\) (d) Use the results of part (c) to sketch the titration curve of \(E_{\text {cell }}\) versus volume of titrant.

Write an equation to represent the oxidation of \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) by \(\mathrm{PbO}_{2}(\mathrm{s})\) in an acidic solution. Will this reaction occur spontaneously in the forward direction if all other reactants and products are in their standard states and (a) \(\left[\mathrm{H}^{+}\right]=6.0 \mathrm{M} ;\) (b) \(\left[\mathrm{H}^{+}\right]=1.2 \mathrm{M}\) (c) \(\mathrm{pH}=4.25 ?\) Explain.

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

Describe how you might construct batteries with each of the following voltages: (a) \(0.10 \mathrm{V} ;\) (b) \(2.5 \mathrm{V} ;\) (c) \(10.0 \mathrm{V}\). Be as specific as you can about the electrodes and solution concentrations you would use, and indicate whether the battery would consist of a single cell or two or more cells connected in series.

A solution containing both \(\mathrm{Ag}^{+}\) and \(\mathrm{Cu}^{2+}\) ions is subjected to electrolysis. (a) Which metal should plate out first? (b) Plating out is finished after a current of \(0.75 \mathrm{A}\) is passed through the solution for 2.50 hours. If the total mass of metal is \(3.50 \mathrm{g},\) what is the mass percent of silver in the product?
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