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\(\mathrm{Ni}^{2+}\) 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

Of the following statements concerning electrochemical cells, the correct ones are: (a) The cathode is the negative electrode in both voltaic and electrolytic cells. (b) The function of a salt bridge is to permit the migration of electrons between the half-cell compartments of an electrochemical cell. (c) The anode is the negative electrode in a voltaic cell. (d) Electrons leave the cell from either the cathode or the anode, depending on what electrodes are used. (e) Reduction occurs at the cathode in both voltaic and electrolytic cells. (f) If electric current is drawn from a voltaic cell long enough, the cell becomes an electrolytic cell. (g) The cell reaction is an oxidationreduction reaction.

Use the Nernst equation and data from Appendix D to calculate \(E_{\text {rell for each of the following cells. }}\) (a) \(\operatorname{Mn}(\mathrm{s}) | \mathrm{Mn}^{2+}(0.40 \mathrm{M}) \| \mathrm{Cr}^{3+}(0.35 \mathrm{M})\) \(\mathrm{Cr}^{2+}(0.25 \mathrm{M}) | \mathrm{Pt}(\mathrm{s})\) (b) \(\operatorname{Mg}\left(\text { s) } | \operatorname{Mg}^{2+}(0.016 \mathrm{M}) \|\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(0.25 \mathrm{M})\right.\) \(\mathrm{OH}^{-}(0.042 \mathrm{M}) | \mathrm{Al}(\mathrm{s})\)

If a chemical reaction is carried out in a fuel cell, the maximum amount of useful work that can be obtained is (a) \(\Delta G ;\) (b) \(\Delta H ;\) (c) \(\Delta G / \Delta H ;\) (d) \(T \Delta S\).

\(E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) for the reaction, \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\) \(\operatorname{sn}^{4+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\) (a) Can a solution be prepared that is \(0.500 \mathrm{M}\) in each of the four ions at \(298 \mathrm{K} ?\) (b) If not, in what direction must a net reaction Occur?

When deciding whether a particular reaction corresponds to a cell with a positive standard cell potential, which of the following thermodynamic properties would you use to get your answer without performing any calculations? Which would you not use? Explain. (a) \(\Delta G^{\circ} ;\) (b) \(\Delta S^{\circ} ;\) (c) \(\Delta H^{\circ} ;\) (d) \(\Delta U^{\circ} ;\) (e) \(K\).

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