Given \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V} ; \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}\) \(E_{\mathrm{Ft}^{3}+i \mathrm{Fe}^{2}}=+0.76 \mathrm{~V} ; E^{\circ} \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=+1.60 \mathrm{~V}\) Which of the following statements is not correct? (a) \(\mathrm{Fe}^{3+}\) does not oxidise \(\mathrm{Ce}^{3+}\). (b) Cu reduces \(\mathrm{Ag}^{+}\)to \(\mathrm{Ag}\). (c) Ag will reduce \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\). (d) \(\mathrm{Fe}^{3+}\) reduces \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\).

Short Answer

Expert verified
Statement (c), 'Ag will reduce Cu^{2+} to Cu', is not correct.

Step by step solution

01

Recall the meaning of electrode potential

The electrode potential, also referred to as the standard reduction potential, indicates the tendency of a species to gain electrons and be reduced. The more positive the reduction potential, the greater the tendency of the species to be reduced.
02

Analyze the statements

To determine if a statement is correct, compare the standard reduction potentials of the species involved. The species with the higher reduction potential will more likely undergo reduction while the species with the lower potential will be oxidized.
03

Assessment of statement (a)

Since the reduction potential of Fe^{3+}/Fe^{2+} (+0.76 V) is less than that of Ce^{4+}/Ce^{3+} (+1.60 V), Fe^{3+} has a lower tendency to be reduced and thus cannot oxidize Ce^{3+} to Ce^{4+}. Statement (a) is correct.
04

Assessment of statement (b)

Comparing the reduction potentials of Ag^{+}/Ag (0.80 V) and Cu^{2+}/Cu (0.34 V), since Ag^{+} has a higher potential, it has a greater tendency to be reduced than Cu^{2+}, so copper can reduce Ag^{+} to Ag. Statement (b) is correct.
05

Assessment of statement (c)

Again, referring to the reduction potentials, because Ag^{+} has a higher potential than Cu^{2+}, Ag will not reduce Cu^{2+} to Cu. Statement (c) is not correct.
06

Assessment of statement (d)

Comparing the potentials of Fe^{3+}/Fe^{2+} (0.76 V) to Cu^{2+}/Cu (0.34 V), we can conclude that Fe^{3+} can reduce Cu^{2+} to Cu because Fe^{3+} has the lower reduction potential and is more likely to be oxidized. Statement (d) is correct.

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Key Concepts

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

Electrochemical Series
The electrochemical series, also known as the activity series, is a list that ranks chemical elements or ions based on their standard reduction potentials. This list is fundamentally important in the study of electrochemistry, as it helps predict the outcome of redox reactions.

Elements with higher reduction potentials are placed higher in the series and are more likely to be reduced, that is, to gain electrons. Conversely, those with lower reduction potentials appear lower in the series and are more likely to lose electrons and be oxidized. The electrochemical series serves as a guide to determine the direction of electron flow, and thus which substance will act as an oxidizing agent and which will act as a reducing agent.
Electrochemical series can:
  • Help determine which metals will corrode first in a given environment

  • Assist in constructing galvanic cells where two different metals can produce electric current

  • Predict how a metal will behave in electroplating processes

  • In the given exercise, comparing the standard reduction potentials was essential to predict reactions between different ions, aligning seamlessly with the electrochemical series.
    Redox Reactions
    Redox reactions, short for reduction-oxidation reactions, are processes that involve the transfer of electrons between two substances. During these reactions, the oxidation state of molecules, atoms, or ions changes. In every redox reaction, two concurrent processes occur: oxidation (loss of electrons) and reduction (gain of electrons).

    Understanding these two processes is critical:
    • Oxidation involves an increase in oxidation state and is accompanied by a loss of electrons.
    • Reduction involves a decrease in oxidation state and is accompanied by a gain of electrons.

    The substances that donate electrons are called reducing agents or reductants, and the substances that accept electrons are known as oxidizing agents or oxidants. An easy way to remember this is by the mnemonic 'LEO says GER' (Lose Electrons Oxidation, Gain Electrons Reduction).

    In the textbook problem, identifying the correct redox behavior of ions based on their standard reduction potentials allowed for the correct evaluation of the given statements.
    Electron Transfer in Redox Reactions
    Electron transfer is the heart of all redox reactions. It is this movement of electrons from one species to another that defines whether a substance is being oxidized or reduced. To predict the direction of electron transfer, one must consider the relative standard reduction potentials of the reactants involved.
  • A species with a higher reduction potential will have a greater tendency to gain electrons and be reduced.

  • A species with a lower reduction potential will be more inclined to lose electrons and be oxidized.


  • For instance, in our textbook problem, the direction of electron transfer between ions was inferred based on their standard reduction potentials. This understanding is crucial for mastering redox reactions and is applicable in designing batteries, understanding corrosion, and many other applications in chemistry and industry.

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    Most popular questions from this chapter

    Which of the following species has an atom with \(+6\) oxidation state? (a) \(\mathrm{MnO}_{4}^{-}\) (b) \(\mathrm{Cr}(\mathrm{CN})_{6}^{3-}\) (c) \(\mathrm{NiF}_{6}^{2-}\) (d) \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\)

    Indicate whether the following conversions represent an oxidation, a reduction or none (neither oxidation \(\begin{aligned}&\text { nor reduction). } \\\&\mathrm{HClO}_{4} & \text { (ii) } \mathrm{NH}_{4}^{+} \text {to } \mathrm{NH}_{3}\end{aligned}\) (i) \(\mathrm{HClO}_{3}\) to \(\mathrm{HClO}_{4}\) (iii) \(\mathrm{NO}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}_{4}\) (iv) \(\mathrm{HSO}_{3}^{-}\)to \(\mathrm{SO}_{4}^{2-}\) (v) \(\mathrm{H}_{2} \mathrm{O}_{2}\) to \(\mathrm{H}_{2} \mathrm{O}\) (i) (ii) (iii) (iv) (v) (a) Oxida- Reduc- None None Oxidation tion tion (b) Oxida- None None Oxida- Reduction tion tion (c) Reduc- Oxida- Reduc- None Reduction tion tion tion (d) Oxida- Reduc- None Reduc- Reduction tion tion tion

    Based on the following reactions, arrange the metals in increasing order of their reduction potentials. $$ \mathbf{Z n}+\mathbf{C u}^{2+} \rightarrow \mathbf{Z n}^{2+}+\mathrm{Cu} ; $$ \(\mathrm{Mg}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Zn}\) \(\mathrm{Cu}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\) (a) \(\quad \mathrm{Mg}>\mathrm{Zn}>\mathrm{Cu}>\mathrm{Ag}\) (b) \(\mathrm{Mg}<\mathrm{Zn}<\mathrm{Cu}<\mathrm{Ag}\) (d) \(\quad \mathrm{Mg}>\mathrm{Cu}>\mathrm{Zn}>\mathrm{Ag}\) (c) \(\quad \mathrm{Zn}<\mathrm{Cu}<\mathrm{Ag}<\mathrm{Mg}\)

    What is the oxidation number of carbon in \(\mathrm{C}_{3} \mathrm{O}_{2}\) (carbon suboxide)? (a) \(+4 / 3\) (b) \(+10 / 4\) (c) \(+2\) (d) \(+2 / 3\)

    Which compound amongst the following has the highest oxidation number of Mn? (a) \(\mathrm{KMnO}_{4}\) (b) \(\mathrm{K}_{2} \mathrm{MnO}_{4}\) (c) \(\mathrm{MnO}_{2}\) (d) \(\mathrm{Mn}_{2} \mathrm{O}_{3}\)

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