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Ch.19 - Electrochemistry
Tro - Chemistry: A Molecular Approach 4th Edition
Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831Not the one you use?Change textbook
All textbooksTro 4th EditionCh.19 - ElectrochemistryProblem 44b,c
Chapter 19, Problem 44b,c

Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the direction of electron flow.
b. 2 H+(aq) + Fe(s) → H2(g) + Fe2+(aq)
c. 2 NO3–(aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)

Verified step by step guidance
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Step 1: Identify the oxidation and reduction half-reactions for each redox reaction. For reaction b, Fe(s) is oxidized to Fe^{2+}(aq), and H^{+}(aq) is reduced to H_{2}(g). For reaction c, Cu(s) is oxidized to Cu^{2+}(aq), and NO_{3}^{-}(aq) is reduced to NO(g).
Step 2: Determine the anode and cathode for each reaction. The anode is where oxidation occurs, and the cathode is where reduction occurs. For reaction b, Fe(s) is the anode, and H^{+}(aq) is the cathode. For reaction c, Cu(s) is the anode, and NO_{3}^{-}(aq) is the cathode.
Step 3: Write the half-reactions for each electrode. For reaction b, the anode half-reaction is Fe(s) → Fe^{2+}(aq) + 2e^{-}, and the cathode half-reaction is 2H^{+}(aq) + 2e^{-} → H_{2}(g). For reaction c, the anode half-reaction is Cu(s) → Cu^{2+}(aq) + 2e^{-}, and the cathode half-reaction is 2NO_{3}^{-}(aq) + 8H^{+}(aq) + 6e^{-} → 2NO(g) + 4H_{2}O(l).
Step 4: Label the species present in each solution. For reaction b, the anode solution contains Fe^{2+}(aq), and the cathode solution contains H^{+}(aq). For reaction c, the anode solution contains Cu^{2+}(aq), and the cathode solution contains NO_{3}^{-}(aq) and H^{+}(aq).
Step 5: Indicate the direction of electron flow. Electrons flow from the anode to the cathode in both reactions. For reaction b, electrons flow from Fe(s) to H^{+}(aq). For reaction c, electrons flow from Cu(s) to NO_{3}^{-}(aq).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Voltaic Cell Structure

A voltaic cell, also known as a galvanic cell, consists of two electrodes: the anode and the cathode, immersed in electrolyte solutions. The anode is where oxidation occurs, releasing electrons, while the cathode is where reduction takes place, accepting electrons. The flow of electrons from the anode to the cathode generates electrical energy, which can be harnessed for work.
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Redox Reactions

Redox reactions involve the transfer of electrons between two species, where one species is oxidized (loses electrons) and the other is reduced (gains electrons). In the provided reactions, identifying the oxidation states of the elements helps determine which species undergoes oxidation and which undergoes reduction. Understanding these processes is crucial for sketching the voltaic cell accurately.

Half-Reactions

Half-reactions represent the individual oxidation and reduction processes occurring in a redox reaction. Each half-reaction shows the species involved, the electrons transferred, and the changes in oxidation states. By writing the half-reactions for the given reactions, one can clearly label the anode and cathode in the voltaic cell and understand the overall electron flow and species present in each half-cell.
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Related Practice
Textbook Question

Balance each redox reaction occurring in basic aqueous solution. c. NO2–(aq) + Al(s) → NH3(g) + AlO2–(aq)

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Textbook Question

Consider the voltaic cell:

d. Indicate the direction of anion and cation flow in the salt bridge

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Textbook Question

Balance each redox reaction occurring in basic aqueous solution. a. MnO4–(aq) + Br–(aq) → MnO2(s) + BrO3–(aq)

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Textbook Question

Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the direction of electron flow.

a. 2 Ag+(aq) + Pb(s) → 2 Ag(s) + Pb2+(aq)

b. 2 ClO2(g) + 2 I–(aq) → 2 ClO2–(aq) + I2(s)

c. O2(g) + 4 H+(aq) + 2 Zn(s) → 2 H2O(l) + 2 Zn2+(aq)

Textbook Question

Use line notation to represent each electrochemical cell in Problem 43.

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