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Ch.20 - Electrochemistry
All textbooksBrown 14th EditionCh.20 - ElectrochemistryProblem 76
Chapter 20, Problem 76

The question is quite comprehensive but could be slightly confusing due to the presentation of chemical equations. Here is a more reader-friendly version: 'Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are: 1. HgO(s) + H2O(l) + 2 e⁻ → Hg(l) + 2 OH⁻(aq) 2. Zn(s) + 2 OH⁻(aq) → ZnO(s) + H2O(l) + 2 e⁻ (b) The value of E°_red for the cathode reaction is +0.098 V. The overall cell potential is +1.35 V. Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction?'

Verified step by step guidance
1
Step 1: Identify the half-reactions and their roles in the electrochemical cell. The given half-reactions are: (1) HgO(s) + H2O(l) + 2 e⁻ → Hg(l) + 2 OH⁻(aq) and (2) Zn(s) + 2 OH⁻(aq) → ZnO(s) + H2O(l) + 2 e⁻. The first reaction is the reduction reaction (cathode), and the second is the oxidation reaction (anode).
Step 2: Understand the relationship between the cell potential (E°_cell), the reduction potential of the cathode (E°_cathode), and the reduction potential of the anode (E°_anode). The overall cell potential is given by the equation: E°_cell = E°_cathode - E°_anode.
Step 3: Substitute the known values into the equation. You are given E°_cell = +1.35 V and E°_cathode = +0.098 V. Substitute these values into the equation: +1.35 V = +0.098 V - E°_anode.
Step 4: Rearrange the equation to solve for E°_anode. This involves isolating E°_anode on one side of the equation.
Step 5: Calculate the value of E°_anode using the rearranged equation. This will give you the standard reduction potential for the anode reaction.

Key Concepts

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

Electrochemical Cells

Electrochemical cells consist of two half-cells where oxidation and reduction reactions occur. In a galvanic cell, the anode is where oxidation takes place, releasing electrons, while the cathode is where reduction occurs, accepting electrons. Understanding the flow of electrons and the reactions at each electrode is crucial for analyzing cell behavior and calculating cell potentials.
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Standard Reduction Potentials

Standard reduction potentials (E°) are measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C) and indicate the tendency of a species to gain electrons and be reduced. A higher E° value means a greater likelihood of reduction. These values are essential for determining the overall cell potential and for comparing the reactivity of different half-reactions.
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Nernst Equation

The Nernst equation relates the cell potential to the standard reduction potentials and the concentrations of the reactants and products. It allows for the calculation of the cell potential under non-standard conditions. In this context, it can be used to derive the standard reduction potential of the anode reaction by knowing the overall cell potential and the cathode potential.
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Related Practice
Textbook Question

Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li(s) + Ag2CrO4(s) → Li2CrO4(s) + 2 Ag(s) (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode?

Textbook Question

Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li(s) + Ag2CrO4(s) → Li2CrO4(s) + 2 Ag(s) (b) Choose the two half-reactions from Appendix E that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions?

Textbook Question

In some applications nickel–cadmium batteries have been replaced by nickel–zinc batteries. The overall cell reaction for this relatively new battery is: 2 H2O(l) + 2 NiO(OH)(s) + Zn(s) → 2 Ni(OH)2(s) + Zn(OH)2(s) (b) What is the anode half-reaction?

Textbook Question

In some applications nickel–cadmium batteries have been replaced by nickel–zinc batteries. The overall cell reaction for this relatively new battery is: 2 H2O(l) + 2 NiO(OH)(s) + Zn(s) → 2 Ni(OH)2(s) + Zn(OH)2(s) (c) A single nickel–cadmium cell has a voltage of 1.30 V. Based on the difference in the standard reduction potentials of Cd2+ and Zn2+, what voltage would you estimate a nickel–zinc battery will produce? (d) Would you expect the specific energy density of a nickel–zinc battery to be higher or lower than that of a nickel–cadmium battery?