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Chapter 18: Problem 126

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is \(\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \quad \mathscr{E}^{\circ}=1.10 \mathrm{~V}\) For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

Short Answer

Expert verified
For every mole of Cd consumed in the cell under standard conditions, the maximum useful work that can be obtained is 212,666 J (Joules).

Step by step solution

01

Write the half-reactions for the overall cell reaction

Before finding the number of moles of electrons transferred, let's write the half-reactions for the given overall cell reaction. This will help us to find the number of electrons exchanged in the process. The given reaction is: Cd(s) + NiO₂(s) + 2 H₂O(l) → Ni(OH)₂(s) + Cd(OH)₂(s) The corresponding half-reactions are: Oxidation half-reaction (Cd is oxidized): Cd(s) → Cd(OH)₂(s) + 2 e⁻ Reduction half-reaction (NiO₂ is reduced): NiO₂(s) + 2 H₂O(l) + 2 e⁻ → Ni(OH)₂(s)
02

Calculate the number of moles of electrons transferred (n)

Observe that in the reduction half-reaction, 2 moles of electrons are gained by the NiO₂(s). In the oxidation half-reaction, 2 moles of electrons are released by the Cd(s). So in the overall reaction, for every mole of Cd consumed, it involves the transfer of 2 moles of electrons. Thus, n = 2 moles of electrons.
03

Use the work formula to compute the maximum useful work

Now that we have the number of moles of electrons transferred (n) and the standard cell potential (E°), we can use the work formula to find the maximum useful work obtainable for every mole of Cd consumed: W = -nFE Here, F is Faraday's constant, which is 96,485 C/mol. Plug the values into the formula: W = -(2 moles) ×(96,485 C/mol) × (1.10 V)
04

Calculate the maximum useful work for every mole of Cd

Now, let's calculate the maximum useful work: W = -(2) × (96,485) × (1.10) W = -212,666 J/mol Since the maximum useful work obtained is negative, it means that when every mole of Cd is consumed in the cell reaction under standard conditions, the cell can perform 212,666 J (Joules) of work on the surroundings.

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Most popular questions from this chapter

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$\mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q)$$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

An electrochemical cell consists of a silver metal electrode im. mersed in a solution with \(\left[\mathrm{Ag}^{+}\right]=1.0 M\) separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of \(5.0 \mathrm{M} \mathrm{NH}_{3}\) that is also \(0.010 \mathrm{M}\) in \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), what is the cell potential at \(25^{\circ} \mathrm{C} ?\) $$\begin{aligned}\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q) & K=1.0 \times 10^{13}\end{aligned}$$

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) \quad K=1.28 \times 10^{83}$$ a. Calculate \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.
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