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Chapter 18: Problem 124

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is $\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ $\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \qquad \mathscr{E}^{\circ}=1.10 \mathrm{V}$ 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 consumed Cd, the maximum useful work that can be obtained at standard conditions is 212,267 Joules.

Step by step solution

01

Write down the given information

We are given the cell potential, \(\mathscr{E}^{\circ} = 1.10 V\), and the balanced chemical equation: \[\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s)\]
02

Calculate the change in Gibbs free energy (ΔG)

We use the equation relating standard cell potential, \(\mathscr{E}^{\circ}\), to the standard Gibbs free energy change, ΔG: \[\Delta G^{\circ} = -nF\mathscr{E}^{\circ}\] Here, \(n\) represents the number of moles of electrons transferred in the reaction, and \(F\) represents Faraday's constant, which is approximately \(96485\, C/mol\).
03

Determine the number of moles of electrons transferred in the reaction (n)

To find \(n\), look at the balanced chemical equation and note that: \(Cd (s) \rightarrow Cd^{2+} + 2e^-\) Thus, 2 moles of electrons are transferred in the given reaction per mole of consumed Cd. Therefore, \(n = 2\).
04

Calculate the standard Gibbs free energy change (ΔG) for the reaction.

Now we have all the necessary values to calculate ΔG. Plug the values into the equation from Step 2: \[\Delta G^{\circ} = -(2\,mol)(96485\, C/mol)(1.10\,V)\] \[\Delta G^{\circ} = -212267\, J/mol\]
05

Calculate the maximum useful work for one mole of consumed Cd

The maximum useful work that can be obtained from a chemical process is equal to the change in Gibbs free energy: Maximum useful work = \(\Delta G^{\circ} = -212267\, J/mol\) So, for every mole of consumed Cd, the maximum useful work that can be obtained at standard conditions is 212,267 Joules.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}{\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)} & {\mathscr{E}^{\circ}=-0.440 \mathrm{V}} \\ {2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)} & {\mathscr{E}^{\circ}=0.000 \mathrm{V}}\end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, $P_{\mathrm{H}_{2}}=1.00 \mathrm{atm},$ and a weak acid, HA, at an initial concentration of 1.00 \(\mathrm{M}\) . If the observed cell potential is 0.333 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

When aluminum foil is placed in hydrochloric acid, nothing happens for the first 30 seconds or so. This is followed by vigorous bubbling and the eventual disappearance of the foil. Explain these observations.

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and $1.0 \times 10^{-4} M(\text { left side })$ a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}$ $\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}$ b. $\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}$ $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}$

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces 1.00 \(\mathrm{kg}\) water at $25^{\circ} \mathrm{C}$ ? Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?
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