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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
The maximum useful work that can be obtained at standard conditions when one mole of Cd is consumed in the nickel-cadmium alkaline battery reaction is approximately -212,270 J/mol.

Step by step solution

01

Find the balanced equation

To find the maximum useful work, we first need to find the balanced equation for the given battery reaction: $$ \mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) $$
02

Find the number of electrons involved in the reaction

Next, we need to find the number of moles of electrons (n) involved in the reaction. We can do this by looking at the oxidation states of the elements involved. Cd changes from 0 (in Cd metal) to +2 (in Cd(OH)\(_2\)). This means that 2 moles of electrons are exchanged for every mole of Cd consumed in the reaction.
03

Find the maximum useful work

Now we can use the formula to calculate the maximum useful work: $$ W = -nFE $$ where n = 2 moles of electrons, F = Faraday's constant = 96485 C/mol, and E = standard cell potential = 1.10 V. By substituting the values, we get: $$ W = -(2)(96485 \,\text{C/mol})(1.10 \,\text{V}) $$
04

Calculate the maximum useful work

Finally, multiply the given values to obtain the maximum useful work in Joules: $$ W = -(2)(96485 \,\text{C/mol})(1.10 \,\text{V}) = -212,270\,\text{rounded} \, \text{J/mol} $$ The maximum useful work that can be obtained at standard conditions when one mole of Cd is consumed in the cell is approximately -212,270 J/mol.

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Key Concepts

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

Standard Cell Potential
The standard cell potential, represented by the symbol \( \mathscr{E}^{\text{o}} \), is a critical quantity in electrochemistry that measures the electromotive force of a cell under standard conditions. These conditions typically mean a temperature of 25°C (298 K), a 1 M concentration for all soluble species, and a 1 atm pressure for any gases involved.

It compares the energy difference between the electrons in the anode and cathode materials. A higher standard cell potential indicates a greater ability of the cell to do work when the electrons flow from the anode to the cathode through an external circuit. For the nickel-cadmium battery in the exercise, the \( \mathscr{E}^{\text{o}} \) value of 1.10 V signifies that electrons have a relatively high potential energy difference to perform work.
Faraday's Constant
Faraday's constant is a fundamental value in electrochemistry and physics, denoted as \( F \). It represents the total charge of one mole of electrons and is approximately equal to 96485 Coulombs per mole (C/mol).

This constant is derived from the elementary charge of an electron and Avogadro's number, providing a link between macroscopic measurements of electric charge in moles to the microscopic characteristics of individual electrons. When calculating the maximum useful work from an electrochemical cell, like in our problem, Faraday's constant helps us convert electrical energy measured in volts to physical work measured in joules.
Electrochemical Cells
Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions, involving the transfer of electrons from one substance to another. These cells consist of two electrodes: an anode (where oxidation occurs) and a cathode (where reduction takes place).

The flow of electrons from the anode to the cathode through an external circuit is what we harness as electricity. In rechargeable batteries, like the nickel-cadmium battery mentioned, these cells can operate in reverse during charging, storing energy as chemical potential to be released later as electrical energy.
Oxidation States
In chemistry, an oxidation state is the degree of oxidation of an atom within a compound. It represents the hypothetical charge that an atom would have if all bonds to atoms of different elements were purely ionic. By comparing the oxidation states of the elements before and after the reaction, scientists can determine how many electrons are transferred in a redox process.

For example, in our exercise, cadmium (Cd) moves from an oxidation state of 0 in metallic form to +2 when it's in the form of Cd(OH)\(_2\). This change indicates that two electrons per Cd atom are being transferred during the reaction. Understanding oxidation states is vital for balancing chemical equations and for calculating the theoretical charge transfer in electrochemical cells.

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Most popular questions from this chapter

Specify which of the following equations represent oxidationreduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\) c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

The compound with the formula TII \(_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{T}^{3+}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Tl}^{+} & & \mathscr{E}^{\circ}=1.25 \mathrm{V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} & \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{E}^{\circ}=0.55 \mathrm{V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

Consider the following galvanic cell: A 15.0 -mole sample of \(\mathrm{NH}_{3}\) is added to the Ag compartment (assume \(1.00 \mathrm{L}\) of total solution after the addition). The silver ion reacts with ammonia to form complex ions as shown: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \\\& K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \\\\\ & K_{2}=8.2 \times 10^{3} \end{aligned}$$ Calculate the cell potential after the addition of 15.0 moles of \(\mathrm{NH}_{3}\)

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

It took \(150 .\) s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+.\)

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