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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) + NiO鈧�(s) + 2 H鈧侽(l) 鈫� Ni(OH)鈧�(s) + Cd(OH)鈧�(s) The corresponding half-reactions are: Oxidation half-reaction (Cd is oxidized): Cd(s) 鈫� Cd(OH)鈧�(s) + 2 e鈦� Reduction half-reaction (NiO鈧� is reduced): NiO鈧�(s) + 2 H鈧侽(l) + 2 e鈦� 鈫� Ni(OH)鈧�(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 NiO鈧�(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掳), 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) 脳(96,485 C/mol) 脳 (1.10 V)
04

Calculate the maximum useful work for every mole of Cd

Now, let's calculate the maximum useful work: W = -(2) 脳 (96,485) 脳 (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

Sketch a cell that forms iron metal from iron(II) while changing chromium metal to chromium(III). Calculate the voltage, show the electron flow, label the anode and cathode, and balance the overall cell equation.

An unknown metal \(\mathrm{M}\) is electrolyzed. It took \(74.1 \mathrm{~s}\) for a current of \(2.00 \mathrm{~A}\) to plate out \(0.107 \mathrm{~g}\) of the metal from a solution containing \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{3}\). Identify the metal.

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{~mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{~g}\) blood plasma, determine the mass percent of alcohol in the blood.
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