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Chapter 20: Problem 104

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{aligned} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ} &=+0.49 \mathrm{~V} \end{aligned} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Short Answer

Expert verified
(a) The discharge and charge reactions for a nickel-cadmium (nicad) battery are: Discharge: \(\mathrm{Cd}(s)+2\mathrm{NiO}(\mathrm{OH})(s)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}(s)+2\mathrm{Ni}(\mathrm{OH})_{2}(s)\) Charge: \(\mathrm{Cd}(\mathrm{OH})_{2}(s)+2\mathrm{Ni}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Cd}(s)+2\mathrm{NiO}(\mathrm{OH})(s)+2\mathrm{H}_{2}\mathrm{O}(l)\) (b) The standard emf of the cell can be calculated as \(E_\text{cell}^\circ = (+0.49 \,\text{V}) - (-0.76 \,\text{V}) = +1.25 \,\text{V}\). (c) The difference between the experimental emf value (+1.30 V) and the calculated emf value (+1.25 V) can be attributed to real-world factors such as specific compositions of the battery components, temperature, and impurities. (d) The equilibrium constant for the overall nicad reaction, based on the given typical emf value, is approximately \(K_\text{eq} \approx 1.5 \times 10^{18}\).

Step by step solution

01

Discharge Reaction

The discharge reaction involves the conversion of chemical energy stored in the battery to electrical energy. For a nickel-cadmium battery, the discharge reaction can be written as: $$ \mathrm{Cd}(s)+2\mathrm{NiO}(\mathrm{OH})(s)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}(s)+2\mathrm{Ni}(\mathrm{OH})_{2}(s) $$
02

Charge Reaction

The charge reaction involves recharging the battery and converting electrical energy back into chemical energy. For a nickel-cadmium battery, the charge reaction is the reverse of the discharge reaction: $$ \mathrm{Cd}(\mathrm{OH})_{2}(s)+2\mathrm{Ni}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Cd}(s)+2\mathrm{NiO}(\mathrm{OH})(s)+2\mathrm{H}_{2}\mathrm{O}(l) $$ #b) Standard Emf of the Cell#
03

Calculating Standard Emf

We can calculate the standard emf of the cell using the given reduction potentials. Remember that the overall cell potential (\(E_\text{cell}^\circ\)) is the difference between the reduction potential of the cathode (\(E_\text{red,cathode}^\circ\)) and the reduction potential of the anode (\(E_\text{red,anode}^\circ\)): $$ E_\text{cell}^\circ = E_\text{red,cathode}^\circ - E_\text{red,anode}^\circ $$ Using the given reduction potentials: $$ E_\text{cell}^\circ = (+0.49 \,\text{V}) - (-0.76 \,\text{V}) = +1.25 \,\text{V} $$ #c) Difference in Experimental and Calculated Emf#
04

Explaining the Difference

The experimental emf value given for a typical nicad battery is +1.30 V, which is different from the calculated emf value of +1.25 V. This difference can be attributed to real-world factors such as the specific compositions of the battery components, temperature, and impurities that can affect the reduction potentials and overall cell potential. These factors are usually not accounted for in the theoretical calculation of standard emf. #d) Equilibrium Constant#
05

Calculating the Equilibrium Constant

We can calculate the equilibrium constant (\(K_\text{eq}\)) for the overall nicad reaction based on the given typical emf value, using the Nernst equation: $$ E_\text{cell} = E_\text{cell}^\circ - \frac{RT}{nF} \ln K_\text{eq} $$ In this exercise, we are given the experimental value of the cell potential (\(E_\text{cell} = 1.30 \,\text{V}\)) and we know the number of moles of electrons transferred (\(n=2\)) in the reaction. We can rearrange the Nernst equation to solve for \(K_\text{eq}\): $$ K_\text{eq} = e^{\frac{nFE_\text{cell}}{RT}} $$ Using \(E_\text{cell} = 1.30 \,\text{V}\), \(n=2\), and standard values for the gas constant (\(R = 8.314 \,\text{J/mol}\,\text{K}\)) and Faraday's constant (\(F =96485 \,\text{C/mol}\)) at room temperature (298 K): $$ K_\text{eq} = e^{\frac{(2)(96485\,\text{C/mol})(1.30\,\text{V})}{(8.314\,\text{J/mol}\,\text{K})(298\,\text{K})}} $$ $$ K_\text{eq} \approx 1.5 \times 10^{18} $$ So the equilibrium constant for the overall nicad reaction is approximately \(1.5 \times 10^{18}\).

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Key Concepts

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

Reduction Potentials
In electrochemistry, reduction potentials are fundamental in determining how likely a substance is to gain electrons, which is referred to as reduction. Each electrode in a battery has a specific reduction potential, reflecting the efficiency with which electrons are transferred.
For the nickel-cadmium battery, the reduction potentials are:
  • Nickel: \( ext{NiO(OH)} + ext{H}_2 ext{O} + ext{e}^- \rightarrow ext{Ni(OH)}_2 + ext{OH}^- \), with \( E_{red}^\circ = +0.49 ext{ V} \)
  • Cadmium: \( ext{Cd(OH)}_2 + 2 ext{e}^- \rightarrow ext{Cd} + 2 ext{OH}^- \), with \( E_{red}^\circ = -0.76 ext{ V} \)
The substance with the higher reduction potential acts as the cathode, attracting electrons, while the other functions as the anode, relinquishing electrons. In a nickel-cadmium cell, nickel serves as the cathode and cadmium as the anode.
Reduction potentials are measured under standard conditions (1 M concentration, 1 atm pressure, and 25掳C), providing a reference for potential calculations.
Standard Emf
The standard electromotive force (emf) of a cell is a measure of the voltage between the two electrodes when no current is flowing. It is calculated using the reduction potentials of the cathode and anode. The formula to calculate the standard emf ( \( E_{cell}^\circ \) ) is:
\[ E_{cell}^\circ = E_{red,cathode}^\circ - E_{red,anode}^\circ \]In the nickel-cadmium battery, substituting the given values:
  • Cathode (nickel): \( +0.49 ext{ V} \)
  • Anode (cadmium): \( -0.76 ext{ V} \)
The calculation gives:\[ E_{cell}^\circ = +0.49 ext{ V} - (-0.76 ext{ V}) = +1.25 ext{ V} \]This theoretical value assumes ideal conditions, accounting solely for the inherent ability of materials to gain or lose electrons. It excludes effects like temperature and impurities, thus sometimes differing from real-life measurements.
Equilibrium Constant
The equilibrium constant, denoted as \(K_{eq}\), relates to the position of equilibrium in reactions and is tightly linked with the emf of a cell through the Nernst equation. The Nernst equation helps calculate \(K_{eq}\) using:
\[ E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln K_{eq} \]To solve for \( K_{eq} \), rearrange the equation:\[ K_{eq} = e^{\frac{nFE_{cell}}{RT}} \]In a nickel-cadmium battery, using \(E_{cell} = 1.30 ext{ V}\), \( n = 2\) (moles of electrons), and standard constants like the gas constant \(R = 8.314 ext{ J/mol} ext{ K}\) and Faraday's constant \(F = 96485 ext{ C/mol}\) at 298 K:\[ K_{eq} = e^{\frac{(2)(96485)(1.30)}{(8.314)(298)}} \]Which results in approximately \(1.5 \times 10^{18}\). This high \(K_{eq}\) indicates a strong tendency for the forward reaction that generates electric energy, demonstrating the efficiency of the nickel-cadmium battery in producing a stable voltage.
Discharge and Charge Reactions
Nickel-cadmium batteries operate through a cycle of discharge and charge reactions. During the discharge process, chemical energy is transformed into electrical energy. For this battery, the discharge reaction is:
  • \( ext{Cd} + 2 ext{NiO(OH)} + 2 ext{H}_2 ext{O} \rightarrow ext{Cd(OH)}_2 + 2 ext{Ni(OH)}_2 \)
The charge reaction is simply the reverse process, where electrical energy is used to restore the original chemical states:
  • \( ext{Cd(OH)}_2 + 2 ext{Ni(OH)}_2 \rightarrow ext{Cd} + 2 ext{NiO(OH)} + 2 ext{H}_2 ext{O} \)
In essence, the discharge and charge reactions exhibit the reversible nature of nickel-cadmium batteries, symbolizing their reusability. By alternating between storing and releasing electrons, these batteries efficiently manage energy conversion, highlighting their utility in both portable electronics and industrial applications. Understanding these reactions, therefore, is crucial to optimizing the lifespan and performance of the batteries.

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Most popular questions from this chapter

Li-ion batteries used in automobiles typically use a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode in place of the \(\mathrm{LiCoO}_{2}\) cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each electrode material. (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathodes deliver less power on discharging? (c) In a battery that uses a \(\mathrm{LiCoO}_{2}\) cathode, approximately \(50 \%\) of the lithium migrates from the cathode to the anode on charging. In a battery that uses a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

In a Li-ion battery the composition of the cathode is \(\mathrm{LiCoO}_{2}\) when completely discharged. On charging, approximately \(50 \%\) of the \(\mathrm{Li}^{+}\) ions can be extracted from the cathode and transported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the \(\mathrm{LiCoO}_{2}\) cathode has a mass of \(10 \mathrm{~g}\) (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Cu}\) to \(\mathrm{Cu}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\) ), \((\mathbf{b})\) reduction of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}\) by \(\mathrm{H}_{2}\) (to form \(\mathrm{H}^{+}\) ), \(\left(\mathbf{c}\right.\) ) reduction of \(\mathrm{I}_{2}\) to \(\mathrm{I}^{-}\) by \(\mathrm{H}_{2} \mathrm{O}_{2},(\mathbf{d})\) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{Sn}^{2+}\left(\right.\) to form \(\left.\mathrm{Sn}^{4+}\right)\).

In the Br酶nsted-Lowry concept of acids and bases, acidbase reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases?

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{aligned} $$ (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q),\) by \(\mathrm{N}_{2} \mathrm{O}(a q),\) and by \(\mathrm{VO}_{2}^{+}(a q)\) (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).
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