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Chapter 18: Problem 163

The nickel-iron battery has an iron anode, an \(\mathrm{NiO}(\mathrm{OH})\) cathode, and a KOH electrolyte. This battery uses the following halfreactions and has an \(E^{\circ}\) value of \(1.37 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}:\) $$ \begin{gathered} \mathrm{Fe}(s)+2 \mathrm{OH}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \\ \mathrm{NiO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}(a q) \end{gathered} $$ (a) Write a balanced equation for the cell reaction. (b) Calculate \(\Delta G^{\circ}\) (in kilojoules) and the equilibrium constant \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). (c) What is the cell voltage at \(25^{\circ} \mathrm{C}\) when the concentration of \(\mathrm{KOH}\) in the electrolyte is \(5.0 \mathrm{M} ?\) (d) How many grams of \(\mathrm{Fe}(\mathrm{OH})_{2}\) are formed at the anode when the battery produces a constant current of \(0.250 \mathrm{~A}\) for \(40.0 \mathrm{~min}\) ? How many water molecules are consumed in the process?

Short Answer

Expert verified
(a) \(\mathrm{Fe}(s) + 2\mathrm{NiO}(s) + 2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{Fe(OH)}_2(s) + 2\mathrm{Ni(OH)}_2(s)\) (b) \(\Delta G^{\circ} \approx -264.13\ \mathrm{kJ/mol}, K \approx 1.16 \times 10^{46}\) (c) Cell voltage \(\approx 1.364\mathrm{V}\) (d) Mass of \(\mathrm{Fe(OH)}_2\approx 0.279\ \mathrm{g},\) \( \approx 1.87\times10^{21}\ \mathrm{water\ molecules} \)

Step by step solution

01

Combine Half-Reactions for Balanced Equation

To find the balanced equation for the cell reaction, combine the given half-reactions. The first half-reaction is the oxidation of iron: \( \mathrm{Fe}(s) + 2 \mathrm{OH}^- (aq) \rightarrow \mathrm{Fe(OH)}_2(s) + 2e^- \). The second is the reduction of Nickel(III) oxide: \( \mathrm{NiO}(s) + \mathrm{H}_2\mathrm{O}(l) + e^- \rightarrow \mathrm{Ni(OH)}_2(s) + \mathrm{OH}^-(aq) \). Multiply the nickel reaction by 2 to balance the electrons and add the two reactions.Final balanced equation:\[ \mathrm{Fe}(s) + 2\ \mathrm{NiO}(s) + 2\ \mathrm{H_2O}(l) \rightarrow \mathrm{Fe(OH)}_2(s) + 2 \ \mathrm{Ni(OH)}_2(s) \]
02

Calculate \(\Delta G^{\circ}\)

Use the relation \(\Delta G^{\circ} = -nFE^{\circ}\), where \(n=2\) (as it involves 2 moles of electrons), \(F = 96485\ \mathrm{C/mol}\) is Faraday's constant, and \(E^{\circ} = 1.37\ \mathrm{V}\). Substitute to find \(\Delta G^{\circ}\).\[ \Delta G^{\circ} = -(2)(96485) (1.37) = -264,129 \ \mathrm{J/mol} \approx -264.13 \ \mathrm{kJ/mol}\]
03

Calculate Equilibrium Constant \(K\)

Use \(\Delta G^{\circ} = -RT\ln K\) to find \(K\), where \(R=8.314\ \mathrm{J/(mol\cdot K)}\) and \(T = 298\ \mathrm{K}\). Rearrange the equation: \(K = e^{-\Delta G^{\circ}/RT}\).\[\ln K = \frac{-(-264129)}{(8.314)(298)} \approx 106.57\]\[K = e^{106.57} \approx 1.16 \times 10^{46}\]
04

Calculate the Cell Voltage with 5.0 M KOH

Use the Nernst equation: \(E = E^{\circ} - \frac{RT}{nF} \ln Q\). Assume \(Q\) reflects only concentrations of aqueous species. Given all solids and liquids have \(a=1\), \(Q=[\mathrm{OH}^-]^2\).Substitute \(R=8.314\ \mathrm{J/(mol\cdot K)}\), \(T=298\ \mathrm{K}\), \(n=2\), \(F=96485\ \mathrm{C/mol}\), \([\mathrm{OH}^-]=5\ \mathrm{M}\). Compute:\[E = 1.37 - \frac{8.314 \times 298}{2 \times 96485} \ln (5^2)\]\[E = 1.37 - \frac{0.0257}{2} \times \ln 25\]\[E \approx 1.37 - 0.006 \approx 1.364 \ \mathrm{V}\]
05

Calculate Mass of \( \mathrm{Fe(OH)}_2 \) Formed

Determine moles of electrons using \(Q=It\), where \(I=0.25\ \mathrm{A}\) and \(t=40\ \times 60\ \mathrm{s}\). \[Q = 0.25 \times 2400 = 600 \ \mathrm{C}\]\[\mathrm{mol\ \text{e}^-} = \frac{600}{96485} \approx 0.00622 \ \mathrm{mol}\]Each mole of electrons corresponds to \(0.5\) moles of \(\mathrm{Fe(OH)}_2\), giving \(0.00311\) moles. Calculate mass: \[\text{mass} = 0.00311 \times [55.8 \ (Fe) + 2 \times (16 + 1 + 16) \ (OH)]\ g/mol\]\[\approx 0.00311 \times 89.7 \approx 0.279 \ \mathrm{g}\]
06

Determine Water Molecules Consumed

From the overall balanced equation, each mole of \(\mathrm{Fe(OH)}_2\) corresponds to one mole of \(\mathrm{H_2O}\). Calculate moles of water:\[0.00311\ \mathrm{mol\ Fe(OH)}_2 \approx 0.00311 \ \mathrm{mol\ H_2O}\]Convert to molecules using Avogadro's number:\[0.00311 \times 6.022 \times 10^{23} \approx 1.87 \times 10^{21} \mathrm{molecules\ of\ water}\]

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Key Concepts

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

Electrochemical Cells
Electrochemical cells, like the nickel-iron battery, convert chemical energy into electrical energy. They consist of two half-cells that have separate reactions. The nickel-iron battery features an iron anode and a nickel oxide cathode. At the anode, iron oxidizes, losing electrons. Meanwhile, at the cathode, nickel oxide is reduced by gaining electrons. These reactions create a flow of electrons through an external circuit, providing electricity.
This flow is driven by the difference in potential energy between the two electrodes, measured as cell voltage.
Gibbs Free Energy
Gibbs Free Energy (\( \Delta G \)) helps us understand if a reaction will occur spontaneously. In electrochemical cells, the change in Gibbs Free Energy is related to the cell potential by the equation \( \Delta G^{\circ} = -nFE^{\circ} \). Here, \( n \) is the number of moles of electrons, \( F \) is Faraday's constant, and \( E^{\circ} \) is the standard cell potential. A negative \( \Delta G \) means that the reaction is spontaneous.
In the nickel-iron battery, the calculated \( \Delta G^{\circ} \approx -264.13 \ \text{kJ/mol} \) signifies a spontaneous reaction that can provide energy.
Equilibrium Constant
The equilibrium constant (\( K \)) describes the ratio of concentrations of products to reactants at equilibrium. For electrochemical processes, \( \Delta G^{\circ} = -RT\ln K \) connects Gibbs Free Energy with the equilibrium constant. R is the gas constant and T is the temperature in Kelvin. A large equilibrium constant (\( K \)) indicates that the products are favored in the reaction at equilibrium.
In the nickel-iron battery, the very high \( K \approx 1.16 \times 10^{46} \) means the reaction strongly favors product formation, aligning with spontaneous reaction behavior.
Nernst Equation
The Nernst Equation adjusts the standard cell potential for non-standard conditions. It takes the form: \( E = E^{\circ} - \frac{RT}{nF} \ln Q \), where \( Q \) is the reaction quotient. This equation shows how cell voltage changes with concentration.
In the nickel-iron battery, when the KOH concentration is 5.0 M, the cell voltage is slightly reduced to 1.364 V. This adjustment illustrates how different concentrations of reactants and products can influence the output voltage of a battery.

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Most popular questions from this chapter

Consider a Daniell cell with \(1.0 \mathrm{M}\) ion concentrations: Does the cell voltage increase, decrease, or remain the same when each of the following changes is made? Explain. (a) \(5.0 \mathrm{M} \mathrm{CuSO}_{4}\) is added to the cathode compartment. (b) \(5.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is added to the cathode compartment. (c) \(5.0 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) is added to the anode compartment. (d) \(1.0 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) is added to the anode compartment.

Gold metal is extracted from its ore by treating the crushed rock with an aerated cyanide solution. The unbalanced equation for the reaction is $$ \mathrm{Au}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q) $$ (a) Balance the equation for this reaction in basic solution. (b) Use any of the following data at \(25^{\circ} \mathrm{C}\) to calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}: K_{\mathrm{f}}\) for \(\mathrm{Au}(\mathrm{CN})_{2}^{-}=6.2 \times 10^{38}, K_{\mathrm{a}}\) for \(\mathrm{HCN}=4.9 \times 10^{-10}\), and standard reduction potentials are $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & E^{\circ}=1.229 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) & E^{\circ}=1.498 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Au}^{+}(a q) & E^{\circ}=1.401 \mathrm{~V} \end{aligned} $$

Consider the following half-reactions and \(E^{\circ}\) values: $$ \begin{array}{ll} \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & E^{\circ}=0.34 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s) & E^{\circ}=-0.13 \mathrm{~V} \end{array} $$ (a) Which of these metals or ions is the strongest oxidizing agent? Which is the strongest reducing agent? (b) The half-reactions can be used to construct three different galvanic cells. Tell which cell delivers the highest voltage, identify the anode and cathode, and tell the direction of electron and ion flow. (c) Write the cell reaction for part (b), and calculate the values of \(E^{\circ}, \Delta G^{\circ}\) (in kilojoules), and \(K\) for this reaction at \(25^{\circ} \mathrm{C}\). (d) Calculate the voltage for the cell in part (b) if both ion concentrations are \(0.010 \mathrm{M}\).

Aluminum, titanium, and several other metals can be colored by an electrochemical process called anodizing. Anodizing oxidizes a metal anode to yield a porous metal oxide coating that can incorporate dye molecules to give brilliant colors. In the oxidation of aluminum, for instance, the electrode reactions areCathode (reduction): \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow 3 \mathrm{H}_{2}(g)\) Anode (oxidation): \(2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\frac{\mathrm{Al}_{2} \mathrm{O}_{3}(s)+6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-}}{\text {Overall reaction: }} 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\) The thickness of the aluminum oxide coating that forms on the anode can be controlled by varying the current flow during the electrolysis. How many minutes are required to produce a \(0.0100 \mathrm{~mm}\) thick coating of \(\mathrm{Al}_{2} \mathrm{O}_{3}\left(\right.\) density \(\left.3.97 \mathrm{~g} / \mathrm{cm}^{3}\right)\) on a square piece of aluminum metal \(10.0 \mathrm{~cm}\) on an edge if the current passed through the piece is \(0.600 \mathrm{~A} ?\)

A mercury battery uses the following electrode half-reactions: \(\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\) \(\mathrm{Hg}(l)+2 \mathrm{OH}(a q) \quad E^{\circ}=0.098 \mathrm{~V}\) \(\mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow\) \(\mathrm{Zn}(s)+2 \mathrm{OH}(a q) \quad E^{\circ}=-1.260 \mathrm{~V}\) (a) Write a balanced equation for the overall cell reaction. (b) Calculate \(\Delta G^{\circ}\) (in kilojoules) and \(K\) at \(25^{\circ} \mathrm{C}\) for the cell reaction. (c) What is the effect on the cell voltage of a tenfold change in the concentration of KOH in the electrolyte? Explain.
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