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Chapter 20: Problem 74

During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from Zn to \(\mathrm{MnO}_{2} ?\)

Short Answer

Expert verified
(a) Mass of \(\mathrm{MnO}_{2}\) reduced = Moles of Zn consumed × 86.94 g/mol = \( \frac{4.50}{65.38} \) moles × 86.94 g/mol ≈ 5.92 g (b) Coulombs transferred = moles of electrons transferred × 96,485 C/mol = \( ( \frac{4.50}{65.38} \) moles × 2) × 96,485 C/mol ≈ 13294 C

Step by step solution

01

Convert the mass of Zn consumed to moles

To convert the mass of Zn consumed (4.50 g) to moles, we'll use the formula: Moles = mass / molar mass where the molar mass of Zn is 65.38 g/mol. Moles of Zn consumed = \( \frac{4.50}{65.38} \) moles
02

Find the number of moles of \(\mathrm{MnO}_{2}\) reduced

In the alkaline battery, the balanced half-reactions are as follows: Anode: \(Zn(s) + 2OH^-(aq) \rightarrow Zn(OH)_2(s) + 2e^-\) Cathode: \(MnO_{2}(s) + H_{2}O(l) + 2e^- \rightarrow MnOOH(s) + 2OH^-(aq)\) By comparing the anode and cathode half-reactions, we can see that the ratio of Zn to \(\mathrm{MnO}_{2}\) is 1:1. This means that the moles of \(\mathrm{MnO}_{2}\) reduced will be equal to the moles of Zn consumed. Moles of \(\mathrm{MnO}_{2}\) reduced = moles of Zn consumed
03

Convert moles of \(\mathrm{MnO}_{2}\) reduced to mass

To convert the moles of \(\mathrm{MnO}_{2}\) reduced to mass, we'll use the formula: Mass = moles × molar mass where the molar mass of \(\mathrm{MnO}_{2}\) is 86.94 g/mol. Mass of \(\mathrm{MnO}_{2}\) reduced = Moles of \(\mathrm{MnO}_{2}\) reduced × 86.94 g/mol
04

Calculate the number of coulombs transferred

Now that we know the number of moles of Zn consumed, we can calculate the number of electrons transferred during the discharge. From the anode half-reaction, we know that for each mole of Zn consumed, 2 moles of electrons are transferred. Therefore: Moles of electrons transferred = moles of Zn consumed × 2 To find the number of coulombs transferred, we need to convert the moles of electrons to coulombs using Faraday's constant: Coulombs = moles of electrons transferred × Faraday's constant where Faraday's constant is 96,485 C/mol. Coulombs transferred = moles of electrons transferred × 96,485 C/mol

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Most popular questions from this chapter

In some applications nickel-cadmium batteries have been replaced by nickel- zinc batteries. The overall cell reaction for this relatively new battery is: $$ \begin{aligned} 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{Zn}(s) & \\ & \longrightarrow 2 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Zn}(\mathrm{OH})_{2}(s) \end{aligned} $$ (a)What is the cathode half-reaction? (b)What is the anode half-reaction? (c) A single nickel-cadmium cell has a voltage of 1.30 \(\mathrm{V}\) . Based on the difference in the standard reduction potentials of \(\mathrm{Cd}^{2+}\) and \(\mathrm{Zn}^{2+},\) what voltage would you estimate a nickel-zinc battery will produce? (d) Would you expect the specific energy density of a nickel-zinc battery to be higher or lower than that of a nickel-cadmium battery?

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(S n\) to \(S n^{2+}\) by \(I_{2}(\) to form I \(),\) (b) reduction (a) oxidation of \(\mathrm{Sn}\) to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) \(( \text { to form } \mathrm{I})\); (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}(\) to form \(\mathrm{I}_{2}),(\mathbf{c})\) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\) (d) reduction of \(\mathrm{Cu}^{2+}\) to Cu by \(\operatorname{Sn}^{2+}(\) to form \( \mathrm{Sn}^{4+} )\).

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C}\) . (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent?

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l}{\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)} \\ {\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{znO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}}\end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text { red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{V}\) . The overall cell potential is \(+1.35 \mathrm{V}\) . Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?
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