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Chapter 20: Problem 76

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) 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 \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2}\) ?

Short Answer

Expert verified
In conclusion, (a) the mass of MnO2 reduced at the cathode during the discharge is approximately 120 g, and (b) approximately 13,300 coulombs of electrical charge are transferred from Zn to MnO2.

Step by step solution

01

Write the balanced reaction equation

The balanced equation for the given reaction is: \(Zn + 2MnO_{2} + 2H_{2}O \longrightarrow Zn(OH)_{4}^{2-} + 2MnOOH \) Note that 1 mole of Zn reacts with 2 moles of MnO2.
02

Calculate the number of moles of Zn consumed

To find the number of moles of Zn consumed, we can use the provided mass and the molar mass of Zn. \(moles~of~Zn = \frac{mass~of~Zn}{molar~mass~of~Zn} \) The molar mass of Zn is 65.38 g/mol. Given that 4.50 g of Zn is consumed: \(moles~of~Zn = \frac{4.50~g}{65.38~g/mol} = 0.06884~mol \)
03

Calculate the number of moles of MnO2 reduced

Using the stoichiometry of the balanced equation (1 mole of Zn reacts with 2 moles of MnO2): \(moles~of~MnO_{2} = 2 \times moles~of~Zn \) \(moles~of~MnO_{2} = 2 \times 0.06884~mol = 0.1377~mol \)
04

Calculate the mass of MnO2 reduced

Using the number of moles and the molar mass of MnO2 (Mn: 54.94 g/mol, O: 16.00 g/mol): \(mass~of~MnO_{2} = moles~of~MnO_{2} \times molar~mass~of~MnO_{2} \) \(mass~of~MnO_{2} = 0.1377~mol \times (54.94~g/mol + 2 \times 16.00~g/mol)\) \(mass~of~MnO_{2} = 0.1377~mol \times 86.94~g/mol = 120~g \) So, the mass of MnO2 reduced is approximately 120 g.
05

Calculate the number of electrons transferred per mole of Zn

From the balanced equation: 1 mole of Zn loses 2 electrons (as it forms Zn(OH)4 with a charge of 2-), and 2 moles of MnO2 gain 2 electrons to form 2 moles of MnOOH. So, the number of electrons transferred per mole of Zn is 2.
06

Calculate the total charge transferred

Using Faraday's law: Charge (Coulombs) = moles of Zn × number of electrons transferred per mole of Zn × Faraday constant (96,485 C/mol) Charge (C) = 0.06884 mol × 2 × 96,485 C/mol Charge (C) = 13,300 C Around 13,300 coulombs of electrical charge are transferred from Zn to MnO2. In conclusion, (a) The mass of MnO2 reduced at the cathode during the discharge is approximately 120 g. (b) Approximately 13,300 coulombs of electrical charge are transferred from Zn to MnO2.

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Most popular questions from this chapter

A cell has a standard cell potential of \(+0.177 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction (a) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

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