During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO 4 (s). (a) What mass of PbO 2 (s) is reduced at the cathode during this same period?

Hi everyone here we have a question that tells us an alkaline battery has the following overall reaction to manganese dioxide solid plus sink solid plus two water liquid forms manga to manga night plus zinc hydroxide solid. In a certain period of discharge, 131 g of zinc from the anad was converted into zinc hydroxide solid, calculate the amount in grams of manganese dioxide solid that is reduced at the cathode at the same time. So our more mass of zinc is .39 g per mole. Our more mass of manganese dioxide is 54 . g per mole plus 16 zero g per mole times two Equals 86. g per mole. Our mass of mega knees dioxide equals g of zinc Times one mole of zinc over 65.39 grams of zinc, times two moles of manganese dioxide Over one mole of zinc times 86 . g of manganese dioxide over one mole of manganese dioxide. So our grams of zinc are canceling out our molds of zinc are canceling out our moles of manganese docks that are canceling out, leaving us with 348 g of manganese dioxide. And that is our final answer. Thank you for watching. Bye.

Ch.20 - Electrochemistry

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Brown 14th Edition

Ch.20 - Electrochemistry

Problem 73a

Chapter 20, Problem 73a

During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO₄(s). (a) What mass of PbO₂(s) is reduced at the cathode during this same period?

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Identify the balanced chemical reaction for the discharge of a lead-acid battery: \[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \]

Determine the molar mass of Pb and PbSO_4 using the periodic table: \[ \text{Molar mass of Pb} = 207.2 \, \text{g/mol} \] and \[ \text{Molar mass of PbSO}_4 = 303.3 \, \text{g/mol} \]

Calculate the moles of Pb converted to PbSO_4 using the given mass: \[ \text{Moles of Pb} = \frac{402 \, \text{g}}{207.2 \, \text{g/mol}} \]

Use stoichiometry from the balanced equation to find the moles of PbO_2 reduced. The reaction shows a 1:1 mole ratio between Pb and PbO_2.

Calculate the mass of PbO_2 reduced using its molar mass: \[ \text{Molar mass of PbO}_2 = 239.2 \, \text{g/mol} \] and \[ \text{Mass of PbO}_2 = \text{Moles of PbO}_2 \times 239.2 \, \text{g/mol} \]

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Redox Reactions

Redox reactions involve the transfer of electrons between two species, where one species is oxidized (loses electrons) and the other is reduced (gains electrons). In the context of a lead-acid battery, lead (Pb) at the anode is oxidized to lead sulfate (PbSO4), while lead oxide (PbO2) at the cathode is reduced during discharge. Understanding this electron transfer is crucial for calculating the mass changes in the battery components.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows us to determine the relationships between the amounts of substances involved. In this case, knowing the mass of Pb oxidized helps us use stoichiometric ratios to find the mass of PbO2 reduced at the cathode, as both processes are linked through the overall reaction of the battery.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is essential for converting between mass and moles in chemical calculations. For this problem, knowing the molar masses of lead (Pb) and lead oxide (PbO2) allows us to calculate the mass of PbO2 reduced based on the mass of Pb oxidized, facilitating the stoichiometric calculations needed to answer the question.