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Chapter 17: Problem 7

The anodic half-cell of lead-acid battery is recharged using electricity of \(0.05\) Faraday. The amount of \(\mathrm{PbSO}_{4}\) electrolyzed in g during the process is: (Molar mass of \(\mathrm{PbSO}_{4}=303 \mathrm{~g} \mathrm{~mol}^{-1}\) ) [Main Jan. 9, 2019 (I)] (a) \(22.8\) (b) \(15.2\) (c) \(7.6\) (d) \(11.4\)

Short Answer

Expert verified
7.6 g, option (c) is correct.

Step by step solution

01

Understanding the Problem

The problem is asking for the mass of lead sulfate (\(\mathrm{PbSO}_4\)) that is electrolyzed when 0.05 Faraday of electricity is used. The equation to consider in this process is:\[\mathrm{PbSO}_4 + 2e^- \rightarrow \mathrm{Pb} + \mathrm{SO}_4^{2-}\]which shows that one mole of \(\mathrm{PbSO}_4\) is involved in the reaction for every two moles of electrons (faradays) used.
02

Identify the Faraday to Mole Relationship

From the given information, we know that 1 mole of \(\mathrm{PbSO}_4\) corresponds to 2 Faradays because it accepts 2 electrons in the reaction. Given 0.05 Faraday, determine how many moles of \(\mathrm{PbSO}_4\) this corresponds to.
03

Calculate Moles of PbSO4 Electrolyzed

Since 2 Faradays are required to electrolyze 1 mole of \(\mathrm{PbSO}_4\), 0.05 Faraday is equivalent to:\[\text{moles of } \mathrm{PbSO}_4 = \frac{0.05}{2} = 0.025\text{ moles}\]
04

Calculate Mass of PbSO4 Electrolyzed

Multiply the moles of \(\mathrm{PbSO}_4\) by its molar mass to find the mass in grams:\[\text{Mass} = 0.025 \text{ moles} \times 303 \text{ g/mol} = 7.575 \text{ g}\]Rounded to one decimal place, this is 7.6 g.
05

Final Selection

Compare this calculated value with the given options. The mass of \(\mathrm{PbSO}_4\) electrolyzed is 7.6 g, which matches option (c).

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Key Concepts

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

Lead-acid battery
A lead-acid battery is a type of rechargeable battery that has been around for over 150 years. It is widely used in vehicles for starting, lighting, and ignitions because of its capability to deliver high surge currents. The lead-acid battery consists of two electrodes, a lead dioxide (PbO鈧�) cathode and a sponge lead (Pb) anode, submerged in a sulfuric acid (H鈧係O鈧�) electrolyte.
When the battery discharges, the chemical reaction generates electrical energy by converting lead dioxide and sponge lead into lead sulfate (PbSO鈧�), releasing energy. During recharging, this process is reversed by supplying external electricity, which converts lead sulfate back to lead dioxide and sponge lead.
Key components and benefits include:
  • Long History: With over a century of use, it's a proven technology.
  • Recyclable: Most lead-acid batteries are recyclable, making them eco-friendly.
  • Cost-Effective: While not the most energy-dense, they are relatively affordable.
  • High Power Output: Ideal for applications needing strong bursts of energy.
Understanding this cycle is crucial for applications that depend on stored power, like cars and backup energy systems.
Faraday's laws of electrolysis
Faraday's laws of electrolysis are fundamental in understanding how electrochemical reactions occur when an electric current passes through a substance. The first law states that the amount of chemical change (e.g., substance deposited or dissolved) is directly proportional to the amount of electricity used, measured in Faradays.
In simple terms, one Faraday (unit of electric charge) deposits the gram equivalent weight of a substance. The second law states that the quantities of different substances liberated by the same quantity of electricity are proportional to their equivalent weights.
These laws are applicable to:
  • Quantifying Reactions: They help to calculate the mass of the products and reactants in an electrochemical cell.
  • Electroplating: Used to determine how much metal will deposit on a surface.
  • Battery Chemistry: Helps understand and calculate reactions within batteries, like the lead-acid battery.
To apply these laws, one must understand the Faraday constant, approximately 96,485 coulombs per mole, which is the charge of one mole of electrons. In the example, only 0.05 Faraday is used, indicating the partial reaction and subsequent computation of material change.
Molar mass calculation
Molar mass calculation is a crucial step in chemistry that involves determining the mass of one mole of a substance. It is the sum of the atomic masses of all the atoms in a given formula and is expressed in grams per mole (g/mol). To calculate the molar mass of a compound like lead sulfate (PbSO鈧�), you add up the average atomic weights of its constituent elements:
For PbSO鈧�:
  • Lead (Pb): 207 g/mol
  • Sulfur (S): 32 g/mol
  • Oxygen (O), with four atoms in the sulfate group: 16 g/mol each, for a total of 64 g/mol
Adding these together, the molar mass of PbSO鈧� comes out to approximately 303 g/mol.
Accurate molar mass calculation is essential for stoichiometry, converting moles to grams, determining yield, and predicting the quantities of reactants and products involved in chemical reactions. In the context of electrochemistry, it enables precise determination of the material electrolyzed or deposited during reactions, as seen in calculating the electrolyzed mass of PbSO鈧� using given Faraday values.

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Most popular questions from this chapter

Identify the correct statement : [Main Online April 10, 2016] (a) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential (b) Iron corrodes in oxygen free water (c) Corrosion of iron can be minimized by forming an impermeable barrier at its surface (d) Iron corrodes more rapidly in salt water because its electrochemical potential is higher

How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing silver in \(+1\) oxidation state for a period of \(8.0\) hours at a current of \(8.46\) amperes? What is the area of the tray if the thickness of the silver plating is \(0.00254 \mathrm{~cm}\) ? Density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\).

A variable, opposite external potential \(\left(\mathrm{E}_{\text {ex }}\right.\) ) is applied to the cell \(\mathrm{Zn} \mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M}) \mid \mathrm{Cu}\), of potential \(1.1 \mathrm{~V}\). When \(\mathrm{E}_{\mathrm{ext}}<1.1 \mathrm{~V}\) and \(\mathrm{E}_{\text {ext }}>1.1 \mathrm{~V}\), respectively electrons flow from : [Main Online April 10, 2015] (a) anode to cathode in both cases (b) cathode to anode and anode to cathode (c) anode to cathode and cathode to anode (d) cathode to anode in both cases

During the discharge of a lead storage battery, the density of sulphuric acid fell from \(1.294\) to \(1.139 \mathrm{~g} / \mathrm{mL}\). Sulphuric acid of density \(1.294\) \(\mathrm{g} / \mathrm{mL}\) is \(39 \%\) by weight and that of \(1.139 \mathrm{~g} / \mathrm{mL}\) is \(20 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) by weight. The battery holds \(3.5\) litres of the acid and the volume remained practically constant during the discharge. Calculate the number of ampere- hours for which the battery must have been used. The charging and discharging reactions are : [1986-5 Marks] Anode : \(\mathrm{Pb}+\mathrm{SO}_{4}^{2-}=\mathrm{PbSO}_{4}+2 \mathrm{e}^{-}\)(discharging) Cathode : \(\mathrm{PbO}_{2}+4 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}+2 \mathrm{e}^{-}=\mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) (discharging) Note : Both the reactions take place at the anode and cathode respectively during discharge. Both reaction get reverse during charging.

How many electrons would be required to deposit \(6.35 \mathrm{~g}\) of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper \(=63.5 \mathrm{u}, \mathrm{N}_{\mathrm{A}}=\) Avogadro's constant): (a) \(\frac{\mathrm{N}_{\mathrm{A}}}{20}\) (b) \(\frac{\mathrm{N}_{\mathrm{A}}}{10}\) (c) \(\frac{\mathrm{N}_{\mathrm{A}}}{5}\) (d) \(\frac{\mathrm{N}_{\mathrm{A}}}{2}\)
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