Problem 52 - Chemistry Matter and Change

Question

Half-reactions for a lead-acid storage battery are below. $$ \begin{array}{c}{\mathrm{PbO}_{2}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+4 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow} \\\ {\mathrm{PbSO}_{4}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad E^{0}=+1.685 \mathrm{V}} \\ {\mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \quad E^{0}=-0.356 \mathrm{V}}\end{array} $$ What is the standard cell potential for one cell in a car battery?

Step-by-Step Solution

Answer

The standard cell potential is 2.041 V.

Step 1: Identify the half-reactions

There are two half-reactions given for the lead-acid storage battery: 1. The oxidation half-reaction: \( \mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \) with a standard potential \( E^{0}=-0.356 \mathrm{V} \). 2. The reduction half-reaction: \( \mathrm{PbO}_{2}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+4\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{PbSO}_{4}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \) with a standard potential \( E^{0}=+1.685 \mathrm{V} \).

Step 2: Determine the standard cell potential

The standard cell potential \( E^0_{\text{cell}} \) is calculated by summing the standard potentials of the two half-reactions: - First, write the cell potential equation: \( E^0_{\text{cell}} = E^0_{\text{reduction}} - E^0_{\text{oxidation}} \). - Substitute the given values: \( E^0_{\text{cell}} = 1.685\,\text{V} - (-0.356\,\text{V}) \).

Step 3: Perform the calculation

Now, compute the standard cell potential: - \( E^0_{\text{cell}} = 1.685 + 0.356 \). - Calculate: \( E^0_{\text{cell}} = 2.041\, \text{V} \).

Step 4: Interpret the result

The standard cell potential \( E^0_{\text{cell}} \) represents the voltage or electric potential difference of one cell under standard conditions. A positive value indicates that the overall reaction is spontaneous. In this case, \( E^0_{\text{cell}} = 2.041 \text{V} \) means the lead-acid battery cell can effectively generate electricity.

Key Concepts

Lead-acid battery

A lead-acid battery is a type of rechargeable battery that is commonly used in vehicles. It functions through electrochemical reactions between lead dioxide (PbO₂) and lead (Pb) plates within a sulfuric acid (H₂SO₄) electrolyte. These components aid in the conversion of chemical energy into electrical energy.
Over 150 years old, the lead-acid battery remains popular due to its low cost and reliable performance for starting engines.

• The battery has two electrodes: the positive electrode made from lead dioxide, and the negative electrode made from lead.
• The electrolyte is usually a diluted sulfuric acid solution.
• During discharge, both the electrodes react with the sulfuric acid to generate lead sulfate (PbSO₄) and water, releasing electrical energy in the process.

When charging, the external electrical supply forces the reactions in the opposite direction, converting lead sulfate back to lead dioxide, lead, and regenerating the sulfuric acid.

Half-reactions

In electrochemistry, half-reactions are expressions that depict either the oxidation or reduction processes in a redox reaction. For a lead-acid battery, there are two primary half-reactions:

• Reduction half-reaction: This involves the gain of electrons. In the lead-acid battery, lead dioxide reacts with sulfuric acid and water to form lead sulfate and water, along with electrons being gained: \[ \mathrm{PbO}_{2} + \mathrm{SO}_{4}^{2-} + 4\mathrm{H}_{3} \mathrm{O}^{+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{PbSO}_{4} + 6 \mathrm{H}_{2} \mathrm{O} \]
• Oxidation half-reaction: This involves the loss of electrons. Here, lead sulfate loses electrons to form lead and sulfate ions: \[ \mathrm{PbSO}_{4} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb} + \mathrm{SO}_{4}^{2-} \]

These half-reactions clearly distinguish the electron movement, helping to balance overall chemical equations and understand electricity production in battery cells.

Standard cell potential

Standard cell potential (\(E^0_{\text{cell}}\)) is a measure of the voltage of a cell under standard conditions (1 M concentration, 1 atm pressure, and 25°C). This potential indicates the cell's ability to produce an electric current. In a lead-acid battery, it is determined by the equation:\[E^0_{\text{cell}} = E^0_{\text{reduction}} - E^0_{\text{oxidation}}\]In our case:\[E^0_{\text{cell}} = 1.685 \, \text{V} - (-0.356 \, \text{V})\]Which leads to a calculation of:\[E^0_{\text{cell}} = 2.041 \, \text{V} \]The positive value of the standard cell potential implies that the reaction is energetically favorable and occurs spontaneously under standard conditions.
This also means a single cell in a lead-acid battery can efficiently produce electrical energy.

Oxidation-reduction reactions

Oxidation-reduction reactions, or redox reactions, are chemical processes where oxidation and reduction occur simultaneously. These reactions are vital in energy production as they involve the transfer of electrons between chemical species.
For the lead-acid battery:

• Oxidation: The lead from lead sulfate undergoes oxidation, losing electrons to form lead ions.
• Reduction: Lead dioxide is reduced by gaining electrons forming lead sulfate and water.

The flow of electrons from anodic to cathodic regions in a circuit forms a current, harnessed to power various devices.
This fundamental principle is what powers cars, ensuring they start efficiently and reliably. Understanding these reactions helps explain the efficiency and mechanics behind battery operation.