krit.club logo

Electrochemistry - Dry cell and lead accumulator

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The Dry Cell (Leclanché Cell) is a primary cell where the chemical energy is converted into electrical energy via irreversible reactions. The container itself acts as the anode (ZnZn container), while a carbon rod (graphite) surrounded by powdered MnO2MnO_2 and carbon serves as the cathode.

Schematic of a Dry Cell showing zinc anode and carbon cathode with electrolytes.
•

In a dry cell, the electrolyte is not a liquid but a moist paste of NH4ClNH_4Cl and ZnCl2ZnCl_2. The zinc anode undergoes oxidation: Zn(s)→Zn2+(aq)+2e−Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-. The ammonia gas produced at the cathode reacts with Zn2+Zn^{2+} to form the complex ion [Zn(NH3)4]2+[Zn(NH_3)_4]^{2+}, preventing pressure buildup.

•

The Lead Accumulator is a secondary cell (rechargeable). During discharge, it acts as a galvanic cell where lead (PbPb) is the anode and lead dioxide (PbO2PbO_2) is the cathode, both immersed in approximately 38%38\% H2SO4H_2SO_4.

Lead storage battery setup showing lead and lead dioxide electrodes in sulfuric acid.
•

Recharging the lead accumulator involves reversing the electrochemical process by applying an external voltage. During charging, PbSO4PbSO_4 deposited on the electrodes is converted back to PbPb and PbO2PbO_2, and the density of H2SO4H_2SO_4 increases as water is consumed and acid is regenerated.

📐Formulae

Zn(s)→Zn2+(aq)+2e− (Anode reaction in Dry Cell)Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \text{ (Anode reaction in Dry Cell)}

2MnO2+2NH4++2e−→Mn2O3+2NH3+H2O (Cathode reaction in Dry Cell)2MnO_2 + 2NH_4^+ + 2e^- \rightarrow Mn_2O_3 + 2NH_3 + H_2O \text{ (Cathode reaction in Dry Cell)}

Pb(s)+SO42−(aq)→PbSO4(s)+2e− (Anode reaction during discharge)Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^- \text{ (Anode reaction during discharge)}

PbO2(s)+SO42−(aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l) (Cathode reaction during discharge)PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l) \text{ (Cathode reaction during discharge)}

Pb(s)+PbO2(s)+2H2SO4(aq)⇌2PbSO4(s)+2H2O(l) (Overall cell reaction)Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightleftharpoons 2PbSO_4(s) + 2H_2O(l) \text{ (Overall cell reaction)}

💡Examples

Problem 1:

Write the overall cell reaction for a lead storage battery during the discharging process and state what happens to the density of the electrolyte.

Solution:

The overall reaction is: Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l). The density of the electrolyte decreases.

Explanation:

During discharging, H2SO4H_2SO_4 is consumed to form water and solid PbSO4PbSO_4. Since H2SO4H_2SO_4 is denser than water, its consumption and the production of H2OH_2O lead to a decrease in the overall density/specific gravity of the electrolyte.

Problem 2:

Why is the dry cell not truly 'dry'?

Solution:

It contains a moist paste of NH4ClNH_4Cl and ZnCl2ZnCl_2.

Explanation:

For ions to move and conduct electricity between the electrodes, a liquid or moisture-rich medium is necessary. If the cell were completely dry, the internal resistance would be infinite, and no current would flow.

Problem 3:

Calculate the amount of PbSO4PbSO_4 produced if 0.50.5 Faradays of electricity is drawn from a lead accumulator.

Solution:

0.5 moles of PbSO40.5 \text{ moles of } PbSO_4 are produced at each electrode, totaling 1.0 mole1.0 \text{ mole} for the whole cell.

Explanation:

From the half-reactions, 11 mole of PbPb produces 11 mole of PbSO4PbSO_4 by releasing 2e−2e^-. Thus, 2F2F of charge produces 11 mole of PbSO4PbSO_4 at the anode and 11 mole at the cathode. Therefore, 0.5F0.5F will produce 0.250.25 moles at each electrode based on the stoichiometry of the 22-electron process.

Problem 4:

During the discharging of a lead storage battery, the density of the H2SO4H_2SO_4 electrolyte falls from 1.294 g/mL1.294 \text{ g/mL} to 1.139 g/mL1.139 \text{ g/mL}. If the total volume of the electrolyte is 800 mL800 \text{ mL}, and the initial percentage of H2SO4H_2SO_4 by mass is 39%39\%, explain the chemical change at the anode and visualize the cell setup.

Discharging lead acid battery showing consumption of sulfate ions at the electrodes.

Solution:

During discharge, the anode reaction is: Pb(s)+SO42−(aq)→PbSO4(s)+2e−Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^- The density decreases because H2SO4H_2SO_4 is consumed to form PbSO4PbSO_4 and water (H2OH_2O) is produced as a byproduct in the overall reaction: Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l) This dilution of the acid results in the drop in density.

Explanation:

The lead anode releases electrons and reacts with sulfate ions from the acid to form solid lead sulfate, which adheres to the plate.

Problem 5:

Identify the components and the direction of electron flow in a standard Leclanché dry cell during operation.

Dry cell diagram highlighting electron flow from Zn outer shell to the central carbon rod.

Solution:

In a dry cell, electrons flow from the Zinc cup (anode) through the external circuit to the Carbon rod (cathode). The Zinc undergoes oxidation: Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^- and these electrons are consumed at the cathode by NH4+NH_4^+ and MnO2MnO_2.

Explanation:

The zinc container serves as the negative terminal (source of electrons), while the graphite rod serves as the positive terminal where reduction occurs.