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Electrochemistry - Dry cell and lead accumulator

Grade 12ICSEChemistry

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

🔑Concepts

Primary Cells (Dry Cell): These are cells where the redox reaction occurs only once and the cell becomes dead after a period of use as the reaction cannot be reversed. The Leclanché cell is the most common example.

Dry Cell Components: The anode consists of a Zinc (ZnZn) container. The cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide (MnO2MnO_2) and carbon. The electrolyte is a moist paste of ammonium chloride (NH4ClNH_4Cl) and zinc chloride (ZnCl2ZnCl_2).

Dry Cell Reactions: At the anode: Zn(s)ightarrowZn2+(aq)+2eZn(s) ightarrow Zn^{2+}(aq) + 2e^-. At the cathode: 2MnO2(s)+2NH4+(aq)+2eightarrowMn2O3(s)+2NH3(g)+H2O(l)2MnO_2(s) + 2NH_4^+(aq) + 2e^- ightarrow Mn_2O_3(s) + 2NH_3(g) + H_2O(l).

Role of ZnCl2ZnCl_2: The NH3NH_3 gas produced at the cathode combines with Zn2+Zn^{2+} ions from ZnCl2ZnCl_2 to form a complex ion [Zn(NH3)4]2+[Zn(NH_3)_4]^{2+}, which prevents the buildup of gas pressure inside the cell.

Secondary Cells (Lead Accumulator): These cells can be recharged by passing current through them in the opposite direction. They can be used through several cycles of discharging and charging.

Lead Accumulator Components: The anode is a lead (PbPb) plate. The cathode is a grid of lead packed with lead dioxide (PbO2PbO_2). The electrolyte is a 38%38\% solution of sulphuric acid (H2SO4H_2SO_4) with a density of 1.30g/cm31.30 \, g/cm^3.

Discharging of Lead Accumulator: During discharge, chemical energy is converted to electrical energy. H2SO4H_2SO_4 is consumed, and PbSO4PbSO_4 is deposited on both electrodes.

Recharging of Lead Accumulator: An external source of EMFEMF is applied. The cell reactions are reversed: 2PbSO4(s)+2H2O(l)ightarrowPb(s)+PbO2(s)+2H2SO4(aq)2PbSO_4(s) + 2H_2O(l) ightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq).

📐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++2eMn2O3+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)+2ePbSO4(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 2e2e^-. 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.

Dry cell and lead accumulator - Revision Notes & Key Formulas | ICSE Class 12 Chemistry