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Electrochemistry - Batteries and Fuel Cells

Grade 12CBSEChemistry

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

🔑Concepts

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Primary Batteries: These are non-rechargeable cells where the reaction occurs only once. Examples include the Dry Cell (Leclanché cell) used in torches and the Mercury cell used in hearing aids. In a dry cell, the anode is ZnZn and the cathode is a carbon rod surrounded by MnO2MnO_2.

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Secondary Batteries: These can be recharged by passing an electric current through them in the opposite direction. The Lead-storage battery (used in automobiles) and the Nickel-Cadmium (Ni−CdNi-Cd) cell are key examples. In a lead-storage battery, the anode is PbPb and the cathode is a grid of lead packed with PbO2PbO_2, using 38%38\% H2SO4H_2SO_4 as the electrolyte.

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Fuel Cells: Galvanic cells designed to convert the energy of combustion of fuels like H2H_2, CH4CH_4, or CH3OHCH_3OH directly into electrical energy. The most successful is the H2−O2H_2-O_2 fuel cell used in the Apollo space program. They are highly efficient (approx 70%70\%) compared to thermal plants (40%40\%) and are pollution-free.

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Corrosion: An electrochemical process where a metal is oxidized by loss of electrons to oxygen and moisture. For rusting of iron, the anode reaction is Fe(s)→Fe2+(aq)+2e−Fe(s) \rightarrow Fe^{2+}(aq) + 2e^- and the cathode reaction is O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l).

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Prevention of Corrosion: Methods include barrier protection (painting), sacrificial protection (galvanization with ZnZn), and using anti-rust solutions like alkaline phosphates.

📐Formulae

ΔG=−nFEcell\Delta G = -nFE_{cell}

Efficiency (η) of Fuel Cell=ΔGΔH×100\text{Efficiency (}\eta\text{) of Fuel Cell} = \frac{\Delta G}{\Delta H} \times 100

Dry Cell Overall: Zn(s)+2MnO2(s)+2NH4+(aq)→Zn2+(aq)+Mn2O3(s)+2NH3(g)+H2O(l)\text{Dry Cell Overall: } Zn(s) + 2MnO_2(s) + 2NH_4^+(aq) \rightarrow Zn^{2+}(aq) + Mn_2O_3(s) + 2NH_3(g) + H_2O(l)

Lead Storage (Discharging): Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)\text{Lead Storage (Discharging): } Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)

Fuel Cell (Overall): 2H2(g)+O2(g)→2H2O(l)\text{Fuel Cell (Overall): } 2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

💡Examples

Problem 1:

Write the cathode and anode reactions that occur during the discharging of a Lead-storage battery.

Solution:

Anode: Pb(s)+SO42−(aq)→PbSO4(s)+2e−Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^- Cathode: PbO2(s)+SO42−(aq)+4H+(aq)+2e−→PbSO4(s)+2H2O(l)PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)

Explanation:

During discharging, the lead-storage battery acts as a galvanic cell. Lead is oxidized at the anode, and Lead dioxide is reduced at the cathode. Both reactions produce PbSO4PbSO_4, which sticks to the electrodes.

Problem 2:

Calculate the theoretical efficiency of a H2−O2H_2-O_2 fuel cell if ΔGf∘\Delta G_f^{\circ} for H2O(l)H_2O(l) is −237.2 kJ mol−1-237.2 \text{ kJ mol}^{-1} and ΔHf∘\Delta H_f^{\circ} is −285.8 kJ mol−1-285.8 \text{ kJ mol}^{-1}.

Solution:

Efficiency η=ΔGΔH×100=−237.2−285.8×100≈83%\eta = \frac{\Delta G}{\Delta H} \times 100 = \frac{-237.2}{-285.8} \times 100 \approx 83\%

Explanation:

The efficiency of a fuel cell is defined as the ratio of the maximum useful work (Gibbs Free Energy change) to the total heat of combustion (Enthalpy change).

Batteries and Fuel Cells - Revision Notes & Key Formulas | CBSE Class 12 Chemistry