Industrial Cells

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Leclanche Cell

Oxidation at Anode (source of electrons)(-): $Z{n}_{(s)}\to Z{n}_{(aq)}^{2+}+2e^{-}$

Reduction at Cathode(+): $2Mn{O}_{2_{(}s)}+2H_{(aq)}^{+}+2e^{-}\to M{n}_2{O}_{3(s)}+H_2{O}_{(l)}$

• $H^{+}$ is from $N{H}_4^{+}$
• $N{H}_4^{+}$ turns into $N{H}_3$

THIS IS A PRIMARY CELL, I.E. NOT RECHARGEABLE Must happen under acidic conditions

Alkaline Cell

• Use hydroxide ions to neutralise hydrogen ions

Lead Acid Battery/Accumulator

Secondary cell, therefore rechargeable

emf of cell is 2.05 V

Possible if reactants remain in proximity of electrodes (Lead sulphate coats both electrodes during discharge)

Discharging

Anode(-), site of oxidation: $P{b}_{(s)}+S{O}_{4(aq)}^{2-}\to PbS{O}_{4(s)}+2e^{-}$

Cathode(+): $Pb{O}_{2(s)}+S{O}_4^{2-}+4H_{(aq)}^{+}+2e^{-}\to PbS{O}_{4(s)}+2H_2{O}_{(l)}$

Overall: $P{b}_{(s)}+Pb{O}_{2(s)}+2S{O}_{4(aq)}^{2-}\to 2PbS{O}_{4(s)}+2H_2{O}_{(l)}+4H_{(aq)}^{+}$

Recharging

Anode(+), site of oxidation: $PbS{O}_{4(s)}+2H_2{O}_{(l)}\to Pb{O}_{2(s)}+S{O}_4^{2-}+4H_{(aq)}^{+}+2e^{-}$

Cathode(-): $PbS{O}_{4(s)}\to P{b}_{(s)}+S{O}_{4(aq)}^{2-}+2e^{-}$

Overall: $2PbS{O}_{4(s)}+2H_2{O}_{(l)}+4H_{(aq)}^{+}\to P{b}_{(s)}+Pb{O}_{2(s)}+2S{O}_{4(aq)}^{2-}$

Hydrogen Fuel Cell

Fuel cell: galvanic cells requiring a continuous source of fuel to sustain the redox reaction (removal of products) Not just Hydrogen, e.g. methane works too More efficient and controlled than combustion (gives available chemical energy without combustion)

Anode: $H_2\to 2H^{+}+2e^{-}$

Cathode: $4e^{-}+O_2+4H^{+}\to 2H_{2}O$

Full equation: $2H_2+O_2\to 2H_{2}O$ Cell potential: +1.23 V