Commercial Cells, Fuel Cells & Corrosion

CHEApplied Electrochemistry

This subtopic covers real-world electrochemistry: primary cells (single-use), secondary cells (rechargeable, e.g. lead storage battery), fuel cells, and the electrochemical corrosion of iron.

Exam questions ask for electrode reactions, distinctions between cell types, and the mechanism of rusting plus methods of protection.

Lead storage cell (overall, discharge)

Pb + PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O

discharge runs left to right; charging reverses it. Both electrodes turn to

PbSO_4

and

H_2SO_4

is consumed.

H_2

O_2

fuel cell (overall)

2H_2 + O_2 \rightarrow 2H_2O

alkaline (

KOH

) cell; the only product is water.

Rusting (anode and overall)

Fe \rightarrow Fe^{2+} + 2e^- \qquad 2Fe + \tfrac{3}{2}O_2 + xH_2O \rightarrow Fe_2O_3\cdot xH_2O

Fe^{2+}

is further oxidised by

O_2

to hydrated ferric oxide (rust).

Feasibility of displacement

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} > 0 \implies \Delta G^\circ = -nFE^\circ_{cell} < 0

a positive

E^\circ_{cell}

means the displacement is spontaneous.

Primary cells (e.g. dry cell, $Zn\text{-}MnO_2$ ) are non-rechargeable because the cell reaction is not easily reversible; secondary cells (e.g. lead storage battery, Ni-Cd) can be recharged by passing current in the reverse direction.
Dry (Leclanche) cell: anode is the $Zn$ container ( $Zn \rightarrow Zn^{2+} + 2e^-$ ); cathode is a graphite rod surrounded by $MnO_2$ and $NH_4Cl$ paste; $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$ ; EMF about $1.5\,\text{V}$ , falling in use because products are not removed.
Lead storage battery on discharge: anode $Pb + SO_4^{2-} \rightarrow PbSO_4 + 2e^-$ ; cathode $PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4 + 2H_2O$ ; $H_2SO_4$ is consumed so its concentration (and density) falls. Charging reverses these reactions.
During charging the lead storage cell acts as an electrolytic cell: $2PbSO_4 + 2H_2O \rightarrow Pb + PbO_2 + 2H_2SO_4$ , regenerating the acid (density rises); the state of charge can be checked with a hydrometer.
$H_2\text{-}O_2$ fuel cell (alkaline, $KOH$ ): anode $2H_2 + 4OH^- \rightarrow 4H_2O + 4e^-$ ; cathode $O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$ ; net $2H_2 + O_2 \rightarrow 2H_2O$ . Advantages: high efficiency, pollution-free (only water as product), continuous operation while fuel is supplied.
A fuel cell differs from an ordinary cell in that reactants (fuel and oxidant) are supplied continuously from outside rather than stored inside; porous carbon electrodes contain catalysts (finely divided Pt/Pd) to speed the electrode reactions. Used in spacecraft (Apollo).
Rusting is electrochemical: an impure iron surface acts as tiny galvanic cells. Anode $Fe \rightarrow Fe^{2+} + 2e^-$ ( $E^\circ = -0.44\,\text{V}$ ); cathode $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ ( $H^+$ from dissolved $CO_2$ /water). $Fe^{2+}$ is then oxidised by $O_2$ to hydrated $Fe_2O_3 \cdot xH_2O$ (rust); water and dissolved $O_2$ are both required.
Factors accelerating rusting: presence of moisture, dissolved $O_2$ , electrolytes (acids, salts, $CO_2$ ) which improve conductivity, and impurities/strained regions in the metal that set up local cells.
Galvanising (zinc coating) protects by sacrificial (cathodic) protection: $Zn$ ( $E^\circ = -0.76\,\text{V}$ ) is more easily oxidised than $Fe$ , so zinc corrodes preferentially and protects iron even when the coating is scratched.
Tin-plating gives only barrier protection: $Sn$ ( $E^\circ(Sn^{2+}/Sn) = -0.14\,\text{V}$ ) is less reactive than $Fe$ , so once the tin layer is scratched the exposed iron becomes the anode and corrodes faster (a galvanic couple accelerates rusting).
Other protection methods: cathodic protection with a more active sacrificial metal (Mg/Zn blocks bolted to ships and pipelines), barrier coatings (paint, grease, electroplating with Cr/Ni), and alloying (stainless steel with Cr/Ni).
Displacement test via the electrochemical series: a metal displaces ions of a less reactive metal. For $Fe + 2Ag^+ \rightarrow Fe^{2+} + 2Ag$ , $E^\circ_{cell} = 0.80 - (-0.44) = +1.24\,\text{V}$ ( $>0$ ), so the reaction is spontaneous and Fe displaces Ag.
Quantitative links: $\Delta G^\circ = -nFE^\circ$ and at $298\,\text{K}$ , $E^\circ_{cell} = \dfrac{0.0591}{n}\log K$ ; Faraday's law connects deposited mass with $Q = It$ (one mole of electrons $= 1\,F = 96500\,\text{C}$ ).

Where the marks go

Confusing galvanising and tin-plating: zinc protects even when scratched (sacrificial), but a scratched tin coating makes iron corrode FASTER — examiners test this reversal directly.
Writing that $H_2SO_4$ concentration rises during discharge of the lead cell — it falls (acid is consumed); it rises during charging.
Forgetting that BOTH water and oxygen are essential for rusting; iron does not rust in dry air or in air-free water alone.
Stating the wrong cathode reaction for rusting — at the cathode oxygen is reduced ( $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ ); iron is oxidised at the anode, not reduced.
Confusing a fuel cell with a primary cell: a fuel cell is fed reactants continuously from outside and does not 'run down' as long as fuel is supplied.

How the board asks it

Give reasonselectrochemical rusting; water and dissolved oxygen both required
Give reasons: iron does not rust in dry air or in air-free water, but rusts rapidly in moist air containing dissolved $CO_2$ .
Distinguishsacrificial protection of zinc versus barrier protection of tin
Distinguish between galvanising and tin-plating of iron with reference to the protection given when the coating is scratched.
Conversionanode, cathode and overall reactions of the alkaline fuel cell
Write the reactions occurring at the anode and cathode of a $H_2$ - $O_2$ fuel cell working in $KOH$ solution, and give the overall cell reaction.
Give reasonslead storage cell; $H_2SO_4$ consumed on discharge, regenerated on charging
Account for the fact that the density of the electrolyte in a lead storage battery falls during discharge but rises again on charging.
Numerical $\Delta G^\circ = -nFE^\circ$ and feasibility from the electrochemical series
For the reaction $Fe + 2Ag^+ \rightarrow Fe^{2+} + 2Ag$ , given $E^\circ_{Ag^+/Ag} = +0.80\,\text{V}$ and $E^\circ_{Fe^{2+}/Fe} = -0.44\,\text{V}$ , calculate $E^\circ_{cell}$ and $\Delta G^\circ$ and state whether iron displaces silver.
Define / statefuel cell contrasted with a primary cell
Define a fuel cell and state two ways in which it differs from a primary cell such as the dry (Leclanche) cell.

Practise this topic

PreviousProducts of Electrolysis & Reasoning

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.