Corrosion, fuel cells and applications
Walk along the seafront in Galway or Bray and you will see iron railings flaking away, even though nobody has connected them to anything. No visible circuit is needed for a spontaneous redox reaction to happen. Rusting is a tiny galvanic cell that the metal builds against itself in damp air. In this page you will see how it works, how to stop it, and how the same ideas explain the batteries in an e-scooter and the hydrogen fuel cell.
Rusting of iron as a galvanic cell
Put a drop of water on a clean iron sheet. Iron is oxidised (it rusts away) at the centre of the drop, where the least dissolved oxygen reaches. Oxygen is reduced at the edge, where it is plentiful. Why does the metal wear away in the oxygen-poor zone rather than where the oxygen is?
The anode is the oxygen-poor zone and the cathode is the oxygen-rich zone. The electrons flow through the metal from the centre of the drop (anode, oxidation) to the edge (cathode, reduction, where oxygen is plentiful). The Fe²⁺ and OH⁻ ions meet in the water and the iron(II) hydroxide is further oxidised by oxygen to rust. The oxygen-rich edge is where oxygen is reduced, so it acts as the cathode and is not attacked. The oxygen-poor centre has no easily reduced species, so it becomes the anode and the iron there dissolves. This is why the damage appears at the centre of the drop, away from where oxygen enters.
Atmospheric corrosion of iron in the presence of damp air takes place through the formation of microcells or local galvanic couples on the surface of the metal.…
Atmospheric corrosion of iron in the presence of damp air takes place through the formation of microcells or local galvanic couples on the surface of the metal. Considering that the aqueous medium on the metal has a neutral pH:
- a) Write the half-equations corresponding to the anodic zone and the cathodic zone, identifying the chemical species that acts as the oxidising agent and the one that acts as the reducing agent.
- b) Deduce the net chemical equation of the initial corrosion process and calculate the ideal electromotive force under the stated conditions of the galvanic cell formed. Data at $298\,\mathrm{K}$: $E^\circ(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44\,\mathrm{V}$; $E(\mathrm{O}_2/\mathrm{OH}^-,\,\mathrm{pH}=7) = +0.81\,\mathrm{V}$. Use the potential of iron with unit activity of its ions and the conditional potential of oxygen given at pH 7; the latter is not a standard potential of the half-equation written with hydroxide. Neglect polarisation and precipitation when estimating the initial voltage.
Preventing rusting
Rust-proofing has two strategies, and the electrochemical series tells you which one works when the surface is scratched.
To protect iron sheets against environmental corrosion, metallic coatings such as zinc galvanising or copper coatings are used. Consider a galvanised iron sheet and…
To protect iron sheets against environmental corrosion, metallic coatings such as zinc galvanising or copper coatings are used. Consider a galvanised iron sheet and another coated with copper, each subjected to an accidental scratch that exposes the base iron to damp air:
- a) Justify which of the two metals in contact will preferentially oxidise in the scratched area of each piece, using the standard potentials: $E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\,\mathrm{V}$; $E^\circ(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44\,\mathrm{V}$; $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\,\mathrm{V}$.
- b) State the protection mechanism that acts in each case (barrier protection or sacrificial cathodic protection).
To protect a steel pipe buried in damp soil, which acts as an electrolyte, it is planned to connect a metal block buried beside it by means of a conducting cable. The…
To protect a steel pipe buried in damp soil, which acts as an electrolyte, it is planned to connect a metal block buried beside it by means of a conducting cable. The following diagram illustrates the two options considered: connecting a magnesium block or a copper block.
Standard reduction potential data at $298\,\mathrm{K}$: $E^\circ(\mathrm{Mg}^{2+}/\mathrm{Mg}) = -2.37\,\mathrm{V}$; $E^\circ(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44\,\mathrm{V}$; $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\,\mathrm{V}$. Faraday constant: $F = 96\,485\,\mathrm{C}\times\mathrm{mol}^{-1}$. Molar mass: $M(\mathrm{Mg}) = 24.31\,\mathrm{g}\times\mathrm{mol}^{-1}$.
- a) Select, with reasoning, which of the two metal blocks will work effectively as a sacrificial anode to protect the iron pipe, and write the oxidation half-equation that takes place in it.
- b) If the average galvanic protection current supplied continuously by this anode is $0.050\,\mathrm{A}$, calculate the mass in grams of the protective metal consumed after $1.0\,\mathrm{year}$ of service ($365\,\mathrm{days}$). Assume that all the current comes from the oxidation of the protective metal, with no additional self-corrosion.
Primary and secondary cells
The battery in a phone or an e-scooter delivers energy for hours and can then be plugged in to put it all back. How can an outside current make atoms undo a spontaneous redox reaction?
In both kinds of cell the naming is the same: anode = oxidation, cathode = reduction. What changes between discharge and recharge is the role of each electrode: because each electrode reaction is reversed, the same electrode swaps from anode to cathode (or the other way round).
The lithium-ion cell and its life-cycle
A lithium-ion cell works on the same principle: galvanic on discharge, electrolytic on recharge, with lithium ions moving between the electrodes through the electrolyte. Its life-cycle matters: mining of lithium and cobalt, use in phones, e-scooters and electric cars, then disposal or recycling. Recycling recovers the metals and avoids waste, which is why the deposit return scheme and battery collection points exist.
Chemical cells and fuel cells compared
In every cell the anode is where oxidation happens, the cathode is where reduction happens, and the electrolyte lets ions move between them. A primary cell is used once. A secondary cell can be recharged. In both, the reactants are stored inside the cell and run out. A fuel cell is different: reactants such as hydrogen and oxygen are supplied continuously from outside, so it works for as long as they are fed in.
An electrochemical accumulator can operate in discharge mode (galvanic cell) and in recharge mode (electrolytic cell): a) Compare the terms for the polarity of the…
An electrochemical accumulator can operate in discharge mode (galvanic cell) and in recharge mode (electrolytic cell):
- a) Compare the terms for the polarity of the electrodes (positive and negative) with the functional terms (anode and cathode) in both stages of operation.
- b) State the fundamental difference in the thermodynamic spontaneity of the processes that take place during discharge compared with recharge, justifying the sign of the change in Gibbs free energy ($\Delta G$) and of the electromotive force ($E_{\mathrm{celda}}$) in each stage. Define the sign of the electromotive force for the reaction written in the direction of each stage, and distinguish it from the positive voltage that the charger must supply.
In the lead–acid battery of a vehicle, the overall discharge process is: $$\mathrm{Pb}(\mathrm{s}) + \mathrm{PbO}2(\mathrm{s}) + 2\,\mathrm{H}2\mathrm{SO}4(\mathrm{aq})…
In the lead–acid battery of a vehicle, the overall discharge process is:
$$\mathrm{Pb}(\mathrm{s}) + \mathrm{PbO}_2(\mathrm{s}) + 2\,\mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow 2\,\mathrm{PbSO}_4(\mathrm{s}) + 2\,\mathrm{H}_2\mathrm{O}(\mathrm{l})$$Standard data at $298\,\mathrm{K}$: $E^\circ(\mathrm{PbO}_2/\mathrm{PbSO}_4) = +1.69\,\mathrm{V}$; $E^\circ(\mathrm{PbSO}_4/\mathrm{Pb}) = -0.36\,\mathrm{V}$. Molar masses: $M(\mathrm{PbSO}_4) = 303.3\,\mathrm{g}\times\mathrm{mol}^{-1}$; $M(\mathrm{H}_2\mathrm{SO}_4) = 98.08\,\mathrm{g}\times\mathrm{mol}^{-1}$; $F = 96\,485\,\mathrm{C}\times\mathrm{mol}^{-1}$.
- a) Write the half-equations that take place at each electrode during the forced recharge process, identifying the anode and the cathode, and determine the minimum theoretical potential difference required to reverse the cell under standard conditions.
- b) To recharge a cell that has consumed $49.04\,\mathrm{g}$ of $\mathrm{H}_2\mathrm{SO}_4$, a constant current of $4.00\,\mathrm{A}$ is applied with a faradaic efficiency of $85.0\%$. Calculate the time required, in hours, to complete the regeneration of the acid and the mass in grams of $\mathrm{PbSO}_4$ that will have been transformed. The minimum theoretical voltage is understood as the reversible limit; for a real current to flow, an excess voltage is required.
Hydrogen fuel cells
Protons cross the membrane, but electrons cannot, so they must travel round the external circuit. That current is the useful electrical output.
A hydrogen–oxygen fuel cell operates with a polymer proton exchange membrane (PEM) in aqueous acidic medium at $298\,\mathrm{K}$. The overall net reaction produces…
A hydrogen–oxygen fuel cell operates with a polymer proton exchange membrane (PEM) in aqueous acidic medium at $298\,\mathrm{K}$. The overall net reaction produces liquid water:
- a) Write the half-equations that take place at the anode and at the cathode, and state, with reasoning, the standard potential of the cell, given that $\Delta G^\circ_{\mathrm{f}}[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = -237.1\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. Faraday constant: $F = 96\,485\,\mathrm{C}\times\mathrm{mol}^{-1}$.
- b) Determine the volume of $\mathrm{H}_2(\mathrm{g})$, measured at $298\,\mathrm{K}$ and $1.00\times 10^5\,\mathrm{Pa}$, needed to supply the cell and maintain a constant net electrical power of $250\,\mathrm{W}$ for $2.00\,\mathrm{horas}$, if the actual operating voltage between electrodes is $0.700\,\mathrm{V}$. Gas constant: $R = 8,314\,\mathrm{J}\times\mathrm{K}^{-1}\times\mathrm{mol}^{-1}$. Assume complete use of the hydrogen, unit faradaic efficiency and no auxiliary electrical consumption.
Certain hydrogen fuel cells use a concentrated aqueous alkaline electrolyte (such as aqueous $\mathrm{KOH}$ at $298\,\mathrm{K}$): a) Write the anodic, cathodic and…
Certain hydrogen fuel cells use a concentrated aqueous alkaline electrolyte (such as aqueous $\mathrm{KOH}$ at $298\,\mathrm{K}$):
- a) Write the anodic, cathodic and overall half-equations in this basic medium, checking that the overall reaction is the same as in acidic medium. Standard data at $298\,\mathrm{K}$: $E^\circ(\mathrm{O}_2/\mathrm{OH}^-) = +0.40\,\mathrm{V}$; $E^\circ(\mathrm{H}_2\mathrm{O}/\mathrm{H}_2,\,\mathrm{OH}^-) = -0.83\,\mathrm{V}$. Calculate the standard electromotive force of the cell.
- b) The ideal thermodynamic efficiency of the fuel cell is defined as $\eta = \frac{\Delta G^\circ}{\Delta H^\circ}$. At $298\,\mathrm{K}$, the combustion of hydrogen to give $\mathrm{H}_2\mathrm{O}(\mathrm{l})$ has $\Delta H^\circ = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta G^\circ = -237.1\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. Calculate $\eta$ and explain why the $100\%$ of the enthalpy cannot be converted into usable electrical work under standard isothermal conditions.
Hydrogen as an energy carrier
Hydrogen is not a fuel you dig out of the ground: it is an energy carrier. You must spend energy to make it, either by reforming natural gas ("grey" hydrogen, which releases CO₂) or by the electrolysis of water using renewable electricity ("green" hydrogen), for example from Atlantic wind farms. The electrolysis uses the reverse of the fuel-cell reaction: $\mathrm{2\,H_2O \rightarrow 2\,H_2 + O_2}$. A fair comparison with other fuels has to look at the whole life-cycle: production, compression, transport and use.
Back to the start: the railings rust because iron and oxygen form a spontaneous galvanic cell. Galvanising stops it because zinc becomes the anode instead.
The carbon footprint associated with the use of hydrogen as an energy carrier is analysed for a fuel-cell bus that generates $1.00\,\mathrm{kWh}$ of electrical energy.…
The carbon footprint associated with the use of hydrogen as an energy carrier is analysed for a fuel-cell bus that generates $1.00\,\mathrm{kWh}$ of electrical energy. The cell operates with an electrical efficiency of $60.0\%$ with respect to the higher heating value of hydrogen ($\Delta H^\circ = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ of $\mathrm{H}_2$, equivalent to $39.7\,\mathrm{kWh}\times\mathrm{kg}^{-1}$ of $\mathrm{H}_2$):
- a) Calculate the mass of $\mathrm{H}_2$ in grams consumed by the cell to produce $1.00\,\mathrm{kWh}$ of useful electrical energy.
- b) If the hydrogen is "grey", obtained by steam reforming of natural gas ($\mathrm{CH}_4 + 2\,\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CO}_2 + 4\,\mathrm{H}_2$) with an emission of $10.0\,\mathrm{kg}$ of $\mathrm{CO}_2$ per kilogram of $\mathrm{H}_2$ produced at the plant, determine the mass of $\mathrm{CO}_2$ in grams emitted per $\mathrm{kWh}$ of electrical energy generated. Compare this value qualitatively with the use of "green" hydrogen obtained by electrolysis powered by photovoltaic solar energy. Distinguish direct operating emissions from life-cycle emissions; without a manufacturing and supply emissions factor, no numerical total footprint can be assigned to green hydrogen, nor can it be claimed to be zero.
An iron plate of dimensions $10.0\,\mathrm{cm} \times 10.0\,\mathrm{cm}$ and uniform thickness is immersed in aerated seawater at $298\,\mathrm{K}$. A differential…
An iron plate of dimensions $10.0\,\mathrm{cm} \times 10.0\,\mathrm{cm}$ and uniform thickness is immersed in aerated seawater at $298\,\mathrm{K}$. A differential aeration corrosion process is established, in which a steady corrosion current of $12.0\,\mathrm{mA}$ is measured experimentally, attributed to the loss of iron from one face of the plate only:
- a) Write the half-equations in the zones of higher and lower concentration of $\mathrm{O}_2$ that make up the differential aeration couple, and justify which of them undergoes anodic dissolution.
- b) Given that the density of iron is $\rho = 7.87\,\mathrm{g}\times\mathrm{cm}^{-3}$, calculate the loss of iron mass in grams after $60.0\,\mathrm{días}$ and determine the equivalent mean corrosion rate in penetration, averaged over that whole face, expressed in millimetres per year ($\mathrm{mm}\times\mathrm{año}^{-1}$). Data: $M(\mathrm{Fe}) = 55.85\,\mathrm{g}\times\mathrm{mol}^{-1}$; $F = 96\,485\,\mathrm{C}\times\mathrm{mol}^{-1}$; $1\,\mathrm{año} = 365\,\mathrm{días}$. The calculated mean penetration does not represent the maximum depth of local pitting.
Topic challenge. Decarbonisation requires managing renewable surpluses. Your mission is to design a technical proposal to store household energy using green hydrogen and…
Topic challenge. Decarbonisation requires managing renewable surpluses. Your mission is to design a technical proposal to store household energy using green hydrogen and ensure the integrity of the installation.
-
Real data and electrolysis: Go to the REE (RedOS) website and look up the current national photovoltaic solar power. Assume that a household feeds a surplus of $3.5\text{ kWh}$ into an electrolytic cell ($V=2.0\text{ V}$). Balance the decomposition of water in acidic medium using the ion-electron method and, applying Faraday's laws, calculate the moles and the volume of $\text{H}_2(g)$ generated under standard conditions.
-
Fuel cell and spontaneity: At night, this gas feeds a $\text{H}_2/\text{O}_2$ cell. Identify the oxidation numbers, write the half-equations and calculate the standard emf ($E^\circ$) using the tabulated potentials ($E^\circ(\text{O}_2/\text{H}_2\text{O}) = 1.23\text{ V}$). Determine the spontaneity by calculating $\Delta G^\circ$.
-
Corrosion and final verdict: The hydrogen is stored in an iron boiler. Choose, with justification, between $\text{Zn}$ and $\text{Cu}$ a sacrificial anode to prevent its corrosion. Write a short report that includes the stoichiometric calculations, the energy balance and the cathodic protection scheme.
Assessment will consider:
-
Accuracy in the ion-electron balancing and the stoichiometric application of Faraday's laws.
-
Rigour in the calculation of potentials, spontaneity ($\Delta G^\circ$) and assignment of oxidation numbers.
-
Appropriate chemical justification of the anticorrosion method and clarity of presentation of the proposal.