Corrosion, Fuel Cells and Applications
A chain-link fence in Houston rusts after a few humid summers, even though nobody ever hooked it up to a circuit. A reaction is happening anyway. Corrosion is a tiny galvanic cell that the metal builds against itself in damp air, slowly using itself up. In this section you will see why, and how engineers stop it.
Corrosion as a spontaneous galvanic cell
Put a drop of water on a clean steel plate. Rust shows up under the center of the drop, not at the edge, where oxygen from the air gets in most easily. Why does the metal get destroyed right where less dissolved oxygen arrives?
The answer to the drop puzzle: the center of the drop has less dissolved $\mathrm{O_2}$, so it becomes the anode and the iron dissolves there. The edge, rich in $\mathrm{O_2}$, becomes the cathode. Electrons travel through the metal between the two zones. This is called differential aeration.
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 oxidizing 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 polarization and precipitation when estimating the initial voltage.
Protecting metals from corrosion
To protect iron sheets against environmental corrosion, metallic coatings such as zinc galvanizing or copper coatings are used. Consider a galvanized iron sheet and…
To protect iron sheets against environmental corrosion, metallic coatings such as zinc galvanizing or copper coatings are used. Consider a galvanized 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 oxidize 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 $1.0\,\mathrm{año}$ of service ($365\,\mathrm{días}$). Assume that all the current comes from the oxidation of the protective metal, with no additional self-corrosion.
Batteries: discharge and recharge
Your phone battery delivers energy for hours while its chemicals are used up. Then a few hours on a charger restores them. How can an outside current force atoms to run a spontaneous redox reaction backward?
An electrochemical accumulator (rechargeable battery) can operate in discharge mode (galvanic cell) and in recharge mode (electrolytic cell): a) Compare the terms for…
An electrochemical accumulator (rechargeable battery) 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.
Fuel cells
The electrons cannot cross the membrane, only the protons can. That forces the electrons through the outside circuit, and that flow is the useful current.
This is a galvanic cell, so $\Delta G^\circ = -nFE^\circ_{\text{cell}}$. For $\mathrm{H_2O(l)}$ with $n = 2$ per mole: $E^\circ_{\text{cell}} = \dfrac{237{,}100}{2 \times 96{,}485} = 1.23\ \mathrm{V}$.
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
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 millimeters 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.
Corrosion rate from current
A steady corrosion current tells you how fast metal is lost. Use $q = It$, then moles of electrons $= q/F$, then the half-reaction ratio.
Back to the start: a fence rusts because the metal itself forms galvanic microcells in damp air. The iron is the anode and dissolved oxygen is reduced at the cathode.