Chapter 2 · Electrochemistry

2.4 Electrolysis, Batteries, Fuel Cells and Corrosion

Push electrons the other way and electricity drives chemistry: Faraday's laws $m=\dfrac{M\,I\,t}{nF}$ say how much is made. The same redox ideas power batteries and fuel cells, and explain why iron rusts and how to stop it.

Every aluminium can, every tonne of chlorine for water treatment and nearly all the pure copper in electrical wiring is made by pushing an electric current through a liquid. Chrome-plated taps, gold-plated connectors and the "silver" on cheap jewellery are all laid down the same way. In these electrolytic cells a power supply forces a non-spontaneous redox reaction to happen, the reverse of what a galvanic cell does. Michael Faraday found in the 1830s that the amount of chemical change is fixed exactly by the amount of charge passed, which makes electrolysis one of the most precisely controllable processes in industry.

The second half of this section returns to galvanic cells in their most useful forms. Batteries pack a galvanic cell into a sealed can: some are used once, others can be recharged by electrolysis. Fuel cells are fed continuously with fuel and oxygen. And corrosion is a galvanic cell nobody wants: tiny cells on the surface of iron that quietly turn steel into rust. Understanding it as electrochemistry shows how to prevent it.

Electrolysis and Faraday's Laws

📘 Definition 2.8 — Electrolytic cell and electrolysis

An electrolytic cell consists of two electrodes dipping into an electrolyte (a solution or a molten ionic compound) and connected to an external source of direct current. Electrolysis is the chemical decomposition brought about by passing the current.

  • The electrode joined to the negative terminal of the supply is the cathode. Electrons are pushed into it, and cations (or water) are reduced there.
  • The electrode joined to the positive terminal is the anode. Electrons are pulled out of it, and anions, water or the anode metal itself are oxidised there.

So in both kinds of cell, oxidation happens at the anode and reduction at the cathode; only the signs differ. Electrodes that take no part in the reaction (platinum, graphite) are inert; those that do (a copper anode that dissolves) are active.

Faraday measured how much metal or gas was produced by different currents running for different times. His results, stated in modern language, are the two laws of electrolysis.

📐 Law 2.11 — Faraday's first law of electrolysis

The mass $m$ of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $Q$ passed through the electrolyte:

$$m=Z\,Q=Z\,I\,t,$$

where $Q=It$ (charge in coulombs = current in amperes × time in seconds) and the constant $Z$ is the electrochemical equivalent of the substance: the mass deposited by $1\ \mathrm{C}$ of charge (unit $\mathrm{g\,C^{-1}}$).

📐 Law 2.12 — Faraday's second law of electrolysis

When the same quantity of electricity is passed through different electrolytes (for example, cells connected in series), the masses of the substances liberated are proportional to their equivalent masses, $\dfrac{\text{molar mass}}{\text{number of electrons needed per formula unit}}$:

$$\frac{m_{1}}{m_{2}}=\frac{M_{1}/n_{1}}{M_{2}/n_{2}}.$$

Both laws follow at once from counting electrons. One mole of electrons carries a charge of $N_{A}\,e\approx6.022\times10^{23}\times1.602\times10^{-19}\ \mathrm{C}$ $\approx9.65\times10^{4}\ \mathrm{C}$; the value used in this course is $96\,487\ \mathrm{C}$. This quantity is one faraday, $1\ F$. A half-reaction such as $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$ needs $2\ \mathrm{mol}$ of electrons, i.e. $2F$ of charge, per mole of copper.

📐 Relation 2.13 — Faraday's laws in one formula
$$\text{moles of electrons}=\frac{Q}{F}=\frac{I\,t}{F},\qquad m=\frac{Q}{nF}\times M=\frac{M\,I\,t}{nF},\qquad Z=\frac{M}{nF},$$

where $n$ is the number of electrons in the electrode half-reaction per formula unit of product and $M$ its molar mass. We use $F=96\,487\ \mathrm{C\,mol^{-1}}$; many exam problems state $96\,500\ \mathrm{C\,mol^{-1}}$, which changes answers only in the fourth significant figure. Always convert time to seconds.

✏️ Example 1 — Copper plating
A current of $2.00\ \mathrm{A}$ is passed through copper(II) sulphate solution for $30.0\ \mathrm{min}$. What mass of copper is deposited on the cathode? ($\mathrm{Cu}=63.5\ \mathrm{g\,mol^{-1}}$, $F=96\,487\ \mathrm{C\,mol^{-1}}$.)
Charge $Q=It=2.00\ \mathrm{A}\times(30.0\times60)\ \mathrm{s}=3600\ \mathrm{C}$.
Electrons $\dfrac{3600}{96\,487}=0.03731\ \mathrm{mol}\ e^{-}$.
Copper $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$: $\dfrac{0.03731}{2}=0.01866\ \mathrm{mol}$ Cu, i.e. $0.01866\times63.5=1.18\ \mathrm{g}$.
Check In one line: $m=\dfrac{63.5\times2.00\times1800}{2\times96\,487}=1.18\ \mathrm{g}$.
✏️ Example 2 — Silver plating time, and three cells in series
(a) How long must a current of $0.500\ \mathrm{A}$ flow through silver nitrate solution to plate $1.50\ \mathrm{g}$ of silver onto a spoon? ($\mathrm{Ag}=108\ \mathrm{g\,mol^{-1}}$.)
(b) Three cells containing aqueous $\mathrm{AgNO_{3}}$, aqueous $\mathrm{CuSO_{4}}$ and molten $\mathrm{AlCl_{3}}$ are connected in series. When $1.08\ \mathrm{g}$ of silver has been deposited, what masses of copper and aluminium are deposited? ($\mathrm{Cu}=63.5$, $\mathrm{Al}=27.0\ \mathrm{g\,mol^{-1}}$.)
(a) Charge needed $\dfrac{1.50}{108}=0.01389\ \mathrm{mol}$ Ag needs $0.01389\ \mathrm{mol}\ e^{-}$: $Q=0.01389\times96\,487=1340\ \mathrm{C}$.
(a) Time $t=\dfrac{Q}{I}=\dfrac{1340}{0.500}=2680\ \mathrm{s}=44.7\ \mathrm{min}$.
(b) Same charge $1.08\ \mathrm{g}$ Ag $=0.0100\ \mathrm{mol}$, so $0.0100\ \mathrm{mol}\ e^{-}$ passed through each cell. Cu: $\dfrac{0.0100}{2}\times63.5=0.318\ \mathrm{g}$. Al: $\dfrac{0.0100}{3}\times27.0=0.0900\ \mathrm{g}$.
Second law The masses are in the ratio of the equivalent masses $108:31.75:9.00$, i.e. $\dfrac{108}{1}:\dfrac{63.5}{2}:\dfrac{27.0}{3}$ ✓.
✏️ Example 3 — How many faradays?
How much charge, in faradays and in coulombs, is needed (a) to reduce $1\ \mathrm{mol}$ of $\mathrm{Cr_{2}O_{7}^{2-}}$ to $\mathrm{Cr^{3+}}$; (b) to oxidise $0.50\ \mathrm{mol}$ of $\mathrm{H_{2}O}$ to $\mathrm{O_{2}}$; (c) to produce $27.0\ \mathrm{g}$ of aluminium from molten $\mathrm{Al_{2}O_{3}}$?
(a) $\mathrm{Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}}$ $\mathrm{\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)}$: each Cr goes from $+6$ to $+3$, two Cr atoms, so $6F=6\times96\,487=5.79\times10^{5}\ \mathrm{C}$.
(b) $\mathrm{2H_{2}O(l)\rightarrow O_{2}(g)+4H^{+}(aq)+4e^{-}}$: $2\ \mathrm{mol}\ e^{-}$ per mole of water, so $0.50\ \mathrm{mol}$ needs $1F=9.65\times10^{4}\ \mathrm{C}$.
(c) $\mathrm{Al^{3+}(l)+3e^{-}\rightarrow Al(l)}$ and $27.0\ \mathrm{g}=1.00\ \mathrm{mol}$: $3F=2.89\times10^{5}\ \mathrm{C}$.
Note That is why aluminium smelters are built next to cheap hydroelectric power: about $3F$ per $27\ \mathrm{g}$ is a huge charge per kilogram.

What Forms at Each Electrode?

Electrolysis of a molten salt is simple: the only ions present are the salt's own. Molten $\mathrm{NaCl}$ gives sodium at the cathode, $\mathrm{Na^{+}(l)+e^{-}\rightarrow Na(l)}$, and chlorine at the anode, $\mathrm{2Cl^{-}(l)\rightarrow Cl_{2}(g)+2e^{-}}$. In aqueous solution there is a competitor: water itself can be reduced or oxidised.

$$\text{reduction of water:}\quad \mathrm{2H_{2}O(l)+2e^{-}\rightarrow H_{2}(g)+2OH^{-}(aq)},\qquad E^{\circ}=-0.83\ \mathrm{V}$$ $$\text{oxidation of water:}\quad \mathrm{2H_{2}O(l)\rightarrow O_{2}(g)+4H^{+}(aq)+4e^{-}},\qquad E^{\circ}_{\mathrm{O_{2}/H_{2}O}}=+1.23\ \mathrm{V}$$
📐 Rules 2.14 — Predicting the products of electrolysis
  1. At the cathode, of all the species that could be reduced (metal cations, $\mathrm{H^{+}}$, water), the one with the highest (most positive) reduction potential is reduced first. $\mathrm{Cu^{2+}}$ ($+0.34$) and $\mathrm{Ag^{+}}$ ($+0.80$) beat water; $\mathrm{Na^{+}}$ ($-2.71$), $\mathrm{K^{+}}$, $\mathrm{Mg^{2+}}$ and $\mathrm{Al^{3+}}$ do not, so hydrogen forms instead.
  2. At an inert anode, the species with the lowest reduction potential (the easiest to oxidise) is oxidised first. On $E^{\circ}$ alone, water ($+1.23$) should beat $\mathrm{Cl^{-}}$ ($+1.36$). But oxygen evolution is slow and needs an extra voltage, the overpotential, so from concentrated chloride solutions chlorine is formed instead. Sulphate and nitrate ions are very hard to oxidise, so with them oxygen forms.
  3. At an active anode (copper, silver, nickel), the anode metal itself is oxidised, $\mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}}$, because this needs less energy than oxidising water or the anion.
  4. Concentration also matters: very dilute $\mathrm{NaCl}$ gives mainly $\mathrm{O_{2}}$ at the anode; concentrated $\mathrm{H_{2}SO_{4}}$ gives peroxodisulphate, $\mathrm{2SO_{4}^{2-}(aq)\rightarrow S_{2}O_{8}^{2-}(aq)+2e^{-}}$.
Electrolyte (electrodes)Cathode productAnode productLeft in solution
molten $\mathrm{NaCl}$ (inert)$\mathrm{Na(l)}$$\mathrm{Cl_{2}(g)}$—
conc. aqueous $\mathrm{NaCl}$, brine (inert)$\mathrm{H_{2}(g)}$$\mathrm{Cl_{2}(g)}$$\mathrm{NaOH}$
aqueous $\mathrm{CuSO_{4}}$ (Pt)$\mathrm{Cu(s)}$$\mathrm{O_{2}(g)}$$\mathrm{H_{2}SO_{4}}$
aqueous $\mathrm{CuSO_{4}}$ (Cu)$\mathrm{Cu(s)}$ depositsCu anode dissolvesunchanged $\mathrm{CuSO_{4}}$
dilute $\mathrm{H_{2}SO_{4}}$ (Pt)$\mathrm{H_{2}(g)}$$\mathrm{O_{2}(g)}$more concentrated acid
conc. $\mathrm{H_{2}SO_{4}}$ (Pt)$\mathrm{H_{2}(g)}$$\mathrm{S_{2}O_{8}^{2-}(aq)}$—
✏️ Example 4 — Brine, copper sulphate and copper refining
Write the electrode reactions and the overall change for the electrolysis of (a) concentrated aqueous $\mathrm{NaCl}$ with inert electrodes; (b) aqueous $\mathrm{CuSO_{4}}$ with platinum electrodes; (c) aqueous $\mathrm{CuSO_{4}}$ with an impure copper anode and a pure copper cathode.
(a) Cathode: $\mathrm{2H_{2}O(l)+2e^{-}\rightarrow H_{2}(g)+2OH^{-}(aq)}$ ($\mathrm{Na^{+}}$, at $-2.71\ \mathrm{V}$, cannot compete). Anode: $\mathrm{2Cl^{-}(aq)\rightarrow Cl_{2}(g)+2e^{-}}$ (overpotential for $\mathrm{O_{2}}$). Overall: $$\mathrm{2NaCl(aq)+2H_{2}O(l)\rightarrow 2NaOH(aq)+H_{2}(g)+Cl_{2}(g)}$$ This is the chlor-alkali process, the source of most of the world's chlorine and sodium hydroxide.
(b) Cathode: $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$. Anode: $\mathrm{2H_{2}O(l)\rightarrow O_{2}(g)+4H^{+}(aq)+4e^{-}}$ (sulphate is not oxidised). Overall: $\mathrm{2CuSO_{4}(aq)+2H_{2}O(l)\rightarrow 2Cu(s)+O_{2}(g)+2H_{2}SO_{4}(aq)}$. The blue colour fades and the solution becomes acidic.
(c) Anode: $\mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}}$. Cathode: $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$. Copper simply moves from anode to cathode, and the solution is unchanged.
Why it purifies More reactive impurities (Fe, Zn) dissolve at the anode but stay in solution; less reactive ones (Ag, Au) do not dissolve and fall as "anode mud". Only copper deposits on the cathode, at better than $99.9\%$ purity.
Interactive 1 — Electrolysis Calculator

Choose an electrolyte and set the current and the time. The cell shows which product forms at each electrode, and the readouts use Faraday's laws, $Q=It$ and $\text{mol}\ e^{-}=Q/F$ with $F=96\,487\ \mathrm{C\,mol^{-1}}$, to give the amount of each product. Press Play to run the electrolysis: cations drift to the cathode ($-$), anions to the anode ($+$), and the deposit or gas grows.

2.030
charge $Q=It$—
electrons $Q/F$—
at the cathode ($-$)—
at the anode ($+$)—

Gas volumes at $298\ \mathrm{K}$ and $1\ \mathrm{bar}$ ($24.8\ \mathrm{L\,mol^{-1}}$); molar masses Na 23.0, Cu 63.5, Ag 108 $\mathrm{g\,mol^{-1}}$.

Batteries and Fuel Cells

A practical battery must be light and compact, give a steady voltage for most of its life and be safe to carry. Any galvanic cell can supply a current, but only a few chemistries meet these demands. One or more cells joined in series make a battery.

📘 Definition 2.9 — Primary and secondary batteries; fuel cells
  • In a primary battery the cell reaction occurs only once; when the reactants are used up the battery is dead and cannot be recharged (dry cell, mercury cell).
  • A secondary battery can be recharged: after use, a current is passed through it in the opposite direction, which reverses the cell reaction by electrolysis. It can go through many charge–discharge cycles (lead storage battery, nickel–cadmium cell).
  • A fuel cell is a galvanic cell that converts the energy of combustion of a fuel (hydrogen, methane, methanol) directly into electrical energy, with the reactants supplied continuously from outside.

Dry cell (Leclanché cell), the ordinary torch cell, primary, about $1.5\ \mathrm{V}$. A zinc container is the anode. A graphite rod in the middle, surrounded by powdered manganese dioxide and carbon, is the cathode. The space between is filled with a moist paste of $\mathrm{NH_{4}Cl}$ and $\mathrm{ZnCl_{2}}$.

$$\text{anode:}\quad \mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$$ $$\text{cathode:}\quad \mathrm{MnO_{2}(s)+NH_{4}^{+}(aq)+e^{-}\rightarrow MnO(OH)(s)+NH_{3}(g)}$$

Manganese is reduced from $+4$ to $+3$. The ammonia formed combines with $\mathrm{Zn^{2+}}$ to give the complex ion $\mathrm{[Zn(NH_{3})_{4}]^{2+}}$. Because ions are consumed and products build up around the electrodes, the voltage drifts down during use.

Mercury cell, primary, $1.35\ \mathrm{V}$, used in hearing aids, watches and older cameras. The anode is zinc–mercury amalgam, the cathode a paste of $\mathrm{HgO}$ and carbon, the electrolyte a paste of $\mathrm{KOH}$ and $\mathrm{ZnO}$.

$$\text{anode:}\quad \mathrm{Zn(Hg)+2OH^{-}(aq)\rightarrow ZnO(s)+H_{2}O(l)+2e^{-}}$$ $$\text{cathode:}\quad \mathrm{HgO(s)+H_{2}O(l)+2e^{-}\rightarrow Hg(l)+2OH^{-}(aq)}$$ $$\text{overall:}\quad \mathrm{Zn(Hg)+HgO(s)\rightarrow ZnO(s)+Hg(l)}$$

No ion appears in the overall reaction, so $Q$ does not change as the cell is used and, by the Nernst equation, the voltage stays constant throughout its life. (Mercury cells are now being phased out because mercury is toxic.)

Lead storage battery, secondary, about $2\ \mathrm{V}$ per cell; six cells in series give the $12\ \mathrm{V}$ car battery. The anode is lead, the cathode a lead grid packed with lead dioxide, and the electrolyte is about $38\%$ sulphuric acid.

$$\text{anode:}\quad \mathrm{Pb(s)+SO_{4}^{2-}(aq)\rightarrow PbSO_{4}(s)+2e^{-}}$$ $$\text{cathode:}\quad \mathrm{PbO_{2}(s)+SO_{4}^{2-}(aq)+4H^{+}(aq)+2e^{-}}$$ $$\mathrm{\rightarrow PbSO_{4}(s)+2H_{2}O(l)}$$ $$\text{overall:}\quad \mathrm{Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)}$$ $$\mathrm{\rightarrow 2PbSO_{4}(s)+2H_{2}O(l)}$$

On discharge, sulphuric acid is used up and water formed, so the density of the acid falls; a mechanic can test the state of charge with a hydrometer. On charging, an external supply drives the reaction backwards:

$$\mathrm{2PbSO_{4}(s)+2H_{2}O(l)}$$ $$\mathrm{\rightarrow Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)}$$

Nickel–cadmium cell, secondary, about $1.2\ \mathrm{V}$, longer-lived than the lead battery but more expensive. Overall discharge reaction: $\mathrm{Cd(s)+2Ni(OH)_{3}(s)}$ $\mathrm{\rightarrow CdO(s)+2Ni(OH)_{2}(s)+H_{2}O(l)}$.

The hydrogen–oxygen fuel cell. Burning hydrogen in a power station turns at best about $40\%$ of its energy into electricity. A fuel cell avoids the heat engine altogether. Hydrogen and oxygen are bubbled through porous carbon electrodes (containing finely divided Pt or Pd as catalyst) into concentrated aqueous $\mathrm{KOH}$:

$$\text{anode:}\quad \mathrm{2H_{2}(g)+4OH^{-}(aq)\rightarrow 4H_{2}O(l)+4e^{-}}$$ $$\text{cathode:}\quad \mathrm{O_{2}(g)+2H_{2}O(l)+4e^{-}\rightarrow 4OH^{-}(aq)}$$ $$\text{overall:}\quad \mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)}$$

It runs as long as fuel is supplied, works at about $70\%$ efficiency, and its only product is water, which the Apollo astronauts drank. Fuel cells now power buses, trucks and back-up generators.

✏️ Example 5 — How much acid does a car battery use?
A lead storage battery delivers a steady current of $1.00\ \mathrm{A}$ for $1.00\ \mathrm{h}$. What masses of (a) $\mathrm{H_{2}SO_{4}}$ and (b) lead are consumed? ($\mathrm{H_{2}SO_{4}}=98.0$, $\mathrm{Pb}=207\ \mathrm{g\,mol^{-1}}$.)
Electrons $Q=1.00\times3600=3600\ \mathrm{C}$, i.e. $\dfrac{3600}{96\,487}=0.03731\ \mathrm{mol}\ e^{-}$.
Stoichiometry The overall reaction transfers $2e^{-}$ and uses $2\ \mathrm{H_{2}SO_{4}}$ and $1\ \mathrm{Pb}$. So $1\ \mathrm{mol}\ \mathrm{H_{2}SO_{4}}$ and $\tfrac12\ \mathrm{mol}$ Pb per mole of electrons.
(a) $0.03731\times98.0=3.66\ \mathrm{g}$ of $\mathrm{H_{2}SO_{4}}$.
(b) $\dfrac{0.03731}{2}\times207=3.86\ \mathrm{g}$ of lead (an equal amount of $\mathrm{PbO_{2}}$ is reduced at the cathode).
✏️ Example 6 — Voltage and efficiency of a hydrogen fuel cell
For $\mathrm{H_{2}(g)+\tfrac12O_{2}(g)\rightarrow H_{2}O(l)}$ at $298\ \mathrm{K}$, $\Delta_{r}G^{\circ}=-237.2\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{r}H^{\circ}=-285.8\ \mathrm{kJ\,mol^{-1}}$. Find (a) $E^{\circ}_{cell}$ of the hydrogen–oxygen fuel cell and (b) its maximum (thermodynamic) efficiency, $\Delta_{r}G^{\circ}/\Delta_{r}H^{\circ}$.
n Each $\mathrm{H_{2}}$ gives $2e^{-}$ ($\mathrm{H}$: $0\rightarrow+1$), so $n=2$ for the equation as written.
(a) $$E^{\circ}_{cell}=-\frac{\Delta_{r}G^{\circ}}{nF}=\frac{237\,200}{2\times96\,487}=1.23\ \mathrm{V}.$$ This equals $E^{\circ}_{\mathrm{O_{2}/H_{2}O}}-E^{\circ}_{\mathrm{H^{+}/H_{2}}}=1.23-0.00$, as it should.
(b) $\dfrac{237.2}{285.8}=0.830$: at most $83\%$ of the reaction's enthalpy can become electrical work.
Meaning Real fuel cells reach about $60$–$70\%$, because of overpotentials and internal resistance, but that is still far above a heat engine. Notice the same $1.23\ \mathrm{V}$ is the minimum voltage for splitting water by electrolysis.

Corrosion and the Hydrogen Economy

Silver tarnishes black, copper roofs turn green and iron rusts. Rusting alone destroys a few per cent of the world's economic output every year, through failed bridges, leaking pipelines and scrapped cars. It is not a simple reaction between iron and oxygen: it is electrochemistry happening on a wet metal surface.

📘 Definition 2.10 — Corrosion
Corrosion is the slow destruction of a metal by its environment, in which the metal is oxidised to its oxide, hydroxide, carbonate or sulphide. For iron in moist air it is called rusting, and the product, rust, is hydrated iron(III) oxide, $\mathrm{Fe_{2}O_{3}\cdot xH_{2}O}$. Corrosion is an electrochemical process: one region of the metal surface acts as an anode and another as a cathode of a tiny galvanic cell, with a film of water as the electrolyte.

The rusting cell. At a spot where the iron is strained or poorly supplied with oxygen (for example, under the centre of a water drop), iron is oxidised and that spot becomes the anode. The electrons travel through the metal to a spot where oxygen is plentiful (the edge of the drop), which acts as the cathode. $\mathrm{H^{+}}$ comes from carbonic acid ($\mathrm{CO_{2}}$ dissolved in the water) or from acid rain.

$$\text{anode:}\quad \mathrm{2Fe(s)\rightarrow 2Fe^{2+}(aq)+4e^{-}},\qquad E^{\circ}_{\mathrm{Fe^{2+}/Fe}}=-0.44\ \mathrm{V}$$ $$\text{cathode:}\quad \mathrm{O_{2}(g)+4H^{+}(aq)+4e^{-}\rightarrow 2H_{2}O(l)},\qquad E^{\circ}=+1.23\ \mathrm{V}$$ $$\text{overall:}\quad \mathrm{2Fe(s)+O_{2}(g)+4H^{+}(aq)\rightarrow 2Fe^{2+}(aq)+2H_{2}O(l)},\qquad E^{\circ}_{cell}=1.67\ \mathrm{V}$$

The $\mathrm{Fe^{2+}}$ ions spread through the water and are oxidised further by air:

$$\mathrm{4Fe^{2+}(aq)+O_{2}(g)+(4+2x)H_{2}O(l)}$$ $$\mathrm{\rightarrow 2Fe_{2}O_{3}\cdot xH_{2}O(s)+8H^{+}(aq)}$$

So rust is deposited away from the pit where the iron is actually eaten. Salt water and acid rain speed rusting greatly, because they make the water a much better electrolyte.

Preventing corrosion.

The hydrogen economy. Fossil fuels are running down and their combustion adds carbon dioxide to the atmosphere. Hydrogen burns, or reacts in a fuel cell, to give only water. In a "hydrogen economy", electricity from sun, wind or water would split water by electrolysis; the hydrogen would be stored and transported and then turned back into electricity in fuel cells, where and when it is needed. The chemistry is the electrochemistry of this chapter; the challenges are engineering ones: making electrolysers cheaper and more efficient, and storing a very light gas safely. India's National Green Hydrogen Mission and similar programmes worldwide aim to make this practical.

⚠️ Common mistakes in electrolysis and batteries

1. Minutes and hours. $Q=It$ needs $t$ in seconds: $30\ \mathrm{min}=1800\ \mathrm{s}$, $1\ \mathrm{h}=3600\ \mathrm{s}$.

2. Electrons per particle. Divide moles of electrons by $n$ from the electrode half-reaction: $2$ for $\mathrm{Cu^{2+}}$, $3$ for $\mathrm{Al^{3+}}$, $4$ for each $\mathrm{O_{2}}$, $2$ for each $\mathrm{H_{2}}$ or $\mathrm{Cl_{2}}$.

3. Signs of electrodes. In an electrolytic cell the cathode is negative (joined to the negative terminal); in a galvanic cell it is positive. Reduction is at the cathode in both.

4. Products in water. Reactive metals (Na, K, Mg, Al) are never deposited from aqueous solution; hydrogen forms instead. Sulphate and nitrate are not oxidised; oxygen forms instead.

5. Primary vs secondary. "Rechargeable" means the cell reaction can be reversed by electrolysis. A dry cell's reaction cannot be reversed usefully.

6. Sacrificial metal. The protecting metal must have a lower (more negative) $E^{\circ}$ than iron.

Interactive 2 — Inside Batteries, a Fuel Cell and a Rusting Drop

Choose a device. The diagram labels the anode and cathode and the readouts give their half-reactions, the overall reaction and the voltage. Press Play to watch electrons flow through the external circuit and ions move through the electrolyte. Compare discharging and charging a lead battery (the electron flow reverses), and see how a water drop on iron becomes a tiny galvanic cell.

anode (oxidation)—
cathode (reduction)—
overall—
voltage · type—

Exercises

I
Faraday's laws ($F=96\,487\ \mathrm{C\,mol^{-1}}$; molar masses in $\mathrm{g\,mol^{-1}}$: Cu 63.5, Zn 65.4, Ni 58.7, Ag 108; gas molar volume $24.8\ \mathrm{L\,mol^{-1}}$ at $298\ \mathrm{K}$, $1\ \mathrm{bar}$).
  • 1 State Faraday's first and second laws of electrolysis.
  • 2 A current of $0.250\ \mathrm{A}$ flows for $2.00\ \mathrm{h}$. Find the charge passed and the amount of electrons.
  • 3 What mass of zinc is deposited from zinc sulphate solution by $1.50\ \mathrm{A}$ in $45.0\ \mathrm{min}$?
  • 4 How long will a current of $3.00\ \mathrm{A}$ take to deposit $2.00\ \mathrm{g}$ of copper?
  • 5 What current is needed to deposit $0.500\ \mathrm{g}$ of nickel from $\mathrm{Ni^{2+}(aq)}$ in $20.0\ \mathrm{min}$?
  • 6 Acidified water is electrolysed with $5.00\ \mathrm{A}$ for $1.00\ \mathrm{h}$. Find the volumes of $\mathrm{H_{2}}$ and $\mathrm{O_{2}}$ produced at $298\ \mathrm{K}$ and $1\ \mathrm{bar}$.
  • 7 Cells of $\mathrm{AgNO_{3}(aq)}$ and $\mathrm{CuSO_{4}(aq)}$ are connected in series. If $0.540\ \mathrm{g}$ of silver is deposited, what mass of copper is deposited?
  • 8 How many coulombs are needed to reduce $1\ \mathrm{mol}$ of $\mathrm{NO_{3}^{-}}$ to $\mathrm{NO}$?
  • 9 Find the electrochemical equivalent $Z$ of copper (deposited from $\mathrm{Cu^{2+}}$).
Check: Faraday calculator
$Q$ (C)—
mol $e^{-}$—
mass (g)—
1 First law: the mass of substance deposited or liberated at an electrode is proportional to the charge passed, $m=ZQ$. Second law: when the same charge passes through different electrolytes, the masses liberated are proportional to their equivalent masses ($M/n$).
2 $Q=0.250\times7200=1800\ \mathrm{C}$; $\dfrac{1800}{96\,487}=0.0187\ \mathrm{mol}\ e^{-}$.
3 $Q=1.50\times2700=4050\ \mathrm{C}$; $\dfrac{4050}{96\,487}=0.04197\ \mathrm{mol}\ e^{-}$; Zn $=\dfrac{0.04197}{2}\times65.4=1.37\ \mathrm{g}$.
4 $\dfrac{2.00}{63.5}=0.03150\ \mathrm{mol}$ Cu needs $0.06299\ \mathrm{mol}\ e^{-}$, $Q=6078\ \mathrm{C}$; $t=\dfrac{6078}{3.00}=2026\ \mathrm{s}=33.8\ \mathrm{min}$.
5 $Q=\dfrac{0.500}{58.7}\times2\times96\,487=1644\ \mathrm{C}$; $I=\dfrac{1644}{1200}=1.37\ \mathrm{A}$.
6 $Q=18\,000\ \mathrm{C}$, $0.1866\ \mathrm{mol}\ e^{-}$. $\mathrm{H_{2}}$ ($2e^{-}$ each): $0.09328\ \mathrm{mol}$, $2.31\ \mathrm{L}$. $\mathrm{O_{2}}$ ($4e^{-}$ each): $0.04664\ \mathrm{mol}$, $1.16\ \mathrm{L}$. The 2 : 1 ratio is the formula of water.
7 $\dfrac{0.540}{108}=5.00\times10^{-3}\ \mathrm{mol}\ e^{-}$; Cu $=2.50\times10^{-3}\times63.5=0.159\ \mathrm{g}$.
8 $\mathrm{NO_{3}^{-}(aq)+4H^{+}(aq)+3e^{-}}$ $\mathrm{\rightarrow NO(g)+2H_{2}O(l)}$ (N: $+5\rightarrow+2$): $3F=2.89\times10^{5}\ \mathrm{C}$.
9 $Z=\dfrac{M}{nF}=\dfrac{63.5}{2\times96\,487}=3.29\times10^{-4}\ \mathrm{g\,C^{-1}}$.
II
Products of electrolysis (inert electrodes unless stated).
  • 1 Name the products at each electrode when molten $\mathrm{NaCl}$ is electrolysed, and write the electrode equations.
  • 2 Name the products and write the overall equation for the electrolysis of concentrated aqueous $\mathrm{NaCl}$. Why is sodium not formed?
  • 3 Aqueous $\mathrm{CuSO_{4}}$ is electrolysed with platinum electrodes. What happens to the colour and the pH of the solution?
  • 4 How does the result of part 3 change if copper electrodes are used?
  • 5 Give the products of electrolysis of (a) dilute and (b) concentrated sulphuric acid.
  • 6 Predict the products when aqueous $\mathrm{AgNO_{3}}$ is electrolysed (a) with platinum electrodes, (b) with silver electrodes.
  • 7 On $E^{\circ}$ values alone, oxygen should be released before chlorine at the anode when brine is electrolysed. Explain why chlorine is released instead.
1 Cathode: $\mathrm{Na^{+}(l)+e^{-}\rightarrow Na(l)}$. Anode: $\mathrm{2Cl^{-}(l)\rightarrow Cl_{2}(g)+2e^{-}}$.
2 $\mathrm{H_{2}}$ at the cathode, $\mathrm{Cl_{2}}$ at the anode, $\mathrm{NaOH}$ in solution: $\mathrm{2NaCl(aq)+2H_{2}O(l)\rightarrow 2NaOH(aq)+H_{2}(g)+Cl_{2}(g)}$. Water ($E^{\circ}=-0.83\ \mathrm{V}$) is far easier to reduce than $\mathrm{Na^{+}}$ ($-2.71\ \mathrm{V}$).
3 Copper deposits and oxygen is evolved; $\mathrm{Cu^{2+}}$ is removed, so the blue colour fades, and $\mathrm{H^{+}}$ is produced at the anode, so the pH falls (sulphuric acid forms).
4 The copper anode dissolves ($\mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}}$) instead of oxygen forming; copper deposits on the cathode at the same rate, so colour and pH stay the same. The anode loses as much mass as the cathode gains.
5 (a) $\mathrm{H_{2}}$ (cathode) and $\mathrm{O_{2}}$ (anode); in effect water is split. (b) $\mathrm{H_{2}}$ (cathode) and peroxodisulphate, $\mathrm{2SO_{4}^{2-}(aq)\rightarrow S_{2}O_{8}^{2-}(aq)+2e^{-}}$ (anode).
6 (a) Silver at the cathode ($E^{\circ}=+0.80\ \mathrm{V}$ beats water), oxygen at the anode (nitrate is not oxidised); nitric acid forms. (b) Silver deposits at the cathode and the silver anode dissolves; the solution is unchanged (as in silver plating).
7 The formation of $\mathrm{O_{2}}$ at the electrode surface is kinetically slow and needs a considerable extra voltage (overpotential). Chloride oxidation needs much less, so in practice the lower overall voltage releases chlorine.
III
Batteries and fuel cells ($F=96\,487\ \mathrm{C\,mol^{-1}}$).
  • 1 Distinguish primary and secondary batteries, with one example of each.
  • 2 Write the electrode reactions of the mercury cell. Why does its voltage stay constant during its life?
  • 3 Write the electrode reactions of a lead storage battery during discharge. What happens to the density of the acid, and what reaction occurs on charging?
  • 4 A lead storage battery delivers $2.50\ \mathrm{A}$ for $2.00\ \mathrm{h}$. What masses of $\mathrm{H_{2}SO_{4}}$ ($98.0$) and lead ($207\ \mathrm{g\,mol^{-1}}$) are consumed?
  • 5 Write the electrode reactions in a hydrogen–oxygen fuel cell with a KOH electrolyte and give two advantages of fuel cells.
  • 6 A hydrogen fuel cell delivers $10.0\ \mathrm{A}$ for $1.00\ \mathrm{h}$. What mass of hydrogen ($\mathrm{H_{2}}=2.0\ \mathrm{g\,mol^{-1}}$) is consumed?
  • 7 In a methanol fuel cell, $\mathrm{CH_{3}OH(l)+\tfrac32O_{2}(g)}$ $\mathrm{\rightarrow CO_{2}(g)+2H_{2}O(l)}$ has $\Delta_{r}G^{\circ}=-702\ \mathrm{kJ\,mol^{-1}}$ and $n=6$. Find $E^{\circ}_{cell}$.
1 Primary: the reaction occurs once and the cell cannot be recharged (dry cell, mercury cell). Secondary: the reaction can be reversed by passing current the other way, so the cell is rechargeable (lead storage battery, nickel–cadmium cell).
2 Anode: $\mathrm{Zn(Hg)+2OH^{-}(aq)\rightarrow ZnO(s)+H_{2}O(l)+2e^{-}}$. Cathode: $\mathrm{HgO(s)+H_{2}O(l)+2e^{-}}$ $\mathrm{\rightarrow Hg(l)+2OH^{-}(aq)}$. The overall reaction $\mathrm{Zn(Hg)+HgO(s)\rightarrow ZnO(s)+Hg(l)}$ contains no dissolved species, so $Q$ (and hence $E$) does not change as it runs.
3 Anode: $\mathrm{Pb(s)+SO_{4}^{2-}(aq)\rightarrow PbSO_{4}(s)+2e^{-}}$. Cathode: $\mathrm{PbO_{2}(s)+SO_{4}^{2-}(aq)+4H^{+}(aq)+2e^{-}}$ $\mathrm{\rightarrow PbSO_{4}(s)+2H_{2}O(l)}$. $\mathrm{H_{2}SO_{4}}$ is consumed and water formed, so the density falls. Charging: $\mathrm{2PbSO_{4}(s)+2H_{2}O(l)}$ $\mathrm{\rightarrow Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)}$.
4 $Q=2.50\times7200=18\,000\ \mathrm{C}$, $0.1866\ \mathrm{mol}\ e^{-}$. $\mathrm{H_{2}SO_{4}}$: $0.1866\times98.0=18.3\ \mathrm{g}$. Pb: $0.09328\times207=19.3\ \mathrm{g}$.
5 Anode: $\mathrm{2H_{2}(g)+4OH^{-}(aq)\rightarrow 4H_{2}O(l)+4e^{-}}$. Cathode: $\mathrm{O_{2}(g)+2H_{2}O(l)+4e^{-}\rightarrow 4OH^{-}(aq)}$. Advantages: high efficiency (about 70%, against about 40% for thermal power), no polluting products (only water), continuous operation as long as fuel is supplied, quiet with no moving parts.
6 $Q=36\,000\ \mathrm{C}$, $0.3731\ \mathrm{mol}\ e^{-}$; $\mathrm{H_{2}}=\dfrac{0.3731}{2}=0.1866\ \mathrm{mol}=0.373\ \mathrm{g}$.
7 $E^{\circ}_{cell}=\dfrac{702\,000}{6\times96\,487}=1.21\ \mathrm{V}$.
IV
Corrosion and its prevention.
  • 1 Write the anode, cathode and overall reactions for the rusting of iron, and find $E^{\circ}_{cell}$.
  • 2 Why does iron rust faster in seawater, or where there is acid rain, than in pure water?
  • 3 Explain why a scratched zinc coating still protects iron, while a scratched tin coating speeds up rusting.
  • 4 Engineering. A magnesium block protecting a steel ship hull carries a steady protective current of $0.50\ \mathrm{A}$. Estimate the mass of magnesium ($24.3\ \mathrm{g\,mol^{-1}}$) consumed in one year ($365$ days).
  • 5 What is meant by the "hydrogen economy"? Give one advantage and one practical difficulty.
1 Anode: $\mathrm{2Fe(s)\rightarrow 2Fe^{2+}(aq)+4e^{-}}$. Cathode: $\mathrm{O_{2}(g)+4H^{+}(aq)+4e^{-}\rightarrow 2H_{2}O(l)}$. Overall: $\mathrm{2Fe(s)+O_{2}(g)+4H^{+}(aq)}$ $\mathrm{\rightarrow 2Fe^{2+}(aq)+2H_{2}O(l)}$; $E^{\circ}_{cell}=1.23-(-0.44)=1.67\ \mathrm{V}$. $\mathrm{Fe^{2+}}$ is then oxidised by air to $\mathrm{Fe_{2}O_{3}\cdot xH_{2}O}$.
2 Dissolved salts make the water film a much better electrolyte, so the local cells carry larger currents. Acid rain supplies $\mathrm{H^{+}}$ for the cathode reaction, raising its potential (Nernst) and its rate.
3 Zinc ($-0.76\ \mathrm{V}$) is a stronger reducing agent than iron ($-0.44\ \mathrm{V}$): at a scratch, zinc is the anode and corrodes while iron is the cathode and is protected. Tin ($-0.14\ \mathrm{V}$) is weaker: at a scratch, iron becomes the anode and the large tin surface acts as cathode, so the iron corrodes faster than if it were bare.
4 $t=365\times24\times3600=3.154\times10^{7}\ \mathrm{s}$; $Q=1.577\times10^{7}\ \mathrm{C}$; $163.4\ \mathrm{mol}\ e^{-}$; Mg ($2e^{-}$ each) $=81.7\ \mathrm{mol}=1.99\times10^{3}\ \mathrm{g}\approx2.0\ \mathrm{kg}$ per year.
5 An energy system in which hydrogen, made by electrolysing water with renewable electricity, is stored and transported and then used as fuel (in fuel cells or engines). Advantage: its only product is water, with no $\mathrm{CO_{2}}$. Difficulty: hydrogen is costly to make efficiently, and it is very light and flammable, so storing and transporting it is hard.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks (a) Write the reactions at the anode and the cathode of a lead storage battery while it is in use. (b) What happens when the battery is recharged? (c) Why is it called a secondary battery?

(a) Anode: $\mathrm{Pb(s)+SO_{4}^{2-}(aq)\rightarrow PbSO_{4}(s)+2e^{-}}$. Cathode: $\mathrm{PbO_{2}(s)+SO_{4}^{2-}(aq)+4H^{+}(aq)+2e^{-}\rightarrow PbSO_{4}(s)+2H_{2}O(l)}$.
(b) An external current reverses the reaction: $\mathrm{2PbSO_{4}(s)+2H_{2}O(l)\rightarrow Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)}$, regenerating the electrodes and the acid.
(c) Because it can be recharged and used again through many cycles.

E2. JEE Main Numerical value. A current of $9.65\ \mathrm{A}$ is passed for $1000\ \mathrm{s}$ through molten $\mathrm{Al_{2}O_{3}}$ (dissolved in cryolite). Taking $F=96\,500\ \mathrm{C\,mol^{-1}}$ and $\mathrm{Al}=27\ \mathrm{g\,mol^{-1}}$, the mass of aluminium deposited, in mg, is ______.

Charge $Q=9.65\times1000=9650\ \mathrm{C}=0.100\ F$.
Aluminium $\mathrm{Al^{3+}(l)+3e^{-}\rightarrow Al(l)}$: $\dfrac{0.100}{3}\times27=0.900\ \mathrm{g}$.
Answer $\mathbf{900}$ mg.

E3. NEET During electrolysis of aqueous copper(II) sulphate with platinum electrodes, the products at the cathode and anode respectively are (A) Cu and $\mathrm{O_{2}}$  (B) Cu and $\mathrm{SO_{2}}$  (C) $\mathrm{H_{2}}$ and $\mathrm{O_{2}}$  (D) $\mathrm{H_{2}}$ and $\mathrm{SO_{3}}$

Cathode $\mathrm{Cu^{2+}}$ ($+0.34\ \mathrm{V}$) is reduced in preference to water ($-0.83\ \mathrm{V}$).
Anode Sulphate is not oxidised; water is: $\mathrm{2H_{2}O(l)\rightarrow O_{2}(g)+4H^{+}(aq)+4e^{-}}$.
Answer (A).

E4. JEE Advanced Three electrolytic cells containing aqueous $\mathrm{CuSO_{4}}$, aqueous $\mathrm{AgNO_{3}}$ and dilute $\mathrm{H_{2}SO_{4}}$ (all with Pt electrodes) are connected in series and a steady current of $2.00\ \mathrm{A}$ is passed until $0.635\ \mathrm{g}$ of copper has been deposited. Find (a) the time taken, (b) the mass of silver deposited, and (c) the total volume of gas collected from the sulphuric acid cell at $298\ \mathrm{K}$ and $1\ \mathrm{bar}$ ($24.8\ \mathrm{L\,mol^{-1}}$). ($\mathrm{Cu}=63.5$, $\mathrm{Ag}=108\ \mathrm{g\,mol^{-1}}$, $F=96\,487\ \mathrm{C\,mol^{-1}}$.)

Electrons $0.635\ \mathrm{g}$ Cu $=0.0100\ \mathrm{mol}$, needing $0.0200\ \mathrm{mol}\ e^{-}$; the same charge flows through every cell in series.
(a) $Q=0.0200\times96\,487=1930\ \mathrm{C}$; $t=\dfrac{1930}{2.00}=965\ \mathrm{s}$ (about $16\ \mathrm{min}$).
(b) Ag: $0.0200\ \mathrm{mol}\times108=2.16\ \mathrm{g}$.
(c) $\mathrm{H_{2}}$: $0.0100\ \mathrm{mol}$; $\mathrm{O_{2}}$: $0.00500\ \mathrm{mol}$; total $0.0150\ \mathrm{mol}\times24.8=0.372\ \mathrm{L}=372\ \mathrm{mL}$.

E5. JEE Main NEET Iron can be given sacrificial protection by coating or connecting it with (A) tin  (B) copper  (C) zinc  (D) lead. ($E^{\circ}$/V: $\mathrm{Fe^{2+}/Fe}$ $-0.44$, $\mathrm{Sn^{2+}/Sn}$ $-0.14$, $\mathrm{Cu^{2+}/Cu}$ $+0.34$, $\mathrm{Zn^{2+}/Zn}$ $-0.76$, $\mathrm{Pb^{2+}/Pb}$ $-0.13$.)

Idea The sacrificial metal must be oxidised in preference to iron, so its $E^{\circ}$ must be lower than $-0.44\ \mathrm{V}$.
Answer (C) zinc. Tin, copper and lead all lie above iron and would speed up its corrosion at a break in the coating.

E6. Board · Case study Read the passage and answer. A city bus runs on a hydrogen–oxygen fuel cell stack with an alkaline electrolyte. Hydrogen from a pressurised tank and oxygen from the air reach porous electrodes; the only exhaust is water vapour. For $\mathrm{H_{2}(g)+\tfrac12O_{2}(g)\rightarrow H_{2}O(l)}$, $\Delta_{r}G^{\circ}=-237.2\ \mathrm{kJ\,mol^{-1}}$. (a) Write the anode and cathode reactions. (2 marks) (b) Calculate $E^{\circ}$ of one cell ($F=96\,487\ \mathrm{C\,mol^{-1}}$). (1 mark) (c) Give one reason why the fuel cell is preferred to a diesel engine for city buses. (1 mark)

(a) Anode: $\mathrm{2H_{2}(g)+4OH^{-}(aq)\rightarrow 4H_{2}O(l)+4e^{-}}$. Cathode: $\mathrm{O_{2}(g)+2H_{2}O(l)+4e^{-}\rightarrow 4OH^{-}(aq)}$.
(b) $n=2$: $E^{\circ}=\dfrac{237\,200}{2\times96\,487}=1.23\ \mathrm{V}$.
(c) No polluting exhaust in the city (only water), or: much higher efficiency of converting fuel energy to work; also quiet.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How much electricity does it take to make one kilogram of "green" hydrogen by electrolysing water, and what would a national hydrogen target mean in terms of power generation?

Why it matters. Steel works, fertiliser plants and refineries use huge amounts of hydrogen, today mostly made from natural gas with large $\mathrm{CO_{2}}$ emissions. India, the European Union, Australia and others plan to make it instead by electrolysis powered by solar and wind. India's National Green Hydrogen Mission, for example, aims at about $5$ million tonnes a year by 2030.

What to do.

  1. How many moles of $\mathrm{H_{2}}$ ($2.016\ \mathrm{g\,mol^{-1}}$) are in $1.00\ \mathrm{kg}$? How many coulombs are needed to make them ($\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)}$, $F=96\,487\ \mathrm{C\,mol^{-1}}$)?
  2. Electrical energy $=Q\times V$. Find the energy per kilogram in $\mathrm{kWh}$ ($1\ \mathrm{kWh}=3.6\times10^{6}\ \mathrm{J}$) at the thermodynamic minimum, $1.23\ \mathrm{V}$, and at a typical operating voltage of $1.90\ \mathrm{V}$. Express the operating efficiency as $1.23/1.90$.
  3. Using the $1.90\ \mathrm{V}$ figure, how much electricity (in TWh, $1\ \mathrm{TWh}=10^{9}\ \mathrm{kWh}$) would $5$ million tonnes of hydrogen a year need?
  4. A $1\ \mathrm{GW}$ solar park produces roughly $2\ \mathrm{TWh}$ a year. About how many such parks would the target require?

Tools: Calculator or spreadsheet; Interactive 1 (dilute $\mathrm{H_{2}SO_{4}}$) to check the moles of $\mathrm{H_{2}}$ per coulomb.

Hand in: Your calculations, a one-paragraph opinion on whether the target is realistic, and one research direction (better catalysts to cut overpotential, cheaper membranes, seawater electrolysis…) that would help most.

Going further. Find out why platinum-group and iridium catalysts are used in some electrolysers and what is being done to replace them with cheaper metals such as nickel and iron.

Instructor note: $1.00\ \mathrm{kg}$ is $496.0\ \mathrm{mol}$ $\mathrm{H_{2}}$, needing $9.572\times10^{7}\ \mathrm{C}$. At $1.23\ \mathrm{V}$: $1.177\times10^{8}\ \mathrm{J}=32.7\ \mathrm{kWh\,kg^{-1}}$; at $1.90\ \mathrm{V}$: $50.5\ \mathrm{kWh\,kg^{-1}}$ (efficiency $65\%$). $5\times10^{9}\ \mathrm{kg}\times50.5\ \mathrm{kWh\,kg^{-1}}\approx253\ \mathrm{TWh}$ a year, i.e. roughly $125$–$130$ one-gigawatt solar parks, before counting water purification and compression.

🧪 Self-check quiz

0/5 correct

Q1. One faraday is

Q2. The mass of copper deposited from $\mathrm{Cu^{2+}(aq)}$ by $0.10\ F$ of charge is about ($\mathrm{Cu}=63.5$)

Q3. When concentrated aqueous $\mathrm{NaCl}$ is electrolysed with inert electrodes, the gas formed at the cathode is

Q4. Which of these is a secondary (rechargeable) battery?

Q5. In the rusting of iron, the reaction at the cathodic region is

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