Electrochemistry • Topic 3 of 3

Electrolysis, Batteries & Corrosion

Electrolysis. An electrolytic cell is the reverse of a galvanic cell: electrical energy from an external source drives a non-spontaneous redox reaction. The electrode connected to the positive terminal is the anode (oxidation), and the one connected to the negative terminal is the cathode (reduction). For example, in molten $\text{NaCl}$, $\text{Na}^+$ is reduced to sodium at the cathode and $\text{Cl}^-$ is oxidised to chlorine at the anode.

Faraday's laws of electrolysis. Faraday quantified electrolysis in two laws:

  • First law: the mass of a substance deposited or liberated at an electrode is proportional to the quantity of charge passed: $m \propto Q$, i.e. $m = ZQ = ZIt$, where $Z$ is the electrochemical equivalent, $Q = It$ the charge ($I$ in amperes, $t$ in seconds).
  • Second law: when the same charge passes through different electrolytes, the masses deposited are proportional to their equivalent masses. One mole of electrons ($1\,F = 96500\,\text{C}$) deposits one gram-equivalent of any substance.

So $m = \dfrac{M \times I \times t}{n \times F}$, where $M$ is the molar mass and $n$ the number of electrons per ion.

Products of electrolysis depend on the electrode material and on which species is easier to discharge (its electrode potential and overpotential). In aqueous $\text{NaCl}$ (brine), $\text{H}_2$ is liberated at the cathode and $\text{Cl}_2$ at the anode (overpotential favours chlorine over oxygen), leaving $\text{NaOH}$ — the basis of the chlor-alkali industry.

Batteries. A battery is a galvanic cell (or several in series) used as a practical source of electrical energy.

  • Primary cells cannot be recharged — the reaction is irreversible. The dry cell (Leclanché cell) has a zinc anode, a graphite cathode surrounded by $\text{MnO}_2$ and carbon, with a moist $\text{NH}_4\text{Cl}/\text{ZnCl}_2$ paste; it gives about $1.5\,\text{V}$. The mercury cell ($\text{Zn}$/$\text{HgO}$) gives a steady $1.35\,\text{V}$ and is used in hearing aids and watches.
  • Secondary cells can be recharged. The lead storage battery (car battery) has a $\text{Pb}$ anode and a $\text{PbO}_2$ cathode in $\sim38\%$ $\text{H}_2\text{SO}_4$; on discharge both electrodes form $\text{PbSO}_4$, and charging reverses this. The nickel–cadmium (Ni–Cd) cell uses $\text{Cd}$ and $\text{NiO(OH)}$, has a long life and is rechargeable.

Fuel cells. A fuel cell converts the energy of combustion of a fuel directly into electricity, with no thermal step, so it is highly efficient (~70%) and clean. In the hydrogen–oxygen fuel cell, $\text{H}_2$ is oxidised at the anode and $\text{O}_2$ reduced at the cathode in $\text{KOH}$; the only product is water. These powered the Apollo space programme.

Corrosion. Corrosion is the slow electrochemical eating-away of a metal by its environment — most familiarly the rusting of iron, hydrated iron(III) oxide $\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}$. On a wet iron surface, one spot acts as an anode ($\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$) and another as a cathode where oxygen is reduced ($\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$); the $\text{Fe}^{2+}$ is then oxidised by air to rust. Water and oxygen are both essential.

Prevention works by breaking this mini-cell: barrier methods (painting, greasing, electroplating with tin or chromium), galvanising (a sacrificial zinc coat), and cathodic protection — connecting the iron to a more reactive metal such as magnesium or zinc that corrodes preferentially as a sacrificial anode.

Standard reduction potentials at 298 K (electrochemical series, abridged)
Electrode (reduction half-reaction)E0 / VNature
Li+ + e- → Li-3.05strongest reducing agent
K+ + e- → K-2.93very reactive metal
Zn2+ + 2e- → Zn-0.76anode in Daniell cell
Fe2+ + 2e- → Fe-0.44rusts readily
2H+ + 2e- → H20.00reference (SHE)
Cu2+ + 2e- → Cu+0.34cathode in Daniell cell
Ag+ + e- → Ag+0.80noble metal
F2 + 2e- → 2F-+2.87strongest oxidising agent
1
Worked Example
How much charge is required to deposit $1$ mole of silver from $\text{AgNO}_3$ solution? ($\text{Ag}^+ + e^- \rightarrow \text{Ag}$)
Solution
  1. Step 1: Reduction needs 1 mole of electrons per mole of $\text{Ag}$ ($n = 1$).
  2. Step 2: One mole of electrons carries $1\,F = 96500\,\text{C}$.
  3. Step 3: Charge $Q = n \times F = 1 \times 96500$.
  4. Step 4: $Q = 96500\,\text{C}$.

Answer: $96500\,\text{C}$ (one Faraday) is needed.

2
Worked Example
A current of $5\,\text{A}$ is passed through molten $\text{AlCl}_3$ for $1$ hour. Find the mass of aluminium deposited. ($\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$, $M = 27\,\text{g mol}^{-1}$)
Solution
  1. Step 1: Charge $Q = It = 5 \times 3600 = 18000\,\text{C}$.
  2. Step 2: $m = \dfrac{M \times I \times t}{n \times F}$ with $n = 3$.
  3. Step 3: $m = \dfrac{27 \times 18000}{3 \times 96500} = \dfrac{486000}{289500}$.
  4. Step 4: $m = 1.68\,\text{g}$.

Answer: About $1.68\,\text{g}$ of aluminium is deposited.

3
Worked Example
How long must a current of $2\,\text{A}$ flow to deposit $1.08\,\text{g}$ of silver? ($M_{\text{Ag}} = 108\,\text{g mol}^{-1}$, $n = 1$)
Solution
  1. Step 1: Moles of $\text{Ag} = \dfrac{1.08}{108} = 0.01\,\text{mol}$.
  2. Step 2: Charge needed $Q = nFn_{mol} = 1 \times 96500 \times 0.01 = 965\,\text{C}$.
  3. Step 3: $t = \dfrac{Q}{I} = \dfrac{965}{2}$.
  4. Step 4: $t = 482.5\,\text{s}$ (about $8$ minutes).

Answer: The current must flow for about $482.5\,\text{s}$ ($\approx 8\,\text{min}$).

4
Worked Example
Write the discharge reactions occurring at each electrode of a lead storage battery.
Solution
  1. Step 1: The anode is spongy lead; it is oxidised.
  2. Step 2: Anode: $\text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2e^-$.
  3. Step 3: The cathode is $\text{PbO}_2$; it is reduced.
  4. Step 4: Cathode: $\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O}$.

Answer: Both electrodes form $\text{PbSO}_4$ on discharge; charging reverses the reactions.

5
Worked Example
Explain the electrochemical mechanism of rusting of iron and name two ions/species essential for it.
Solution
  1. Step 1: On wet iron, one region acts as anode: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$.
  2. Step 2: Another region acts as cathode where oxygen is reduced: $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$.
  3. Step 3: The released $\text{Fe}^{2+}$ is further oxidised by air to $\text{Fe}^{3+}$, forming hydrated $\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}$ (rust).
  4. Step 4: Both water and oxygen (dissolved $\text{O}_2$) are essential; rusting stops if either is absent.

Answer: Rusting is a tiny galvanic action on the iron surface; water and oxygen are both essential.

6
Worked Example
What is cathodic protection? Explain how a magnesium block prevents an underground steel pipe from rusting.
Solution
  1. Step 1: Cathodic protection forces the metal to act as a cathode (where no oxidation occurs).
  2. Step 2: Magnesium has a more negative $E^0$ ($-2.37\,\text{V}$) than iron ($-0.44\,\text{V}$), so it is oxidised in preference.
  3. Step 3: Connected to the steel pipe, the Mg block becomes the sacrificial anode and corrodes instead.
  4. Step 4: The iron pipe stays protected as long as the Mg block lasts and is periodically replaced.

Answer: A more reactive sacrificial anode (Mg) corrodes preferentially, keeping the steel pipe as a protected cathode.

Key Points

  • Electrolysis uses electrical energy to drive a non-spontaneous redox reaction; anode (+) oxidises, cathode ($-$) reduces.
  • Faraday's first law: $m = ZIt$ with $Q = It$; mass $m = \frac{MIt}{nF}$, where $1\,F = 96500\,\text{C}$.
  • Primary cells (dry cell, mercury cell) are non-rechargeable; secondary cells (lead storage, Ni–Cd) are rechargeable.
  • A H2–O2 fuel cell converts combustion energy directly to electricity in KOH, giving only water and high efficiency.
  • Rusting is electrochemical, needing water and oxygen; prevented by barriers, galvanising and cathodic (sacrificial-anode) protection.
Tap an option to check your answer0 / 4
Q1.One Faraday of charge is equal to:
Explanation: One Faraday is the charge of one mole of electrons, $\approx 96500\,\text{C mol}^{-1}$.
Q2.Which of the following is a secondary (rechargeable) cell?
Explanation: The lead storage (and Ni–Cd) battery is rechargeable; dry, mercury and Leclanché cells are primary cells.
Q3.The product of combustion in a hydrogen–oxygen fuel cell is:
Explanation: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ — the only product is water, making it a clean source.
Q4.Connecting iron to a block of magnesium to prevent rusting is called:
Explanation: Magnesium, being more reactive, corrodes as a sacrificial anode, protecting the iron cathode.