Electrochemistry

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 2.

What if a reaction had to take the long way round?

Drop a zinc rod into copper sulphate solution and it quickly turns red-brown with copper, while the solution warms slightly. Electrons jump straight from zinc to copper ions, and all the energy is lost as heat. Now put the zinc in one beaker and the copper in another, and let the electrons travel through a wire to get from one to the other. Same reaction, but now the electrons do useful work on the way, and you have a battery.

That is the chapter in one picture. Electrochemistry links chemical reactions to electrical energy in both directions: a spontaneous reaction can produce electricity (a galvanic cell), and electricity can force a non-spontaneous reaction to happen (an electrolytic cell). Everything else (voltages, conductivity, Faraday's laws, batteries, rusting) follows from that.

Galvanic cells: electricity from a reaction

A galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy. The Daniell cell is the standard example. It is based on

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

and is built from two half-cells, each a metal electrode dipping in a solution of its own ions.

Electrons flow through the external wire from the anode (Zn) to the cathode (Cu). The two beakers are joined by a salt bridge, a U-tube holding an inert electrolyte such as KCl or KNO₃. Without it the circuit would stall almost at once, because the zinc beaker would build up positive charge (extra Zn²⁺) and the copper beaker would become negative (Cu²⁺ leaving). The salt bridge lets ions move in the solutions, K⁺ towards the cathode and the negative ions towards the anode, to keep both beakers neutral and complete the circuit. Electrons travel in the wire and ions travel in the solution, never the other way round. The bridge electrolyte must not react with either half-cell, so KNO₃ is used instead of KCl if the cell contains silver ions.

Cell notation (IUPAC convention). Anode on the left, cathode on the right, a single bar for a phase boundary and a double bar for the salt bridge: Zn(s) ∣ Zn²⁺(aq) ‖ Cu²⁺(aq) ∣ Cu(s). The cell potential, measured with no current drawn (the emf), is

$$E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = E_{\text{cathode}} - E_{\text{anode}}$$

and it is positive for a spontaneous cell. With both ion concentrations at 1 mol L⁻¹ the Daniell cell gives 1.1 V.

What if you oppose the cell with an external voltage? The cell is pushing electrons with 1.1 V, so an opposing external voltage (E_ext) changes what happens.

Oxidation is always at the anode and reduction at the cathode. The signs of the electrodes flip between the two cell types, but that rule does not.

Electrode potentials and the table that predicts reactions

A single electrode's potential cannot be measured, only the difference between two. So all potentials are measured against a reference, the standard hydrogen electrode (SHE): a platinum electrode in contact with H₂ gas at 1 bar and H⁺ at 1 mol L⁻¹. It is assigned a potential of zero at all temperatures.

Standard electrode potential (E°) is the potential of a half-cell with all species at unit activity (1 M, 1 bar), connected to SHE. IUPAC defines every E° as a reduction potential, for M^n+ + ne⁻ → M. A positive E° means the species is reduced more easily than H⁺, and a negative E° means less easily.

Read the table from top to bottom: oxidising power falls and reducing power rises. Any species higher in the table will oxidise the reduced form of a species lower in the table. A metal with a lower (more negative) E° displaces a metal with a higher E° from its salt solution, and only metals with negative E° release hydrogen from dilute acid.

Worked example: will it react? (a) Calculate $E^\circ_{\text{cell}}$ for the Daniell cell. (b) Will copper dissolve in dilute HCl? (c) Will iron displace copper from CuSO₄ solution?

The common wrong step: adding the oxidation and reduction potentials, or flipping the sign of the cathode value. Always take reduction potentials of both half-cells and subtract: cathode minus anode.

Beyond the textbook. For the exam: SHE is the reference electrode with E° = 0. In reality: the SHE is awkward to run because it needs pure hydrogen at exactly 1 bar and a freshly platinised electrode, so laboratories mostly use secondary reference electrodes such as the saturated calomel electrode, which are calibrated against the SHE. Why it matters: the value of E° is always against SHE, but you rarely meet a real SHE outside a textbook.

The Nernst equation: when concentrations are not 1 M

E° applies only at standard conditions. As a cell works, reactants are used up and products pile up, and the emf falls. The Nernst equation captures this. For the reaction aA + bB → cC + dD:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \dfrac{RT}{nF}\ln Q \quad\Rightarrow\quad E_{\text{cell}} = E^\circ_{\text{cell}} - \dfrac{0.059}{n}\log Q \ \text{(at 298 K)}$$

where

$$Q = \dfrac{[C]^c\,[D]^d}{[A]^a\,[B]^b}$$

(solids and pure liquids are left out) and $n$ is the number of electrons transferred. Intuitively, a cell with more products than reactants has less driving force, so the emf is lower, and when Q is small the cell is more powerful. For a half-cell, M^n+ + ne⁻ → M, the same equation gives

$$E = E^\circ - \dfrac{0.059}{n}\log\dfrac{1}{[\text{M}^{n+}]}$$

At equilibrium. When the cell is flat, $E_{\text{cell}} = 0$ and $Q$ has become the equilibrium constant $K$, so at 298 K

$$E^\circ_{\text{cell}} = \dfrac{0.059}{n}\log K$$

and the Gibbs energy of the cell reaction is $\Delta_r G = -nFE_{\text{cell}}$, which gives

$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$$

A positive $E^\circ$ means a negative $\Delta G^\circ$, which means a spontaneous reaction. Also, E° is an intensive property (it does not depend on how much material is used), but ΔG° is extensive, so doubling the equation doubles ΔG° but leaves E° unchanged.

Worked example: Nernst equation. (a) Find the electrode potential of a zinc electrode in 0.01 M Zn²⁺. (b) Find $E_{\text{cell}}$ for Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s), with $E^\circ = 3.17$ V.

The common wrong step: forgetting that [Ag⁺] is squared in Q, or using n = 1 because Ag⁺ takes only one electron. The n in the Nernst equation is for the reaction as written, which is 2.

Worked example: equilibrium constant and Gibbs energy. For the Daniell cell, $E^\circ = 1.10$ V and $n = 2$. Find $K$ and $\Delta_r G^\circ$ ($F = 96487\ \text{C mol}^{-1}$).

The common wrong step: using n = 1. Two electrons move for every Zn that reacts.

Pause and predict: a Daniell cell is run until [Zn²⁺] is far larger than [Cu²⁺]. What happens to its emf, and what is it when the reaction reaches equilibrium?

Check your answer: the emf falls steadily as Q grows. At equilibrium $Q = K$, and $E_{\text{cell}} = 0$, so the cell is dead and no more current can be drawn.

Beyond the textbook. For the exam: $\Delta_r G^\circ = -nFE^\circ$, and $K$ is found from $E^\circ = \frac{0.059}{n}\log K$. In reality: E° is the thing you can look up, but a real battery's voltage drifts with use, exactly as Nernst predicts. This is why phone batteries are specified by a nominal voltage and why the percentage display is calibrated rather than read straight from the voltage. Why it matters: the equation is a statement about energy, not just a way to get exam marks.

Conductance: how well does a solution carry current?

An electrolyte carries current by moving ions, not electrons. The resistance of a uniform conductor of length l and area A is

$$R = \rho\,\dfrac{l}{A}$$

where ρ is the resistivity. Its reciprocal is the conductivity (κ), in S m⁻¹. Remember that 1 S cm⁻¹ = 100 S m⁻¹.

Solutions are measured in a conductivity cell, and the geometry is packed into the cell constant, $G^{\ast} = l/A$. You find it by filling the cell with KCl of known κ and measuring $R$, giving $G^{\ast} = \kappa R$. Then for any solution $\kappa = G^{\ast}/R$. An alternating current is used, because a direct current would electrolyse the solution and change its concentration while you measure.

Molar conductivity ($\Lambda_m$) is the conductivity per mole, $\Lambda_m = \kappa / c$. With κ in S cm⁻¹ and concentration $M$ in mol L⁻¹, the practical form is

$$\Lambda_m\ (\text{S cm}^2\ \text{mol}^{-1}) = \dfrac{\kappa \times 1000}{M}$$

Also, $1\ \text{S m}^2\ \text{mol}^{-1} = 10^{4}\ \text{S cm}^2\ \text{mol}^{-1}$.

Worked example: from resistance to molar conductivity. A conductivity cell filled with 0.1 M KCl (κ = 0.0129 S cm⁻¹) has a resistance of 100 Ω. The same cell with 0.02 M NaCl has a resistance of 520 Ω. Find Λm of the NaCl solution.

The common wrong step: skipping the 1000. κ is per cm³ but concentration is per litre (1000 cm³), so the factor is needed to put them on the same footing.

Dilution. Conductivity (κ) always falls on dilution, because there are fewer ions in each cm³. Molar conductivity (Λm) always rises, because the volume holding one mole grows faster than κ falls. The two rises have different shapes.

Λ°m is the limiting molar conductivity, the value at infinite dilution.

Kohlrausch's law of independent migration of ions. At infinite dilution each ion contributes its own fixed amount, whatever the other ion is:

$$\Lambda^\circ_m = \nu_+ \lambda^\circ_+ + \nu_- \lambda^\circ_-$$

where ν is the number of each ion in the formula and λ° is its limiting ionic conductivity. This lets us get Λ°m for a weak electrolyte from strong electrolyte data, and then

$$\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m} \qquad K_a = \dfrac{c\,\alpha^2}{1 - \alpha}$$

Worked example: acetic acid. Λ°m values at 298 K (S cm² mol⁻¹): HCl 425.9, CH₃COONa 91.0, NaCl 126.4. For 0.001 M acetic acid, Λm = 39.05 S cm² mol⁻¹. Find $\Lambda^\circ_m$, α and $K_a$.

The common wrong step: trying to extrapolate Λm to zero concentration for acetic acid. The curve is far too steep.

Pause and predict: you dilute a KCl solution and an acetic acid solution, each ten times. Whose Λm rises more steeply?

Check your answer: acetic acid. The ions of KCl are always fully dissociated, so dilution only reduces their attraction a little, while dilution of the weak acid makes it dissociate much more, creating a flood of new ions.

Electrolysis and Faraday's laws

In an electrolytic cell an external source forces a reaction. The question is which species reacts at each electrode, and the answer depends on the E° values (the species that is easier to reduce goes at the cathode, and the one that is easier to oxidise goes at the anode) and on the nature of the electrode.

Why does aqueous NaCl give chlorine, not oxygen? On E° alone, water (1.23 V for the O₂ reaction) should be oxidised before chloride (1.36 V). But the oxygen evolution is slow and needs an extra voltage, called an overpotential, so chloride is oxidised instead. That overpotential is the answer NTA wants.

Faraday's laws. First law: the amount of substance liberated at an electrode is proportional to the charge passed, $Q = It$. Second law: for the same charge, the amounts of different substances liberated are in the ratio of their equivalent weights. One faraday, F = N_A × e = 96487 C mol⁻¹ (use 96500 in sums), is the charge carried by one mole of electrons.

$$m = \dfrac{M\,I\,t}{nF}$$

where $M$ is molar mass and $n$ is the number of electrons per ion.

Worked example: copper deposition. A current of 1.5 A flows through CuSO₄ solution for 10 minutes. What mass of copper is deposited? (Cu = 63 g mol⁻¹)

The common wrong step: using minutes for time, or n = 1. Time must be in seconds, and n counts electrons per copper ion.

Worked example: aluminium. How much charge deposits 1 mol of aluminium, and what mass of Al is deposited by 5 A passed for 1930 s? (Al = 27 g mol⁻¹)

Batteries and fuel cells

A primary battery is used once. A secondary battery can be recharged by running the reaction backwards with an external current.

Overall reaction in the lead storage battery: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O during discharge. During charging the reaction is reversed. The overall mercury cell reaction is Zn(Hg) + HgO → ZnO + Hg, which involves no ion whose concentration changes, and that is why its voltage stays at 1.35 V throughout its life.

Fuel cells. A fuel cell turns the energy of burning a fuel directly into electricity, with the fuel and oxidant fed in continuously. In the hydrogen–oxygen cell, H₂ and O₂ are bubbled over porous carbon electrodes (with Pt or Pd catalyst) into hot concentrated NaOH.

Efficiency is about 70%, against about 40% for a thermal power plant, and the only product is water, so it is pollution free.

Beyond the textbook. For the exam: the lead storage cell discharges as Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. In reality: sulphuric acid is used up and water is made, so the density of the acid falls as the battery discharges. A hydrometer reading of the density tells a mechanic how charged a car battery is. Why it matters: the equation is not only bookwork. It is a gauge you can read.

Corrosion: a battery you did not want

Rusting is a tiny galvanic cell on the surface of the iron. A film of water with dissolved CO₂ acts as the electrolyte. At one spot iron acts as the anode and at another spot as the cathode.

The positive E°cell of 1.67 V shows that rusting is spontaneous. Prevention works in two ways: coat the iron to keep out air and water (paint, grease, galvanising, tin plating), or attach a sacrificial electrode of a more reactive metal such as Mg or Zn. The sacrificial metal has a lower E°, so it is oxidised in preference and the iron is saved.

Beyond the textbook. For the exam: both zinc coating (galvanising) and tin plating protect iron by covering it. In reality: once the coating is scratched they behave very differently. Zinc (−0.76 V) is more reactive than iron (−0.44 V), so exposed iron stays protected while the zinc corrodes instead. Tin (−0.14 V) is less reactive than iron, so iron becomes the anode and rusts even faster than bare iron. Why it matters: this is why a scratched tin can rusts quickly but a scratched galvanised bucket does not.

Pause and predict: a copper plate is riveted to an iron pipe in a damp basement. Which metal corrodes?

Check your answer: the iron. With E° of −0.44 V against +0.34 V for copper, iron is the more easily oxidised metal, so it becomes the anode and corrodes, while copper is protected. Replacing the copper with zinc would flip the result.

Formula sheet

Exam traps

60-second revision

Where this chapter connects

• Before this: Redox Reactions gives you oxidation, reduction and half-reactions, which are the building blocks of every cell here, and Solutions supplies molarity and the ideas about ions in water.
• Next: Chemical Kinetics asks how fast a reaction goes, which is the question electrochemistry's K and ΔG cannot answer.
• Forward link: Thermodynamics and Equilibrium are the sources of ΔG and K, and Current Electricity explains the resistance and resistivity behind conductance.