2. Electrochemistry

What you will learn

• Identify electrode reactions independently of electrode signs.
• Read cell notation and use standard reduction potentials.
• Apply the Nernst equation and connect potential with Gibbs energy and equilibrium.
• Distinguish conductivity from molar conductivity.
• Calculate electrolysis products from charge.
• Explain batteries, fuel cells and corrosion using electron and ion paths.

Big idea: electrons do useful work when a redox reaction gives them an organised route.

A cell combines a chemical driving force with two charge paths. The electronic route passes through conductors; the ionic route passes through an electrolyte. If either route is broken, sustained current stops even when the reactants remain.

This chapter follows that chain from a single electrode interface to complete cells and practical devices. Keep the written reaction visible. It determines the sign of potential, the electron count and the reaction quotient used in each calculation.

Faraday’s nineteenth-century electrolysis studies connected deposited amounts with transferred charge. Daniell’s separated-electrode cell and Nernst’s composition–potential relation helped turn practical electrical devices into quantitative chemical systems. The key advance was linking an observed voltage or mass change with a balanced reaction.

A reaction can move electrons through a wire only if its oxidation and reduction are arranged into a complete circuit.

Electrochemistry connects chemical change with electrical work. An electrochemical cell has two electrode–electrolyte interfaces and a path for charge to move between them. An electrode conducts electrons. The electrolyte conducts charge through mobile ions. Electrons in a metal wire and ions in a solution are different charge carriers.

A galvanic cell uses a spontaneous overall redox reaction to supply electrical work. An electrolytic cell uses an external power source to drive a nonspontaneous overall reaction. In both, oxidation happens at the anode and reduction at the cathode. These reaction-based definitions never change.

A metal can exchange atoms and ions at its surface. Its electrode potential describes the tendency of that interface to participate in electron transfer relative to a reference. An isolated single-electrode potential cannot be measured directly with an ordinary voltmeter; a second electrode is needed.

An externally powered cell deposits metal ions as solid metal. Name the electrode where deposition occurs.

Metal ions gain electrons during deposition. Electron gain is reduction. The reduction electrode is the cathode in either operating mode.

Result: Deposition occurs at the cathode; its sign depends on whether the cell is galvanic or electrolytic.

1. Can oxidation occur at the cathode of an ordinary electrochemical cell?
2. What carries charge across an electrolyte?

Answers

1. No. The cathode is defined as the reduction electrode.
2. Mobile ions, rather than a flow of free electrons through the bulk solution.

Separating two half-reactions lets electron transfer do useful work on its route from one reactant to another.

Consider a zinc electrode in a zinc-ion solution and a copper electrode in a copper-ion solution. The electrode solutions communicate through a suitable ionic connection. The metal terminals are connected externally. The overall reaction can be understood by adding the half-reactions.

$$\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+} + 2 e^{-}$$

$$\mathrm{Cu}^{2+} + 2 e^{-} \longrightarrow \mathrm{Cu}$$

$$\mathrm{Zn} + \mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+} + \mathrm{Cu}$$

During galvanic operation, electrons move through the wire from zinc anode to copper cathode. Conventional current is opposite to electron flow. The ionic connection completes the circuit and limits the build-up of charge. In a common salt-bridge arrangement, anions migrate towards the anode compartment and cations towards the cathode compartment. Choose a bridge electrolyte that does not strongly react with the cell materials.

Cell notation records interfaces. A single vertical line represents a phase boundary; a double line often represents a salt bridge. By convention, write anode on the left and cathode on the right: $\mathrm{Zn(s)|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu(s)}$. An inert conductor such as platinum is used if a half-cell has no conducting solid reactant.

Standard reduction potentials are tabulated relative to the standard hydrogen electrode, assigned zero by convention. “Standard” refers to specified standard activities and gas pressure, not automatically to room temperature. Tables state the temperature, commonly near room temperature. A pure solid has activity one in the usual reference treatment.

$$E_{\rm cell}^{\circ}=E_{\rm cathode}^{\circ}-E_{\rm anode}^{\circ}$$

A proposed cell has standard reduction potentials $+0.62\,\mathrm V$ on the right and $-0.18\,\mathrm V$ on the left. Test the written galvanic direction.

Subtract the left reduction potential from the right reduction potential. Potentials are intensive; do not multiply them by electron-balancing coefficients.

Result: $E_{\rm cell}^{\circ}=0.8\,\mathrm V$. Its positive sign favours the written reaction under standard conditions.

1. Which side is written first in standard galvanic-cell notation?
2. Why does an ionic bridge matter if a wire is already connected?

Answers

1. The anode, on the left.
2. The wire supplies only the electronic path. Ionic charge transport is also needed for sustained operation.

A cell potential depends on composition. A positive standard potential does not guarantee that every nonstandard mixture will react in the same direction.

The Nernst equation connects cell potential with the reaction quotient $Q$. Write the balanced cell reaction first. Then build $Q$ from activities with powers equal to stoichiometric coefficients. Omit pure solids and pure liquids in their standard states.

$$E=E^{\circ}-\frac{RT}{nF}\ln Q$$

For a sufficiently dilute solution, concentrations divided by their standard concentration are often used as approximations to activities. A gas contribution uses its pressure relative to a standard pressure. These ratios make the reaction quotient dimensionless. The electron number belongs to the balanced overall reaction.

For a two-electron cell at $298\,\mathrm K$, take $E^{\circ}=0.800\,\mathrm V$ and $Q=100$. Use $R=8.314$ and $F=96485$ in consistent SI units.

A quotient above one makes the logarithmic correction positive, so it is subtracted from the standard potential.

Result: $E=0.74087\,\mathrm V$. It is below the standard value but remains positive.

At equilibrium the cell has no net reaction-driving potential, so $E=0$ and $Q=K$. Rearranging gives a direct connection between standard potential and equilibrium constant.

$$\ln K=\frac{nFE^{\circ}}{RT}$$

$$\Delta_rG=-nFE,\qquad\Delta_rG^{\circ}=-nFE^{\circ}$$

A positive cell potential corresponds to negative Gibbs-energy change for the written reaction. The reversible maximum electrical work is connected with Gibbs energy; real devices also lose energy through resistance and other irreversible processes.

Use the same standard cell data to calculate standard Gibbs energy and the base-ten logarithm of the equilibrium constant.

Keep joules, coulombs and volts consistent. Convert natural logarithms to base-ten logarithms using $\ln 10$.

Result: $\Delta_rG^{\circ}=-154376\,\mathrm{J\,mol^{-1}}$ and $\log_{10}K=27.0606$. A large positive logarithm means products are strongly favoured under the defined standard-state model.

1. What are E and Q at equilibrium?
2. If Q increases at fixed temperature, what happens to E in the written Nernst relation?

Answers

1. The net cell potential is zero and the reaction quotient equals the equilibrium constant.
2. It decreases.

A concentrated solution can carry more current per centimetre yet be less effective per mole of solute. Conductivity and molar conductivity answer different questions.

Resistance $R$ measures opposition to current in a specified object. Conductance is $G=1/R$. Resistivity is a material property, while conductivity is its reciprocal. Cell geometry relates the measured resistance to conductivity.

$$\kappa=\frac{l}{RA}=\frac{\text{cell constant}}{R},\qquad\Lambda_m=\frac{1000\kappa}{c}$$

The second expression gives $\Lambda_m$ in $\mathrm{S\,cm^2\,mol^{-1}}$ when $\kappa$ is in $\mathrm{S\,cm^{-1}}$ and $c$ in $\mathrm{mol\,L^{-1}}$. Using SI units removes that particular conversion factor: match the volume unit to the length unit throughout.

Calibrate the cell constant using a reference solution of known conductivity. An alternating signal helps avoid continuous electrolysis and electrode polarisation during a conductivity measurement. A measured resistance alone is insufficient unless the cell geometry is known or calibrated.

A calibrated cell has constant $1.20\,\mathrm{cm^{-1}}$. A $0.0400\,\mathrm M$ electrolyte gives resistance $200\,\Omega$.

Calculate conductivity from cell constant divided by resistance. Then divide by concentration with the matching volume conversion.

Result: $\kappa=0.006\,\mathrm{S\,cm^{-1}}$ and $\Lambda_m=150\,\mathrm{S\,cm^2\,mol^{-1}}$.

On dilution, conductivity usually decreases because there are fewer ions per unit volume. Molar conductivity generally increases because ionic motion becomes less hindered; for a weak electrolyte, a larger degree of ionisation adds a further increase.

For many strong electrolytes at sufficiently low concentration, $\Lambda_m\approx\Lambda_m^{\circ}-A\sqrt c$. Extrapolation towards zero concentration gives the limiting value. For a weak electrolyte the rise can be strongly curved, so the same straight-line extrapolation is unsuitable.

Kohlrausch’s limiting law adds the ionic contributions with their stoichiometric counts: $\Lambda_m^{\circ}=\sum_i\nu_i\lambda_i^{\circ}$. Carefully chosen strong-electrolyte combinations can supply a weak electrolyte’s limiting value.

A weak monoprotic acid has $\kappa=4.00\times10^{-5}\,\mathrm{S\,cm^{-1}}$ at $c=0.00100\,\mathrm M$. Its limiting ionic contributions are $349.6$ and $40.9\,\mathrm{S\,cm^2\,mol^{-1}}$.

Use $\alpha\approx\Lambda_m/\Lambda_m^{\circ}$ for the simple dilute weak-electrolyte model. Then use $K_a\approx c\alpha^2/(1-\alpha)$.

Result: $\Lambda_m^{\circ}=390.5$, $\Lambda_m=40$, $\alpha=0.102433$ and $K_a\approx0.00001169$. The equilibrium calculation uses the concentration-based dilute approximation.

Conductivity can also help estimate the solubility of a sparingly soluble salt. Correct for the conductivity of the solvent, obtain the dissolved concentration from a justified limiting molar conductivity, then use the dissolution stoichiometry to construct the solubility product.

1. Which usually increases on dilution: κ or Λm?
2. What is the limiting molar conductivity of $\mathrm{CaCl_2}$ in terms of ionic contributions?

Answers

1. Molar conductivity usually increases; conductivity usually decreases.
2. $\lambda^{\circ}_{\rm Ca^{2+}}+2\lambda^{\circ}_{\rm Cl^-}$.

In electrolysis, charge counts electrons. Stoichiometry then turns the electron amount into a deposited or released amount of substance.

An external source can make the cathode reduce a species and the anode oxidise a species even when the reverse overall reaction would be spontaneous. Product choice depends on the electrolyte, electrode material, concentration and electrode kinetics. Aqueous solutions also contain water, which may compete as reactant.

$$q=It,\qquad n(e^-)=\frac qF,\qquad m_{\rm product}=\frac{MIt}{zF}$$

Here $z$ is the electron number per formula unit of product in the relevant half-reaction. The same charge passes through cells connected in series. Their product masses need not match because $M/z$ can differ.

A copper-ion solution is electrolysed with $2.00\,\mathrm A$ for $1200\,\mathrm s$. Assume all cathodic charge deposits copper from $\mathrm{Cu^{2+}}$. Use molar mass $63.5\,\mathrm{g\,mol^{-1}}$.

The half-reaction requires two electrons per copper atom. Use seconds, not minutes, with amperes.

Result: $m=0.78976\,\mathrm g$. If competing reactions take some current, the actual deposited mass is lower.

For a molten salt, the available ions often give the main electrode products directly. In an aqueous electrolyte, compare plausible reductions at the cathode and oxidations at the anode, then consider concentration and overpotential. A reactive metal anode may dissolve instead of allowing the oxidation expected at an inert anode.

Aqueous copper sulfate with an inert cathode commonly gives copper deposition. At an inert anode, water oxidation may supply oxygen. With a copper anode, copper dissolution can replace that process. In concentrated chloride solutions, chlorine evolution can compete effectively at a suitable anode even when a simple standard-potential comparison suggests another reaction.

Faraday’s laws specify how much product follows from the charge assigned to a reaction. They do not identify the reaction on their own. Write the half-reaction before choosing the electron number.

1. How much charge passes at $0.750\,\mathrm A$ for $800\,\mathrm s$?
2. Why can changing the anode material change electrolysis products?

Answers

1. $600\,\mathrm C$.
2. A reactive anode may itself oxidise, while an inert anode lets solution species supply the oxidation reaction.

A battery packages a redox system so its reactants and charge paths remain useful during discharge.

A primary cell is designed mainly for one discharge cycle. Its discharge chemistry and construction do not provide convenient repeated regeneration. A conventional zinc–carbon cell uses zinc as the anode and a manganese-dioxide-containing cathodic mixture with an ammonium-chloride/zinc-chloride paste. Its composition and practical reactions are more complex than a single ideal half-reaction.

An alkaline manganese cell uses an alkaline electrolyte and commonly gives better useful performance than the simple zinc–carbon design. A mercury-oxide cell historically supplied a fairly steady potential because its overall reaction does not greatly change the concentration of dissolved ions. Mercury toxicity is a serious disposal concern; this is a chemistry example, not a recommendation to use such cells.

$$\mathrm{Zn} + \mathrm{HgO} \longrightarrow \mathrm{ZnO} + \mathrm{Hg}$$

A secondary cell can be recharged by supplying electrical work in a direction that restores its reactants. A lead–acid cell uses lead and lead dioxide electrodes in sulfuric acid. During discharge both electrode materials form lead sulfate, and the acid composition changes.

$$\mathrm{Pb} + \mathrm{PbO_{2}} + 2 \mathrm{H_{2}SO_{4}} \longrightarrow 2 \mathrm{PbSO_{4}} + 2 \mathrm{H_{2}O}$$

During charging, the external source drives the reverse net chemistry within the cell’s practical operating limits. Rechargeability does not imply that every side reaction reverses. Ageing, gas evolution and material changes limit real performance.

A nickel–cadmium cell is another classical secondary-cell example, using cadmium and nickel oxyhydroxide chemistry in alkaline electrolyte. Cadmium toxicity makes material recovery and appropriate disposal important. Modern rechargeable cells also include lithium-ion systems, where ion transfer and electrode host structures support repeated cycling.

Why can a discharge curve remain fairly flat for one cell but fall markedly for another?

Cell potential reflects the chemical state of the active materials. If the overall reaction keeps the relevant activities nearly constant during part of discharge, the equilibrium contribution to potential can stay fairly steady. Resistance and kinetic losses also matter.

Result: Compare reaction chemistry and activities; “all batteries keep constant voltage” is not a general rule.

1. Which battery class is designed for repeated charging?
2. What common solid product forms at both lead–acid electrodes on discharge?

Answers

1. Secondary batteries.
2. Lead sulfate.

A fuel cell keeps receiving reactants while it operates. It does not need to store its entire fuel supply inside the electrode assembly.

A hydrogen–oxygen fuel cell converts the Gibbs-energy decrease of the overall water-forming reaction into electrical work. The fuel and oxidant are supplied separately to electrodes. A catalyst helps the electrode reactions proceed at useful rates; it does not create additional reaction energy.

$$2 \mathrm{H_{2}} + \mathrm{O_{2}} \longrightarrow 2 \mathrm{H_{2}O}$$

In an alkaline model, hydrogen oxidation supplies electrons at the anode and oxygen reduction consumes them at the cathode. The electrolyte and membrane determine the ions that move internally. Acidic or polymer-electrolyte designs use different half-reaction bookkeeping, while the overall hydrogen–oxygen reaction is the same.

$$2 \mathrm{H_{2}} + 4 \mathrm{OH}^{-} \longrightarrow 4 \mathrm{H_{2}O} + 4 e^{-}$$

$$\mathrm{O_{2}} + 2 \mathrm{H_{2}O} + 4 e^{-} \longrightarrow 4 \mathrm{OH}^{-}$$

Fuel cells can avoid some limitations of a combustion–heat-engine route, but their real efficiency is below the reversible limit. Fuel purity, catalyst performance, water management and internal losses matter. For hydrogen, the environmental impact also depends on how the hydrogen and its supply energy were produced.

A fuel cell is fed hydrogen continuously. Is it automatically a rechargeable battery?

A rechargeable battery normally restores its stored active materials by electrical charging. A fuel cell normally receives fuel and oxidant from outside and removes products.

Result: No. Both are electrochemical devices, but their reactant-supply arrangements differ.

1. What is the overall chemical product of an ideal hydrogen–oxygen fuel cell?
2. Does using hydrogen at the point of operation prove the complete energy chain has no emissions?

Answers

1. Water.
2. No. Fuel production and energy supply must also be considered.

Corrosion is often a small electrochemical cell spread across different regions of the same metal surface.

On wet iron, an anodic region can dissolve while a nearby region supplies a reduction process. Differences in oxygen access, surface condition and dissolved ions can help create these local cells. The metallic body carries electrons between regions; the thin water layer carries ionic charge.

$$\mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+} + 2 e^{-}$$

$$\mathrm{O_{2}} + 2 \mathrm{H_{2}O} + 4 e^{-} \longrightarrow 4 \mathrm{OH}^{-}$$

Further oxidation and precipitation create iron oxides and hydroxides. Rust is a variable hydrated mixture, commonly represented with hydrated iron(III) oxide. It is not a compact protective film, so corrosion can continue beneath and through it.

Barrier coatings limit access of water and oxygen. Galvanising uses a zinc layer that can provide sacrificial protection. A connected metal that oxidises more readily can act as a sacrificial anode, while the protected structure becomes the cathodic part of the coupled system. Proper materials selection and engineering are needed; random metal contact can make corrosion worse.

A hydrogen-based energy system links production, storage, transport and electrochemical use. Electrolysis can store supplied electrical energy in chemical form, and a fuel cell can later return part of it as electrical work. Each stage has losses. Hydrogen is an energy carrier in that chain, not an unlimited primary source of energy.

A coated iron surface develops a scratch. Why can a zinc coating still help after its barrier is locally damaged?

Zinc can oxidise preferentially when it is electrically connected to the exposed iron and an electrolyte completes the circuit. This supplies sacrificial rather than purely barrier protection.

Result: The remaining connected zinc can serve as the anode, helping protect the iron at the defect.

1. What two charge paths are present in wet electrochemical corrosion?
2. Why does rust often fail to stop further corrosion?

Answers

1. An electronic path through metal and an ionic path through the moisture layer.
2. Its porous, poorly protective structure still allows reactants to reach the metal.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

2.1 Define anode and cathode without referring to electrode signs.

2.2 Describe the separate jobs of the wire and salt bridge in a galvanic cell.

2.3 Find standard cell potential from cathode $+0.48\,\mathrm V$ and anode $-0.32\,\mathrm V$.

2.4 Why are half-cell potentials not multiplied when half-reactions are scaled?

2.5 At $298\,\mathrm K$, a two-electron cell has $E^{\circ}=0.500\,\mathrm V$ and $Q=10$. Calculate E.

2.6 Find standard Gibbs energy for $n=3$ and $E^{\circ}=0.400\,\mathrm V$.

2.7 What does a negative potential imply for the written reaction under the stated conditions?

2.8 A cell has constant $0.800\,\mathrm{cm^{-1}}$ and resistance $160\,\Omega$. Find conductivity.

2.9 The preceding solution has concentration $0.0250\,\mathrm M$. Find molar conductivity.

2.10 Why does a weak electrolyte show a large rise in molar conductivity on dilution?

2.11 Given limiting ionic contributions $a$ for a divalent cation and $b$ for a monovalent anion, write the salt limiting conductivity.

2.12 A weak acid has $\Lambda_m=30.0$ and $\Lambda_m^{\circ}=400.0$ in matching units. Estimate its ionised fraction.

2.13 At $1.25\,\mathrm A$ for $960\,\mathrm s$, how many moles of electrons pass?

2.14 How much silver deposits from $\mathrm{Ag^+}$ under that charge? Use molar mass $108\,\mathrm{g\,mol^{-1}}$.

2.15 What changes when an inert anode is replaced by a reactive copper anode in suitable copper-ion electrolysis?

2.16 Compare a primary cell and a secondary cell by chemical regeneration.

2.17 Explain why a fuel cell needs both an external electronic path and an internal ionic path.

2.18 Explain sacrificial protection of an iron structure in electrochemical terms.

Answers

2.1 Oxidation occurs at the anode; reduction occurs at the cathode.

2.2 The wire carries electrons externally; the bridge supplies ionic charge transport internally.

2.3 $0.8\,\mathrm V$.

2.4 Potential is an intensive energy-per-charge quantity. Scaling changes total electron amount, not energy per charge.

2.5 $E=0.47044\,\mathrm V$.

2.6 $-115782\,\mathrm{J\,mol^{-1}}$.

2.7 Its Gibbs-energy change is positive; the reverse reaction is favoured under those conditions.

2.8 $0.005\,\mathrm{S\,cm^{-1}}$.

2.9 $200\,\mathrm{S\,cm^2\,mol^{-1}}$.

2.10 Its ionisation increases, in addition to changes in interionic effects and mobility.

2.11 $a+2b$ for a salt containing one cation and two anions.

2.12 $\alpha=0.075$.

2.13 $0.012437\,\mathrm{mol}$.

2.14 $1.34321\,\mathrm g$, assuming complete current efficiency.

2.15 Copper oxidation can supply the anodic reaction and replenish copper ions.

2.16 A secondary cell is designed for repeated electrical regeneration; a primary cell is mainly designed for a single discharge.

2.17 Both are necessary to complete charge transport while oxidation and reduction occur at separate electrodes.

2.18 A connected, more readily oxidised material acts as an anode and is consumed, while iron is kept cathodic in the coupled system.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

How much silver deposits from $\mathrm{Ag^+}$ under that charge? Use molar mass $108\,\mathrm{g\,mol^{-1}}$.

$1.34321\,\mathrm g$, assuming complete current efficiency.

What changes when an inert anode is replaced by a reactive copper anode in suitable copper-ion electrolysis?

Copper oxidation can supply the anodic reaction and replenish copper ions.

Compare a primary cell and a secondary cell by chemical regeneration.

A secondary cell is designed for repeated electrical regeneration; a primary cell is mainly designed for a single discharge.

Explain why a fuel cell needs both an external electronic path and an internal ionic path.

Both are necessary to complete charge transport while oxidation and reduction occur at separate electrodes.

Explain sacrificial protection of an iron structure in electrochemical terms.

A connected, more readily oxidised material acts as an anode and is consumed, while iron is kept cathodic in the coupled system.