3.2 Electron transfer reactions

Oxidation is loss of electrons and an increase in oxidation state; reduction is gain of electrons and a decrease. The oxidizing agent is reduced, and the reducing agent is oxidized.

Use the oxidation-state rules and total charge to identify which species changed and which agent caused the change.

In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn rises from 0 to +2 and is oxidized, so it is the reducing agent; Cu²⁺ falls from +2 to 0 and is reduced, so it is the oxidizing agent. Name agents from what happens to them, not from the process they cause in the other species.

Separate oxidation and reduction, balance atoms, add H2O and H+ in acidic solution as needed, balance charge with electrons, then multiply to cancel electrons before adding.

A valid full redox equation conserves atoms and charge and contains no uncancelled electrons.

For MnO₄⁻ → Mn²⁺ in acid, balance O with 4H₂O, H with 8H⁺ and charge with 5e⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. After combining halves, cancel electrons and any identical H⁺ or H₂O, then recheck both atoms and net charge.

To adapt an acidic half-equation to neutral or basic conditions, first balance it with H₂O, H⁺ and e⁻. Add the same number of OH⁻ to both sides to neutralize every H⁺, replace H⁺+OH⁻ by H₂O, then cancel water appearing on both sides. Recheck atoms and total charge; do not leave free H⁺ in a stated neutral medium unless the chemistry justifies it.

A more active metal more readily donates electrons to a less active metal ion. A halogen with greater reduction tendency oxidizes the halide of a weaker halogen.

Test a predicted displacement by placing one metal in the other metal's sulfate or comparing supplied electrode data.

Zinc displaces Cu²⁺ because Zn more readily oxidizes: Zn + Cu²⁺ → Zn²⁺ + Cu. Chlorine displaces Br⁻ because Cl₂ more readily reduces. Keep the metal and halogen trends in their correct electron directions instead of using one vague 'more reactive' rule.

A metal above hydrogen in the activity series can donate electrons to acid and release hydrogen gas; a metal below hydrogen, such as copper, does not react with dilute hydrochloric acid.

Balance the electron transfer behind the molecular equation: metal atoms are oxidized and 2H⁺ + 2e⁻ → H₂ is the reduction. Use the metal charge and acid anion to construct the salt rather than assuming every metal forms a 2+ ion.

Oxidation always occurs at the anode and reduction always occurs at the cathode. In a voltaic cell the anode is negative and cathode positive; in an electrolytic cell the anode is positive and cathode negative.

Name electrodes from the half-reactions before assigning signs. Electrons leave the anode and reach the cathode through the external circuit; a power supply reverses the polarities in an electrolytic cell but never changes where oxidation and reduction occur.

A voltaic cell uses a spontaneous redox reaction to convert chemical energy to electrical energy. Electrons flow through the wire from anode to cathode; the salt bridge carries ions to maintain charge neutrality.

Both half-cells connect to the external circuit and the salt bridge must contact both solutions.

In a Zn|Zn²⁺ || Cu²⁺|Cu cell, Zn is oxidized at the negative anode and electrons travel through the wire to the positive Cu cathode, where Cu²⁺ is reduced. Salt-bridge anions migrate toward the anode compartment and cations toward the cathode compartment to prevent charge buildup; electrons do not flow through the bridge.

Write the discharge half-equations first. Charging a secondary cell requires an external potential to drive their reverse, whereas a primary cell is not designed for safe efficient reversal and a fuel cell continues only while reactants are supplied. Rechargeability is a reaction-design property, not simply the presence of a power socket.

In molten salt there is no water: metal ions are reduced to metal at the cathode and anions are oxidized at the anode. For molten chloride, chloride forms chlorine gas.

Molten MgCl₂ contains only Mg²⁺ and Cl⁻: Mg²⁺ + 2e⁻ → Mg at the cathode and 2Cl⁻ → Cl₂ + 2e⁻ at the anode. The melt conducts by ion migration; do not introduce H₂, O₂ or water-based competition into a molten-salt question.

A primary alcohol oxidizes to an aldehyde and then a carboxylic acid; a secondary alcohol oxidizes to a ketone. Reflux supports further oxidation to the acid, while distillation can remove an aldehyde.

In a primary-alcohol experiment, distil the aldehyde as it forms to limit further oxidation; heat under reflux when the carboxylic acid is required. Tertiary alcohols lack the required hydrogen on the carbon bearing –OH and are not oxidized in the same way.

A carboxylic acid can be reduced through an aldehyde to a primary alcohol; a ketone is reduced to a secondary alcohol. Hydride ions supply the reduction equivalent in these transformations.

Track the carbon functional group rather than only the reagent: an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol. Hydride supplies an electron-rich H unit to the carbonyl carbon; named reducing agents and detailed mechanisms are outside this objective.

Hydrogenation adds H2 across π bonds. Continue addition until the required saturated product is formed; nickel, palladium or platinum catalysts with heat or pressure are typical conditions.

Count π bonds to determine hydrogen demand: one mole of H₂ saturates one C=C, while full conversion of one C≡C to C–C needs two moles of H₂. Keep the carbon skeleton unchanged when drawing the product.

Use reduction-form data: a more positive E° means greater tendency to be reduced and stronger oxidizing behaviour. A very negative metal reduction potential indicates ease of reverse oxidation and strong reducing behaviour.

The standard hydrogen electrode uses H₂(g) at 100 kPa in contact with aqueous H⁺ of unit activity (commonly represented as 1 mol dm⁻³) at 298 K on an inert platinum surface, and is assigned E° = 0.00 V. Pair an unknown half-cell with this reference, use polarity to identify reduction, then interpret more positive reduction potential as stronger oxidizing tendency.

A positive E°cell indicates a spontaneous voltaic direction. Reverse the direction if the calculated sign is negative.

Select the more positive reduction potential as the cathode reaction, keep both tabulated values as reduction potentials, and calculate E°cell = E°cathode − E°anode. Do not multiply an electrode potential when a half-equation is scaled.

Worked $E^\circ_{cell}$ example: $E^\circ(\ce{Ag+/Ag})=+0.80\,\mathrm{V}$ and $E^\circ(\ce{Cu^{2+}/Cu})=+0.34\,\mathrm{V}$. Silver is the cathode, so $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.80-0.34=+0.46\,\mathrm{V}$. The positive result predicts the spontaneous reaction $\ce{2Ag+ + Cu -> 2Ag + Cu^{2+}}$ under standard conditions. Do not multiply $E^\circ$ when doubling the silver half-equation.

n is the moles of electrons transferred and F is Faraday's constant. Positive E°cell gives negative ΔG° and a spontaneous reaction.

Find n from the balanced overall redox equation, not from a single unscaled half-equation. With E° in volts and F in C mol⁻¹, ΔG° is obtained in J mol⁻¹; convert to kJ mol⁻¹ only at the end.

Worked $\Delta G^\circ$ example: for $\ce{2H+ + Zn -> Zn^{2+} + H2}$, $n=2$ and $E^\circ_{cell}=+0.76\,\mathrm{V}$. Using $F=9.65\times10^4\,\mathrm{C\,mol^{-1}}$, $\Delta G^\circ=-(2)(9.65\times10^4)(0.76)=-1.47\times10^5\,\mathrm{J\,mol^{-1}}=-147\,\mathrm{kJ\,mol^{-1}}$. Its negative sign agrees with a spontaneous standard-cell reaction.

At each electrode compare the possible aqueous species using reduction or oxidation tendencies. Water may react instead of sulfate or in dilute halide solution; concentrated halide can be oxidized, while molten salt contains no water.

List the solute ion and water as competing possibilities at each electrode, then use electrode-potential data together with stated concentration conditions to select products. In aqueous sulfate, water commonly supplies the anode gas; in concentrated halide, halogen formation may compete. Do not transfer molten-salt products automatically to solution.

The object being coated is the cathode, where coating-metal ions are reduced to metal. Use an electrolyte containing those ions; a coating-metal anode can replenish them.

Trace metal atoms through the circuit: M atoms may oxidize at a soluble anode to maintain Mⁿ⁺, while Mⁿ⁺ gains electrons and deposits on the object. Reversing the object to the anode would remove metal rather than coat it.