3.2 Electron transfer reactions

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.