Exam Revision

Oxidation is the loss of electrons and an increase in oxidation number. Reduction is the gain of electrons and a decrease in oxidation number. Remember OIL RIG.

The oxidising agent accepts electrons and is reduced. The reducing agent donates electrons and is oxidised. Identify the complete reacting species, including its charge.

The sum of oxidation numbers equals the species’ overall charge. An uncombined element is 0; a monatomic ion equals its charge; oxygen is usually −2; hydrogen is usually +1. In Cr₂O₇²⁻, 2x + 7(−2) = −2, so chromium is +6.

Balance atoms other than O and H, then balance O using H₂O, H using H⁺, and charge using electrons. Check atoms and charge. Electrons are products in oxidation and reactants in reduction.

Balance each half-equation as though acidic. Add OH⁻ to both sides to neutralise every H⁺, form water and cancel excess water. Multiply half-equations to equalise electrons, add them and cancel common species.

Oxidation occurs at the negative anode; reduction occurs at the positive cathode. Electrons flow through the external circuit from anode to cathode. Chemical energy is converted into electrical energy.

It allows ion movement, maintaining electrical neutrality and completing the circuit. In a typical metal/metal-ion cell, anions migrate towards the anode compartment and cations towards the cathode compartment. Electrons do not travel through the salt bridge.

Read the half-equations. Metal forming metal ions means the electrode loses mass and metal-ion concentration increases. Metal ions forming metal means the electrode gains mass and metal-ion concentration decreases. An inert electrode conducts electrons without being a reactant.

For the relevant species present, the half-equation with the more positive reduction potential undergoes reduction; reverse the other for oxidation. E°cell = E°cathode − E°anode, using both tabulated reduction potentials. A positive result predicts thermodynamic feasibility under standard conditions.

Never multiply electrode potentials when multiplying equations. Choose reactants actually present. Nonstandard conditions can change predictions, and a feasible reaction may be slow because of activation barriers or a protective surface layer.

Alkanes are hydrocarbons with only single bonds; alkenes contain C=C; haloalkanes have a halogen replacing an alkane hydrogen; alcohols contain –OH; carboxylic acids contain –COOH; esters contain R–C(=O)–O–R′.

Identify the family, choose the longest appropriate parent chain containing the principal functional group or double bond, number to prioritise its position, locate and alphabetise substituents, and add the correct suffix. Carbon stems are meth, eth, prop, but, pent, hex, hept and oct. Check each carbon has four bonds.

Name the group attached to the single-bonded oxygen first as an alkyl group. Name the carbonyl-containing chain second as an alkanoate, counting the carbonyl carbon. CH₃COOCH₂CH₃ is ethyl ethanoate.

Identify polar bonds from electronegativity differences, then consider molecular shape. If the bond dipoles cancel, the molecule is nonpolar; if they do not, it is polar. CO₂ has cancelling dipoles; H₂O does not.

All molecules have dispersion forces. Polar molecules also have dipole–dipole attractions. Suitable molecules with H bonded to N, O or F can also form hydrogen bonds. Identifying the strongest force does not mean the other forces disappear.

A hydrogen covalently bonded to N, O or F is attracted to a lone pair on N, O or F of another molecule. An ester can accept hydrogen bonds from water, but ordinary ester molecules cannot hydrogen-bond to each other because they lack a suitable hydrogen donor.

Identify and compare the intermolecular attractions. Stronger attractions generally give higher boiling points, lower volatility and greater viscosity for comparable substances. Longer chains generally strengthen dispersion forces. Boiling separates molecules rather than breaking their covalent bonds.

Identify attractions it can form with water, especially hydrogen bonding, and compare them with attractions disrupted during mixing. A polar functional group favours water solubility; a larger nonpolar hydrocarbon region generally reduces it. Ethanol hydrogen-bonds with water; hexane does not.

n = m/M; n = cV; N = nNₐ. Nₐ = 6.02 × 10²³ mol⁻¹. Use mass in grams with molar mass in g mol⁻¹, and volume in litres with concentration in mol L⁻¹.

Find moles of the compound, multiply by the number of that atom or ion per formula unit, then multiply by Nₐ. Ca(NO₃)₂ contains 2 N atoms, 6 O atoms and 2 nitrate ions per formula unit. Include waters of crystallisation.

Balance the equation, convert the known quantity to moles, apply the coefficient ratio, and convert to the requested quantity. For gravimetry: precipitate mass → precipitate moles → target substance moles → target mass or concentration. Mass percentage uses the original sample mass.

mol L⁻¹ means moles per litre of solution; g L⁻¹ means grams per litre; %(m/v) means grams per 100 mL; %(v/v) means mL per 100 mL. ppm by mass means mg per kg and approximately equals mg L⁻¹ for dilute aqueous solutions with density about 1 g mL⁻¹. g L⁻¹ = mol L⁻¹ × molar mass.

When the amount of the specified solute stays constant, with no reaction, solute loss or solute addition. Use final solution volume, not just water added. To undo dilution, multiply the diluted concentration by final volume divided by the original aliquot volume.

Identify the ions, use the solubility table, construct the insoluble product with balanced charges, balance the equation and cancel spectator ions. Keep the precipitate intact. Example: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). If all products remain soluble, there is no precipitation reaction.

A Brønsted–Lowry acid donates H⁺; a base accepts H⁺. Conjugate pairs differ by one H⁺. An amphiprotic species can donate or accept H⁺, such as HCO₃⁻. A polyprotic acid can donate more than one proton per molecule in successive steps.

Strength describes the extent of ionisation or reaction with water: strong is essentially complete; weak is partial. Concentration describes the amount of solute per volume of solution. A concentrated weak acid is still weak. HCl is strong, CH₃COOH is weak and NH₃ is a weak base.

A reactive metal with a suitable acid produces a salt and H₂. A carbonate with an acid produces a salt, CO₂ and H₂O. A hydroxide with an acid produces a salt and H₂O. Balance formulas, atoms and charges, include states, and retain solids and weak acids intact in net ionic equations.

pH = −log₁₀[H₃O⁺] and [H₃O⁺] = 10^(−pH). For a strong monoprotic acid, hydronium concentration approximately equals acid concentration when water’s contribution is negligible. One pH unit represents a tenfold hydronium concentration difference.

Find [OH⁻] from its formula and concentration first; for Ba(OH)₂, [OH⁻] = 2c. Then pOH = −log₁₀[OH⁻] and pH = pKᵥ − pOH, where pKᵥ here denotes pKw. At 25°C, pKw ≈ 14. Neutral means [H₃O⁺] = [OH⁻], not necessarily pH 7 at every temperature.

Components have different relative attractions to the stationary and mobile phases. Greater stationary-phase attraction means less travel; greater mobile-phase affinity means more travel. Do not assume that a more polar substance travels less without identifying both phases.

Rf = distance from origin to spot centre ÷ distance from origin to solvent front. It has no units and normally lies between 0 and 1. Compare substances under the same conditions; matching Rf values support identification but do not prove it alone.

Draw the origin in pencil, keep it above the initial solvent level, apply small concentrated spots and mark the solvent front promptly. Multiple spots indicate multiple separated components; one spot alone does not guarantee purity.

Balance the equation, calculate titrant moles using n = cV and the mean titre, apply the mole ratio to find analyte moles, divide by the original aliquot volume, then undo any initial dilution. Do not assume a 1:1 ratio or divide by the total liquid volume in the conical flask.

Each titre equals the final burette reading minus the initial reading. Average the concordant accurate titres within the specified tolerance. Normally exclude the rough titre; do not discard a result merely because it differs from your expected answer.

Equivalence occurs when reactants are present in the exact reacting mole ratio. The endpoint is the observed indicator colour change. Choose an indicator whose transition range lies within the steep pH change near equivalence. Phenolphthalein suits typical weak-acid/strong-base titrations.

Rinse the burette with titrant, the pipette with the solution being transferred, and the conical flask with deionised water. Residual water in the burette or pipette dilutes its solution; water added to the flask does not change the measured analyte moles.

Trace error → measured quantity → calculated result. Wet precipitate gives a mass that is too high, so calculated analyte amount is too high. Lost precipitate gives a mass that is too low, so the result is too low. Explain the chain rather than only saying “inaccurate”.

Accuracy is closeness to the accepted value; precision is agreement between repeated results. Random error causes unpredictable variation and can be reduced through suitable repeats and averaging. Systematic error causes consistent bias and requires correction of its cause. Precise results can still be inaccurate.

Use sufficient precipitating reagent, check precipitation is complete, transfer and filter without loss, wash appropriately and dry to constant mass. The precipitate should be pure, of known composition and sufficiently insoluble. Use the original analysed sample volume when calculating its concentration.

The C=C becomes C–C and atoms or groups add across it. Adding H₂ produces an alkane with a metal catalyst; Br₂ produces a dibromoalkane and decolourises bromine; HX produces a haloalkane; H₂O produces an alcohol under suitable acid-catalysed conditions. Bromine addition does not produce an alcohol.