Topic 8: Redox Chemistry AS and Groups 1, 2 and 7

Assign the fixed values first, multiply each oxidation number by its atom count, set the total equal to the species charge, and solve for the unknown.

Write one unknown for the requested element, include every subscript and coefficient, and make the weighted total equal to 0 for a neutral compound or to the ion charge for a polyatomic ion.

Apply peroxide and metal-hydride exceptions before using the usual O = −2 and H = +1 rules. Reinsert the result and verify the total charge.

Write the numeral immediately after the relevant element name in parentheses. Use I, II, III, IV, V, VI or VII—not an Arabic numeral or a signed ionic charge.

Treat the given oxidation numbers as signed contributions. Choose the smallest whole-number ratio that makes their total zero, then write the electropositive element first and simplify the subscripts.

Keep a polyatomic ion intact and use brackets when more than one is required. Confirm both overall neutrality and the lowest ratio.

For Mg + 2H$^+$$\rightarrow$Mg$^{2+}$+ H$_2$, magnesium changes 0 to +2 and loses two electrons, so it is oxidised. Hydrogen changes +1 to 0 and gains electrons, so it is reduced.

In s- and p-block reactions, assign oxidation numbers before and after, identify each change, then use the electron count to confirm that total electron loss equals total electron gain.

In Mg + 2H$^+$$\rightarrow$Mg$^{2+}$+ H$_2$, H$^+$ gains electrons and is the oxidising agent; Mg loses electrons and is the reducing agent.

Identify the species containing the atom whose oxidation number decreases: that whole reactant species is the oxidising agent. The reactant whose oxidation number increases is the reducing agent.

In Cl$_2$ + H$_2$O $\rightleftharpoons$ HCl + HClO, chlorine starts at 0. It becomes −1 in HCl and +1 in HClO, so the same Cl$_2$ is both reduced and oxidised.

Confirm one starting oxidation number, at least two products containing that element, one increase and one decrease. Then check the balanced equation.

Cr$_2$O$_7^{2-}$$\rightleftharpoons$2CrO$_4^{2-}$changes colour with conditions, but chromium remains +6 in both ions. It is not redox. Chlorine reacting with cold alkali gives Cl$^-$and ClO$^-$, so Cl changes 0 to −1 and +1: disproportionation.

Assign only the oxidation numbers needed to test change, but compare the same element in reactants and products and cite the numerical changes.

Metals generally form positive ions by losing valence electrons. Their oxidation number increases from 0 in the element to a positive value: Na$\rightarrow$Na$^+$+e$^-$ and Mg$\rightarrow$Mg$^{2+}$+2e$^-$.

Non-metals generally form negative ions by gaining electrons. Their oxidation number decreases from 0: Cl$_2$+2e$^-$$\rightarrow$2Cl$^-$and O$_2$+4e$^-$\rightarrow$2O$^{2-}$.

These opposite electron changes allow metals to act as reducing agents and non-metal molecules such as halogens to act as oxidising agents in many s- and p-block reactions.

Write the changing species, balance its atoms, then add electrons to the more positive side until total charge is equal. For a full ionic equation, multiply half-equations so electron numbers match, add them and cancel electrons and any identical species.

Adding the last pair gives Fe(s)+2Fe$^{3+}$(aq)$\rightarrow$3Fe$^{2+}$(aq). Atoms and total charge are both balanced and no electron remains in the overall equation.

The outer electron feels weaker effective attraction and requires less energy to remove. Group 2 values remain generally higher than neighbouring Group 1 values because a Group 2 atom has a greater nuclear charge with a similar shell pattern.

Down the group, an extra shell increases distance and shielding, effective nuclear attraction for the outer electron falls, ionisation energy decreases, and forming M$^+$ or M$^{2+}$ becomes easier. Reactions with water or oxygen therefore become faster and more vigorous.

Group 2 atoms must lose two electrons rather than one, but the down-group trend is governed by the decreasing energies required to remove their outer electrons.

Magnesium reacts much more readily with steam: Mg(s)+H$_2$O(g)$\rightarrow$MgO(s)+H$_2$(g). Calcium, strontium and barium form increasingly alkaline hydroxide solutions or suspensions with cold water.

Typical evidence includes metal burning in oxygen/chlorine, hydrogen effervescence with water, and an alkaline indicator colour when hydroxide forms.

Soluble products give alkaline solutions. With acid, the basic oxide or hydroxide is neutralised to a salt and water; a solid may dissolve and the mixture may warm.

Reaction with water depends on accessibility and solubility: MgO reacts slowly, while heavier Group 2 oxides react more readily. Acid neutralisation is the common chemical pattern.

Increasing hydroxide solubility makes saturated solutions more alkaline down the group. Decreasing sulfate solubility means adding sulfate ions increasingly gives a white precipitate, with BaSO$_4$ especially useful for sulfate testing.

When comparing observations, separate solubility from reaction rate and use the correct anion. A cloudy mixture or precipitate indicates limited solubility, not absence of ions.

A small and/or highly charged cation has high charge density and strongly polarises the electron cloud of NO$_3^-$ or CO$_3^{2-}$. This weakens bonds within the anion and makes thermal decomposition easier. Down a group, cation radius increases, polarising power falls and thermal stability rises.

For similarly sized cations, a 2+ ion polarises more strongly than a 1+ ion. Lithium resembles magnesium because Li$^+$ is unusually small, so both have appreciable polarising power.

Heat excites electrons in atoms or ions to higher energy levels. When they fall to lower levels, they emit photons whose energies match the level differences. Characteristic wavelengths combine to give a diagnostic flame colour.

The observed colour identifies the metal ion because its allowed energy-level separations are characteristic. Sodium contamination can mask other colours because its yellow emission is intense.

Clean a nichrome/platinum wire loop with concentrated HCl and heat until no colour remains. Moisten it with HCl, pick up the sample, place it in a non-luminous blue flame, record the colour, and clean between samples.

Use similar sample amounts, particle sizes, heating positions and flame conditions. Record both the temperature/heating needed and verified gases rather than inferring decomposition from appearance alone.

CO$_2$ is confirmed by Ca(OH)$_2$+CO$_2$$\rightarrow$CaCO$_3$+H$_2$O. Acidifying the sulfate test removes carbonate interference before Ba$^{2+}$ is added.

Methyl orange changes red in acid through orange at the endpoint to yellow in alkali. Phenolphthalein is colourless in acid and pink in alkali; detect the first permanent very pale endpoint colour appropriate to titration direction.

With carbonate in the flask and HCl in the burette, methyl orange changes from yellow towards the first permanent orange endpoint. Read the burette at eye level to the nearest calibrated precision.

Use a volumetric pipette and filler, do not rinse the conical flask with analyte, and wash flask walls down with distilled water without changing moles present.

For quantities multiplied or divided, add percentage uncertainties. For a burette with ±0.05 cm$^3$ per reading, a titre has ±0.10 cm$^3$; percentage uncertainty is $0.10/\text{titre}\times100$%. A larger sensible titre reduces this percentage.

Obtain concordant titres and average them to improve precision. Report a result to precision supported by the apparatus and include dominant systematic limitations separately.

Pipette a known aliquot of standard acid into a conical flask, add the suitable indicator (commonly phenolphthalein for ethanedioic acid/NaOH), titrate with NaOH to concordant endpoints, and use the balanced equation to find NaOH concentration.

For 100.0 cm$^3$ of 0.0500 mol dm$^{-3}$ H$_2$C$_2$O$_4$$\cdot$2H$_2$O, moles required = 0.00500 mol; multiply by the hydrate's molar mass when calculating the solid mass.

The elements also darken down the group: fluorine pale yellow, chlorine pale green, bromine red-brown and iodine grey-black in standard states.

A more reactive halogen oxidises a less reactive halide: X$_2$+2Y$^-$$\rightarrow$2X$^-$+Y$_2$. X$_2$gains electrons and is reduced; Y$^-$ loses electrons and is oxidised.

HClO/chlorate(I) produced in water is an oxidising disinfectant that kills microorganisms, so chlorine is used in water treatment. Dose must be controlled because chlorine chemistry can also be hazardous.

Bromine and iodine can undergo analogous reactions, with the same need to balance atoms and track the halogen from 0 into lower and higher oxidation states.

Reducing ability increases HCl < HBr < HI because the H–X bond weakens and X$^-$ is increasingly easy to oxidise down the group.

Ag$^+$+X$^-$$\rightarrow$AgX(s). Hydrogen halides form white ammonium halide smoke with ammonia, HX(g)+NH$_3$(g)$\rightarrow$NH$_4$X(s), and form acidic solutions in water, HX+H$_2$O$\rightarrow$H$_3$O$^+$+X$^-$.

F$_2$ should displace every lower halide, while At$_2$ should displace none of Cl$^-$, Br$^-$ or I$^-$. Compounds and displacement behaviour are predicted by continuing the same electron-gain, shielding and polarisability trends.

State the observed trend, identify its cause, place F or At beyond the known sequence, and give a directional prediction. Keep state, colour, electronegativity and redox strength as separate claims.