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An increase in oxidation state records oxidation; a decrease records reduction. Thus ${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}$ raises the iron oxidation state by one, while ${\mathrm{MnO}}_4^{-}\to {\mathrm{Mn}}^{2+}$ lowers the manganese oxidation state from $+7$ to $+2$. Five iron ions supply the five-electron inventory required per permanganate ion.

In an acidic chemical solution, the complete half-reactions areFirst balance oxygen with four water molecules, then hydrogen with eight protons, and finally electric charge with five electrons. Multiply the iron half-reaction by five and add; the exchanged electrons cancel:Both sides contain one manganese, four oxygen, eight hydrogen and five iron chemical atoms, with net electric charge $+17e$. The ${\mathrm{H}}^{+}$ notation represents the acidic aqueous proton inventory; it does not imply bare protons moving independently of water. These half-reactions and their sum are the acidic balance used in the documented permanganate-iron titration.^[ref]

The coefficients predict amounts consumed, not a collision of five ions or an elementary reaction. Bulk chemical solution also contains compensating counterions, which are omitted from this net chemical equation. The hydrogen and oxygen oxidation states are unchanged; balancing their inventories remains essential even though they do not account for the exchanged electrons.

The historical operational fraction of Ammonia released by alkaline permanganate after preliminary removal of free Ammonia. It is neither a specific albumen measurement nor necessarily all organic nitrogen.

Sum oxidation states over all chemical atoms of the specified molecule, ion or formula unit, including multiplicities:Here $q$ is the net electric charge and $e$ the positive elementary charge. For a neutral entity the sum is zero; for an ion it equals its signed electric charge number $z$. The reason is simple: the allocation assigns every valence electron exactly once, so the sum is the neutral electron budget minus the actual electron count.

For water, $2(+1)+(-2)=0$; for hydrogen peroxide, $2(+1)+2(-1)=0$. In permanganate, ${\mathrm{MnO}}_4^{-}$, assigning each oxygen $-2$ gives $x+4(-2)=-1$, so manganese has oxidation state $+7$. This is not a free ${\mathrm{Mn}}^{7+}$ ion: the entire permanganate anion has net electric charge $-e$. The sum rule supplies a constraint; without additional assignments it cannot determine every individual oxidation state.

In the ionic assignment, both tetrahedral anions would require Mn(VII), $d^0$. Such a high oxidation state is strongly oxidizing and is best supported by strong bonding to oxygen. Short Mn–O bonds combine good orbital overlap with oxygen's resistance to further oxidation. This stabilizes permanganate relative to a hypothetical all-sulfide analogue.

sulfide is larger, more polarizable and much easier to oxidize than oxide. An attempted Mn(VII) sulfide would therefore combine an exceptionally oxidizing metal center with reducing ligands: internal redox can reduce manganese while oxidizing sulfur, for example toward S–S-bonded or elemental sulfur products. The larger sulfur donors also provide less favorable short multiple bonding to the small, highly charged manganese center. The oxygen compound supports Mn(VII); the sulfide combination is redox-incompatible. This predicts instability without requiring a specific unprovided decomposition product.

Iron in nonoxidizing hydrochloric acid principally gives an iron(II) solution; tin in oxidizing aqua regia principally gives a tin(IV) solution. These are the two preparations paired most directly by “respectively”. Their contrast is clearer when the other acid treatment of each metal is considered as well. “Nitrohydrochloric acid” is the historical name for aqua regia, containing nitric and hydrochloric acids.

For iron in dilute hydrochloric acid,The fresh solution is typically pale green when sufficiently concentrated. Hydroxide gives a pale green precipitate of iron(II) hydroxide, which darkens and turns brown on exposure to air as oxidation proceeds. Ammonia can similarly precipitate a ferrous hydroxide rather than forming a stable soluble excess-ammonia complex under the usual test conditions. Ferricyanide gives a deep blue product; the modern structural account of this blue material should not be replaced by an overly literal formula inferred solely from the reagent names. Ferrous solution also reduces a suitable oxidant such as acidic permanganate:

If instead oxidizing acid treatment produces iron(III), the solution exhibits ferric reactions: hydroxide or Ammonia gives red-brown iron(III) hydroxide; thiocyanate gives a strong red colour; and ferrocyanide gives Prussian blue. Chloride complex formation affects colour and some equilibria, so “yellow” is not a universal iron(III) identity test. Oxidizing aqua regia supplies this ferric comparison rather than merely generating more of the same ferrous salt.

Tin in ordinary hydrochloric acid gives a tin(II) solution with hydrogen evolution:Its characteristic property is reducing power. With mercury(II) chloride, controlled addition can first form white mercury(I) chloride and then, with further reductant, grey metallic Mercury:The chloride formulae are useful net bookkeeping, not a complete statement of all aqueous chloro-complexes.

In aqua regia, tin is oxidized predominantly to tin(IV). A representative balance, choosing nitrogen monoxide as the nitrate-reduction product, isActual gas products depend on the acid conditions. Sufficient chloride and acidity retain tin in solution; dilution can cause hydrolysis and precipitation of hydrated tin(IV) oxide. This differs from the simple expectation that every dilution leaves unchanged molecular ${\mathrm{SnCl}}_4$ in water. Hydrogen sulfide in suitable acidic conditions gives yellow tin(IV) sulfide, compared with brown tin(II) sulfide from a stannous solution. Tin(IV) does not reproduce the characteristic reducing reaction of fresh tin(II) with mercuric chloride; reduction to tin(II) is needed first. Oxidation during storage can therefore change the reactions of a nominally stannous preparation.

Combine investigation of the source with several independent analyses; a clear appearance or a single negative chemical test cannot exclude sewage pollution. No sample results are supplied, so the answer is an examination scheme rather than a finding about an actual water sample.

Record where and when the water was collected, recent rainfall, proximity of drains and cesspits, and possible routes from sewage to the supply. Compare with a credible uncontaminated sample from the same geological setting. Use clean, suitably closed vessels and examine promptly; contamination introduced during sampling and chemical changes during storage can both mislead. Separate portions are needed when one analysis alters the sample for another.

For an 1867 investigation, inspect colour, turbidity and sediment, and examine concentrated sediment by microscopy for suspicious organic debris. These observations can motivate further work but cannot identify all disease-causing agents. Determine residue after evaporating a measured volume; carefully heating that residue may reveal charring or a volatile fraction. Loss on ignition is not a specific measure of sewage: mineral constituents can also change on heating, and some organic material can be lost during evaporation itself.

Test a separate aliquot for chloride, for example by silver nitrate precipitation, and compare its quantity with local background. Elevated chloride can accompany sewage but also comes from natural salts. Examine Ammonia or ammonium and oxidized nitrogen, nitrite and nitrate. Organic nitrogen can be converted during decomposition and nitrification, so an old contamination may have little free Ammonia. Conversely, a nitrogen signal alone does not establish its origin.

An oxidant-consumption experiment with a known amount of potassium permanganate, under controlled reaction time, temperature and acidity, supplies another comparison. Residual oxidant is measured against a reagent blank. A large consumption indicates oxidizable material, but reducing inorganic constituents also consume oxidant, while some organic materials react incompletely. It is therefore an operational test, not a direct count of sewage organisms.

A method actually reported in June–July 1867 first distils off free Ammonia, then uses alkaline permanganate to generate further Ammonia from some organic nitrogen. A boiling flask, condenser and successive receiving vessels separate the fractions, whose Ammonia can be compared by colour with standards using Nessler's reagent. The later fraction became known as albuminoid ammonia. It is neither proof of the presence of albumen specifically nor complete conversion of every organic nitrogen compound. Contemporary discussion already disputed how different substances behaved in the preliminary distillation.^[ref]

A present-day assessment adds validated microbiological water testing, including a specific Escherichia coli test using culture or an accepted equivalent, together with quality control and a sanitary survey. Escherichia coli is an indicator of faecal contamination, not a label for all sewage-derived pathogens. The Australian drinking water guideline specifies none detected in a $100\mathrm{mL}$ sample and notes that the indicator does not reliably represent all enteric viruses or protozoa.^[ref]

The justified conclusion comes from the combined evidence and the detection limits. Chemical indicators can support a sewage-contamination hypothesis, but chemical clarity or a negative result in one aliquot is not a certificate of microbiological safety. This modern distinction should not be attributed to an examiner as a fully developed 1867 laboratory protocol.

The apparatus and motivating measurement are described in permanganate-iron titration; the following is an original account of its possible errors. Assume correct titrant standardization, the acidic permanganate-iron balance, quantitative conversion of the sampled ${\mathrm{Fe}}^{2+}$, and additive contributions expressed as equivalent titrant volumes. WriteHere $V$ is the sample titre, $V_b$ the titration blank, $B_s,B_b$ background reagent consumption, and $D_s,D_b$ signed offsets of the observed titration endpoints from the respective stoichiometry conditions after background consumption. The subscripts distinguish sample and blank. $I$ is additional sample-specific permanganate demand, absent from the blank; $A$ is apparent delivery used to fill an initially air-containing burette tip rather than enter the flask. Assume the blank's tip is already filled; the sample should also have $A=0$ after proper preparation. These terms are a diagnostic model, not measured components of the stipulated $0.10\mathrm{mL}$ blank.

Subtracting the two accounts proves the error relationThe simple inference is valid when this residual is negligible. Matching water and sulfuric acid tests the background terms, but does not by itself establish $D_s=D_b$: flask composition and viewing conditions can differ. A coloured sample can mask residual permanganate, changing the assigned titration endpoint without a corresponding redox consumption. A sample-specific reducing agent instead contributes positive $I$ by consuming permanganate. A reagent titration blank cannot measure either effect just by omitting the sample.

A transient purple streak before mixing is not the mixed-flask titration endpoint. Conversely, the equivalence point is the five-to-one inventory match, without a promised visible flag. In the ideal selective model, a persistent excess colour requires some permanganate beyond that match. No particular drop size, waiting time or endpoint offset is stipulated here; the same stated colour criterion must be applied to sample and blank, and its transferability checked.

The model also predicts error direction. If a tip bubble disappears during sample delivery, $A>0$ makes the inferred amount of substance too high because final minus initial burette reading includes liquid that stayed in the tip. Conditioning the tip before the first reading prevents this particular discrepancy; subtracting a separate titration blank is not an instrument repair. Repeated aliquots test scatter, a known added ${\mathrm{Fe}}^{2+}$ inventory tests the five-to-one response, and an independent titrant standardization tests $c$. These are the proposed controls from permanganate-iron titration, with distinct diagnostic roles; none alone proves absence of all sample interference.

A proposed modern teaching measurement puts a pipette-measured portion of an iron ${\mathrm{Fe}}^{2+}$ chemical solution in a conical flask, acidifies with excess dilute sulfuric acid, and adds standardized potassium permanganate from a clamped burette while swirling. The documented endpoint principle is that added purple permanganate is consumed until the first small excess leaves a pale purple colour; no separate colour indicator is required. Insufficient acidity can produce brown manganese dioxide, invalidating the assumed five-electron half-reaction.^[ref]

Record initial and final burette readings; their difference is the delivered volume $V$. The following controls are proposed additions to test the inference. Run a reagent blank with the same water, sulfuric acid and endpoint criterion but no sample, obtaining $V_b$. Use a separately established potassium permanganate concentration $c$, clear the burette tip of air, and repeat sample portions. A known added amount of substance of ${\mathrm{Fe}}^{2+}$ should increase consumption by one fifth of that amount. A blank estimates reagent demand and the small endpoint excess; it cannot remove an interfering sample reactant.

Assuming complete, selective reaction according to the acidic permanganate-iron balance, the corrected inference isWith $c$ in $\mathrm{mol}{\mathrm{L}}^{-1}$, use $V$ and $V_b$ in litres. As an original stipulated calculation, not an experimental result, take $c=0.02000\mathrm{mol}{\mathrm{L}}^{-1}$, $V=12.60\mathrm{mL}$ and $V_b=0.10\mathrm{mL}$. Then the permanganate consumption is $0.2500\mathrm{mmol}$ and the inferred ${\mathrm{Fe}}^{2+}$ amount is $1.250\mathrm{mmol}$. In a $25.00\mathrm{mL}$ sample portion this corresponds to $0.05000\mathrm{mol}{\mathrm{L}}^{-1}$.

In this example the corrected volume is $12.50\mathrm{mL}=0.01250\mathrm{L}$. The acidic permanganate-iron balance gives $z_A=1$ for ${\mathrm{Fe}}^{2+}$ and $z_T=5$ for permanganate. Thus $0.2500\mathrm{mmol}$ of permanganate accepts $1.250\mathrm{mmol}$ of formal electrons, exactly what $1.250\mathrm{mmol}$ of ${\mathrm{Fe}}^{2+}$ supplies. The amount of substance ratio is five to one, despite equality of the exchanged electron amounts.

The denominator is the original $25.00\mathrm{mL}$ aliquot, not its final volume after adding sulfuric acid, water and titrant. Such dilution changes concentrations in the flask without changing the initial sampled amount of substance of ${\mathrm{Fe}}^{2+}$. Omitting the $0.10\mathrm{mL}$ titration blank would instead give $1.260\mathrm{mmol}$ and $0.05040\mathrm{mol}{\mathrm{L}}^{-1}$, an overestimate of $0.80\%$ relative to the corrected constructed value. The displayed digits express the stipulated calculation, rather than establish its measurement uncertainty.

The endpoint detects residual permanganate, not the iron identity directly. Other sample reactants that consume permanganate raise the apparent ${\mathrm{Fe}}^{2+}$ amount; prior oxidation of sample ${\mathrm{Fe}}^{2+}$ to ${\mathrm{Fe}}^{3+}$ lowers it. A recovery check probes response to the added iron but does not prove absence of every interference. The readout determines the reacting ${\mathrm{Fe}}^{2+}$ inventory under the stated conditions, rather than total iron in an arbitrary mixture. No measured titres are asserted here.

A salt of potassium cations and permanganate anions, supplying permanganate in a redox titration.