Oxidation, Reduction and the Electron That Makes the Picture

Metol develops a film and sodium thiosulfate does not, although thiosulfate carries twice the sulfur, dissolves far more readily and costs a fraction as much. Sodium sulfite protects a developer and sodium sulfate does nothing at all, although the two differ by one oxygen atom. A print developer that has gone brown in the tray no longer reaches black. Every one of those facts is about electrons: which molecule will hand one over, how willingly, and to what.

One idea, and everything else is bookkeeping

Oxidation is the loss of electrons. Reduction is the gain of electrons. They happen together, always, because an electron that leaves one species arrives at another. There is no such thing as an oxidation on its own.

The older definition — oxidation is combination with oxygen — is where the word came from, and it is still the one Kodak’s 1924 primer leads with. Its chapter on elementary chemistry defines oxidation as raising an element “from the level of oxidation of one of its compounds to another level in which it is combined with more oxygen”, and defines a chemical reducer as a substance with “an affinity for oxygen” that “can liberate the metals from their salts”. That description works for the reactions in front of it and fails for the central one, because when a developer turns silver bromide into silver, no oxygen is involved anywhere. The electron definition covers both.

Two words that follow, and that are constantly misused:

• An oxidising agent is the species that takes electrons. It is itself reduced.
• A reducing agent is the species that gives electrons. It is itself oxidised.

These are roles in a particular reaction, not permanent properties of a substance. Sulfite is a reducing agent when it meets quinone and an oxidising agent when it meets a strong enough reductant. Iron(III) oxidises hydroquinone; iron(II) is the reducing agent that makes a cyanotype. The question is never “is this an oxidiser?” but “is this an oxidiser towards that?”

The one transaction, drawn once

1. 1The reducing agent, before — holds the electrons it is about to give up — a developing agent, sulfite, iron(II)
2. 2The electrons in transit — two of them here; the number is the thing that must balance
3. 3The oxidising agent, before — the silver ion, Ag⁺, which can accept one electron each
4. 4Reduced: metallic silver — Ag⁺ + e⁻ → Ag. This is the picture
5. 5Oxidised: the spent reducing agent — quinone from hydroquinone, sulfate from sulfite, iron(III) from iron(II)

Nothing else on this page is a new idea. Oxidation states, half-equations and the electrochemical series are all ways of keeping track of this one transaction.

Oxidation states: counting electrons you cannot see

An oxidation state (or oxidation number) is an accounting device. OpenStax puts it exactly: for an atom in a compound it is “the charge the atom would have in the compound if the compound were ionic”. It is not a real charge on a real atom in a covalent molecule; it is a way of noticing that electrons have shifted.

The rules you need here are short:

1. An atom in an element has oxidation state 0. Silver metal is 0, sulfur is 0, oxygen in O₂ is 0.
2. A simple ion has the oxidation state of its charge. Ag⁺ is +1, Br⁻ is −1, Fe³⁺ is +3.
3. In a compound, oxygen is normally −2 and hydrogen +1.
4. The sum over all the atoms equals the charge on the species. That is the rule that does the work.

Rule 4 lets you find the awkward one by subtraction. In the sulfite ion, SO₃²⁻, the three oxygens contribute 3 × (−2) = −6, and the total must be −2, so sulfur is +4. In sulfate, SO₄²⁻, four oxygens give −8 against a total of −2, so sulfur is +6. The single oxygen that separates sulfite from sulfate is exactly two electrons’ worth of change on the sulfur, which is why sulfite is a reducing agent and sulfate — having already given those electrons away — is nothing at all.

Sulfur, from −2 to +6, with photography's species placed on the axis

1. 1Sulfide, S²⁻, at −2 — the end of the road; silver sulfide is the sepia image
2. 2Sulfur, S, at 0 — the element
3. 3Thiosulfate, S₂O₃²⁻, average +2 — two sulfur atoms that are not alike; +2 is their mean
4. 4Tetrathionate, S₄O₆²⁻, average +2.5 — what thiosulfate becomes when something oxidises it
5. 5Sulfite, bisulfite, metabisulfite, SO₂, all +4 — the preservative family, all at the same state
6. 6Sulfate, SO₄²⁻, at +6 — spent: it has no more electrons to give

Read left to right for oxidation. The whole of a sulfite preservative's working life is the single hop from +4 to +6.

Two of those entries need an honest caveat. Thiosulfate’s +2 is an average. PubChem’s recorded connectivity for the ion shows a central sulfur carrying two oxygens and one terminal sulfur, so the two sulfur atoms sit in visibly different environments and neither of them is really “+2”; the formalism divides the total between them and returns the mean. Tetrathionate’s +2.5 is an average of four atoms and is not a state any atom occupies. Use those numbers to see which way a reaction has moved, not as a description of an atom.

The rest of the course’s redox cast, assigned the same way:

Species	Element	Oxidation state	How you get it
Ag metal	silver	0	an element
Ag⁺ in AgNO₃ or AgBr	silver	+1	its ionic charge
Br⁻ in KBr or AgBr	bromine	−1	its ionic charge
Br₂	bromine	0	an element
Fe²⁺ (“ferrous”)	iron	+2	its ionic charge
Fe³⁺ (“ferric”)	iron	+3	its ionic charge
Cr³⁺	chromium	+3	its ionic charge
Cr₂O₇²⁻, dichromate	chromium	+6	2Cr + 7(−2) = −2, so Cr = +6
O₂	oxygen	0	an element
O in water or in an oxide	oxygen	−2	rule 3

☠Chromium(VI): studied, not used

Dichromates were once the working chemistry of the carbon and gum processes and of several bleaches and intensifiers, and the course explains what chromium(VI) does there. It does not put it in a darkroom. PubChem’s aggregation of ECHA notifications for potassium dichromate carries H340 (may cause genetic defects), H350 (may cause cancer) and H360 (may damage fertility or the unborn child) alongside H272 (oxidiser), H300 and H330 (fatal if swallowed, fatal if inhaled), H314 (severe skin burns and eye damage) and H410 (very toxic to aquatic life with long-lasting effects). Chromium(III), which is what chrome alum contains, is a different substance with a different classification, and the two must never be spoken of as one thing. Where a historical process needs chromium(VI), Part XXVI treats it as chemistry and history.

Half-equations: doing the count out loud

A half-equation shows one side of the transaction with the electrons written in. Reduction has electrons on the left, where they are consumed; oxidation has them on the right, where they are produced.

The photographic reduction is as simple as a half-equation gets:

Ag⁺ + e⁻ → Ag

Reduction: the silver ion becomes silver metal. Atoms balance; charge is +1 − 1 = 0 on the left and 0 on the right.

The developing agent supplies the electron, and is oxidised. Kodak’s 1928 primer names the reactants and products for hydroquinone plainly: hydroquinone is oxidised to quinone, and if you add sulfite to quinone, “the quinone oxidizes the sulphite to sulphate and is itself reduced again to hydroquinone”. The electron bookkeeping below is not quoted from that primer — it is worked here, with the half-reaction method of OpenStax section 17.1, from the reactants and products the primer states, and then checked atom by atom and charge by charge. A developer is alkaline, so the hydroxide form is written.

C₆H₄(OH)₂ + 2 OH⁻ → C₆H₄O₂ + 2 H₂O + 2 e⁻

Oxidation: hydroquinone (C₆H₆O₂) to quinone (C₆H₄O₂), two electrons released

Check it. On the left, C₆H₄(OH)₂ is C₆H₆O₂; adding 2 OH⁻ gives C₆H₈O₄ and a charge of −2. On the right, C₆H₄O₂ plus 2 H₂O is C₆H₈O₄, charge 0, plus two electrons for a total of −2. Atoms match; charge matches.

Now combine. One hydroquinone gives two electrons, and each silver ion takes one, so the silver half-equation is doubled:

2 Ag⁺ + C₆H₄(OH)₂ + 2 OH⁻ → 2 Ag + C₆H₄O₂ + 2 H₂O

Development, as a balanced whole: two silver ions reduced per molecule of developing agent oxidised

The electrons have cancelled, which is the test that you got the multiplication right. Charge is +2 − 2 = 0 on the left and 0 on the right.

📷What that stoichiometry says about a tray

One molecule of hydroquinone reduces two silver ions. A litre of a developer carrying 5 g of hydroquinone (molar mass 110.11, so 0.045 mol) therefore has enough electrons in it for about 0.09 mol of silver — roughly 10 g. That is the ceiling the chemistry sets, before anything about rate, oxidation, restraint or exhaustion is considered, and a real capacity is far below it. The point is not the number but its origin: capacity is an electron count, and Part VIII turns that into the capacities a manufacturer actually publishes.

Redox potential, and how far it can be trusted

The half-equations say what can happen. Standard reduction potential, E°, is the number that says which way it will tend to go. OpenStax section 17.3 is careful about what it is: a single half-cell potential cannot be measured, only a difference, so the standard hydrogen electrode is defined as exactly 0 V and everything is quoted against it, under standard-state conditions of 1 mol/L, 1 bar and 298 K. The interpretation is the useful part: the more positive the potential, the stronger the species on the left is as an oxidising agent — the more it wants those electrons. Arrange the couples in order and you have the electrochemical series.

Couple (written as a reduction)	E° / V	What it means here
Au³⁺ + 3 e⁻ → Au	+1.498	gold takes electrons from almost anything: the basis of gold toning
Cl₂ + 2 e⁻ → 2 Cl⁻	+1.358	chlorine is a powerful oxidiser
O₂ + 4 H⁺ + 4 e⁻ → 2 H₂O	+1.229	oxygen in acid; the developer’s enemy
Cr₂O₇²⁻ + 14 H₃O⁺ + 6 e⁻ → 2 Cr³⁺ + 21 H₂O	+1.232	dichromate in acid
Br₂(aq) + 2 e⁻ → 2 Br⁻	+1.087	why free bromine must be mopped up in an emulsion
Ag⁺ + e⁻ → Ag	+0.7996	the reaction that makes the picture
Fe³⁺ + e⁻ → Fe²⁺	+0.771	the iron couple of cyanotype and kallitype
I₂ + 2 e⁻ → 2 I⁻	+0.536	iodine, the weakest of the halogens here
O₂ + 2 H₂O + 4 e⁻ → 4 OH⁻	+0.401	the same oxygen, in alkali
[Ag(NH₃)₂]⁺ + e⁻ → Ag + 2 NH₃	+0.373	silver held by ammonia
AgCl + e⁻ → Ag + Cl⁻	+0.222	silver held in a chloride lattice
[Ag(S₂O₃)₂]³⁻ + e⁻ → Ag + 2 S₂O₃²⁻	+0.017	silver held by thiosulfate
2 H⁺ + 2 e⁻ → H₂	0.000	the reference, by definition

Every value is from OpenStax Appendix L at 25 °C. (The appendix prints the thiosulfate complex’s charge as 3+; it is 3−, and the slip is worth noticing because it is the kind of thing a table can contain.)

The ladder: who can oxidise whom

1. 1Read upwards for oxidising power — a couple higher up can oxidise the reduced form of one lower down, at standard conditions
2. 2The silver row is where photography sits — +0.7996 V: strong enough to be worth reducing, weak enough that ordinary organic molecules can do it
3. 3The four silver entries are one element — free ion +0.80, ammonia complex +0.37, chloride lattice +0.22, thiosulfate complex +0.02

Values from OpenStax Appendix L at 25 °C and standard-state concentrations. A darkroom is at none of those concentrations, which is the caution the text below spends its time on.

That last row is the most important thing on the ladder, and the complex-formation page is built on it. The silver ion’s potential falls from +0.80 V free in solution to +0.02 V when two thiosulfate ions are attached to it. Binding a metal ion changes how badly it wants an electron — which is another way of saying that solubility, stability and redox behaviour are the same question asked three ways.

!What a tabulated potential does not tell you

Kodak’s own 1928 primer, discussing what it calls the “reduction potential” of developing agents, warns that it “alone does not determine the speed with which the developer develops the image, because this depends chiefly upon the rate at which the developer diffuses into the film and on the quantity of developing agent and other substances in the developer”. A high reduction potential, it says, lets a developer “continue to develop more nearly at a normal rate under adverse circumstances, such as at a low temperature or in the presence of bromide”. So the table ranks tendency, not speed. Four things between it and the darkroom: real concentrations are nothing like 1 mol/L, so the working potential differs from the tabulated one; pH shifts the potential of every couple that involves H⁺ or OH⁻, which is most of them; the reaction happens on an adsorbed surface rather than in free solution; and rate is a separate subject entirely, taken up two pages on. A tabulated potential is a strong hint and a poor prophecy.

Why the exposed grain develops and the one beside it does not

Here is the sentence that contains the whole of development selectivity, from Kodak’s 1928 primer: a developing agent’s “affinity for oxygen must be within certain narrow bounds; it must be a sufficiently strong reducer to reduce the exposed silver salt, and at the same time must not affect that which has not been exposed.”

That is the design constraint, and it is unforgiving in both directions. A reducing agent too weak develops nothing. A reducing agent too strong reduces every crystal it touches, exposed or not, and what you get is not a picture but a uniformly grey sheet — chemical fog. The same primer notes that too much alkali pushes a developer in exactly that direction.

What makes an exposed crystal different is a latent-image speck: a few atoms of silver deposited by light at a particular place on the crystal. Development happens preferentially there. The silver in the rest of the crystal is chemically identical to the silver in the crystal next door — the difference is that one crystal has a site where the reaction will start and the other does not.

Two crystals, one difference

1. 1Latent-image speck — a few atoms of silver, put there by light; invisible by any ordinary means
2. 2Developer arrives at both crystals — the same molecules, the same concentration, the same time
3. 3Exposed: the whole crystal is reduced — the speck is where the reaction can start; from there it takes the lot
4. 4Unexposed: nothing happens in time — not "cannot": will not, at a useful rate — which is the kinetics page

Amplification is the point: a handful of atoms decides the fate of a whole crystal. Part IV puts numbers on the ratio and on how the speck does it.

Two honest qualifications. First, “does not develop” means does not develop appreciably in the time you allow; an unexposed grain would eventually reduce, and that is a rate statement rather than a thermodynamic one. Second, why the speck works is a question about semiconductors and surfaces that this page deliberately stops short of; Part IV takes up the latent image and development as amplification, with the mechanism and the evidence for it.

The developing agents, and what becomes of them

A developing agent is a reducing agent chosen to sit inside those narrow bounds. The three the course uses most are hydroquinone, metol and phenidone; pyrogallol and ascorbic acid appear later, and Foma still sells a phenidone-and-isoascorbate paper developer, so the family is not a historical curiosity.

They do not behave alike, and Kodak’s 1928 primer explains why in terms it calls reduction potential, ranking the agents it stocked from the lowest to the highest: hydroquinone, then Athenon, then pyro, then Kodelon (para-aminophenol), and finally Elon — the name Kodak used for metol. The consequence it draws is a visible one. Agents of high reduction potential “make the image flash up all over at once, because they start development very quickly even in the lesser exposed portions”, while agents of low potential “bring up the highlights of the image first and the shadows do not fully appear until the highlights are somewhat developed.”

The oxidised product is not a bystander. This is the part beginners skip and then cannot explain their negatives.

• Hydroquinone gives quinone, which can go on to oxidise other things — including the sulfite meant to protect it.
• Pyrogallol’s oxidation product is yellow and is deposited in the film along with the silver, so, as the primer puts it, a pyro developer without sulfite gives “a very yellow negative, the image consisting partly of silver and partly of the oxidized pyrogallol”. That stain is not a defect in staining developers; it is the image, and it adds density where the silver is densest.
• Metol’s product is not deposited in a coloured form, which is why the same primer says a sulfite-rich hydroquinone image comes out “almost as blue an image as with Elon”.

So the colour of a negative is a redox statement. Part VIII is where the agents are the subject: superadditivity, the organic mechanisms, and why two agents together do more than the sum of what each does alone.

The sulfur system: preservative, not developer

Sodium sulfite is in almost every developer, and its job is redox. Sulfur at +4 is two electrons short of its comfortable +6, and it gives them up readily. Air will take them:

2 SO₃²⁻ + O₂ → 2 SO₄²⁻

Sulfite oxidised by dissolved oxygen to sulfate, which the 1928 primer writes in its own notation as Na₂SO₃ + O = Na₂SO₄

and so will quinone, which is the mechanism the primer describes explicitly:

C₆H₄O₂ + SO₃²⁻ + H₂O → C₆H₄(OH)₂ + SO₄²⁻

Sulfite reduces quinone back to hydroquinone and is itself oxidised to sulfate

Read that second equation carefully, because it is doing two different jobs. The sulfite is spending itself so the developing agent does not have to — sacrificial protection — and it is also putting the developing agent back, regenerating hydroquinone from quinone. The primer’s practical consequence is that sulfite which has already effloresced and oxidised on the shelf is worth suspecting, “since the sulphate is not a preservative”. The aerial oxidation page next takes that apart in detail and turns it into storage practice.

Bisulfite (HSO₃⁻) and metabisulfite (S₂O₅²⁻) contain sulfur at the same +4. They are not different reducing agents; they are the same reducing agent delivered at a different pH, which is why the buffers page treats the sulfite and bisulfite pair as a buffer system rather than as two chemicals.

And thiosulfate, for all its sulfur, is not a developer. Its sulfur averages +2, so on the face of it there is room to oxidise, and something can indeed take two electrons from it — Mike Ware’s account of Sparling’s iron toner gives the half-equation:

2 S₂O₃²⁻ → S₄O₆²⁻ + 2 e⁻

Thiosulfate oxidised to tetrathionate, the first step with almost any oxidising agent

But those electrons do not go usefully to silver ions inside a grain to make an image. What thiosulfate does to silver instead is bind it, which is the subject of the complex-formation page. When thiosulfate is oxidised in the presence of image silver, the tetrathionate it forms attacks the silver in a quite different way, converting it to silver sulfide:

4 Ag + S₄O₆²⁻ + H₂O → 2 Ag₂S + HSO₃⁻ + HSO₄⁻

Tetrathionate sulfiding metallic silver: the chemistry of the historical 'old hypo' toning bath

That is a toner, and — when it happens by accident in an exhausted fixer or a poorly washed print — it is also a decay mechanism. Part XX takes up both.

Iron, the course’s other redox metal

The Fe³⁺/Fe²⁺ couple sits at +0.771 V, just below silver. Two consequences run through the course.

First, iron(III) is an oxidising agent strong enough to attack a developing agent: the 1924 primer notes that ferric salts added to hydroquinone “will oxidize it to quinone and will themselves be reduced to ferrous salts”. That is one reason iron in a water supply or a rusty vessel is unwelcome near a developer.

Second, and far more important, light reduces iron(III) to iron(II) in the presence of a suitable organic ligand, and that single fact is the basis of every iron-based printing process: cyanotype, Van Dyke brown, kallitype, platinum and palladium. The iron(II) formed where light struck then reduces something else — ferricyanide to make Prussian blue, or a silver or noble-metal salt to make the image metal. Prussian blue is itself a redox curiosity: Mike Ware’s monograph identifies it as ferric ferrocyanide, “quite unusual in containing the element combined in two different states of oxidation”, and attributes its intense colour directly to electrons hopping between the iron(II) and the iron(III).

The details, the ligands and the numbers belong to Part XXI (cyanotype), Part XXIV (Van Dyke and kallitype) and Part XXV (platinum and palladium). What this page contributes is the reading: whenever you see an iron(III) salt on an ingredient list of a printing-out process, it is there to be reduced by light.

Bleaching is development run backwards

If development is reduction, then converting image silver back into a silver salt is oxidation, and that is exactly what a bleach does. Kodak’s own sheet on toners says it in one line: in the sepia sequence, “the bleach bath (Solution A) converts metallic silver in the print to light-sensitive silver bromide”, after which the toner bath “converts the silver image to silver sulfide”.

Ag + Br⁻ → AgBr + e⁻

Bleaching, as a half-equation: image silver oxidised back to a halide, with an oxidising agent taking the electron

The same idea underlies reducers, which remove silver from a negative, and intensifiers, which add density. And it produces the course’s worst piece of vocabulary, which Kodak’s 1928 primer had already complained about a century ago: reducing a negative — weakening it — is chemically an oxidation, and the primer warns that “the use of the word reduction leads to confusion with true chemical reduction”. This course keeps the two apart by never using “reduce” for the darkroom operation without saying so.

Part XX covers toning, reduction and intensification as practice. Here they are only a category: oxidising agents applied to a finished image.

The reading skill this page gives you

Take any published formula. Every ingredient is playing one of a small number of redox roles, and you can usually name it before reading a word of the description.

Role	What it does	How you spot it
Developing agent	gives electrons to silver ions	an organic compound, usually with −OH or −NH₂ groups on a benzene ring; a gram or a few grams per litre
Preservative	is oxidised instead of the developing agent, and regenerates it	sulfite, bisulfite or metabisulfite; tens of grams per litre, far more than the developing agent
Alkali	raises the pH so the agent is in its active form	carbonate, borax, metaborate, hydroxide; no redox role at all
Restrainer	suppresses the free silver ion concentration and slows development of unexposed grains	a soluble bromide, usually well under a gram per litre; also no redox role
Antifoggant	adsorbs on the grain and raises the barrier to fogging	an organic nitrogen compound such as benzotriazole; fractions of a gram

Two of those five are redox players and three are not, and confusing them is the commonest mistake in reading a formula. A restrainer does not “weaken” the developer’s reducing power; it changes the concentration of what is being reduced, and the solubility page has already shown you how, through the common-ion effect. An alkali does not supply electrons; it decides how many of the developing agent’s molecules are in the deprotonated form that can.

Recap

Oxidation is losing electrons, reduction is gaining them, and the two are one transaction seen from each end. Oxidising agent and reducing agent are roles in a given reaction rather than fixed identities. Oxidation states are a bookkeeping device — the charge an atom would carry if the compound were ionic — and the rule that the sum equals the total charge is what lets you find sulfur’s +4 in sulfite and +6 in sulfate by subtraction. Half-equations show the electrons explicitly, and combining them means matching the counts: two silver ions per hydroquinone. Standard reduction potentials rank tendency, not speed, and the four silver entries in the table — free ion at +0.80 V, ammonia complex at +0.37, chloride lattice at +0.22, thiosulfate complex at +0.02 — show that binding an ion changes how much it wants an electron. Development is the reduction of silver ion to silver, selective because the developing agent is strong enough to reduce an exposed crystal at its latent-image speck and not strong enough to reduce an unexposed one; too strong, and you get fog instead of a picture. The oxidised developing agent is a real substance with consequences, yellow in the case of pyro. Sulfite protects by being oxidised instead, and by handing the developing agent back. Thiosulfate is not a developer: it binds silver rather than reducing it, and when oxidised it becomes tetrathionate, which sulfides silver. And bleaching is simply the whole thing run in reverse.

Next: the oxidising agent you did not put in the bottle. Air is in contact with every solution you make, and the page that follows is about how fast it wins.

Check your understanding

Question 1. Assign the oxidation state of sulfur in sodium metabisulfite, Na₂S₂O₅, and say what it tells you about the compound.

• +2, the same as thiosulfate, so it behaves like hypo
• +6, the same as sulfate, so it has no reducing power left
• +4, the same as sulfite and bisulfite, so it is the same reducing agent delivered at a different pH
• +5, which is unique to metabisulfite

Show the answer and why

Answer: +4, the same as sulfite and bisulfite, so it is the same reducing agent delivered at a different pH

Two sodium ions contribute +2 and five oxygens contribute −10, and the compound is neutral, so the two sulfur atoms together must be +8 and each is +4. That is the same state as in sulfite and bisulfite. The three salts are not three different preservatives: they are the same +4 sulfur, and the choice between them is a choice about the pH they produce when they dissolve, which is why the buffers page treats sulfite and bisulfite as a conjugate pair.

Question 2. Balance the reduction of silver ion by a two-electron developing agent and check the charge. Starting from Ag⁺ + e⁻ → Ag and C₆H₄(OH)₂ + 2 OH⁻ → C₆H₄O₂ + 2 H₂O + 2 e⁻, what is the combined equation?

• Ag⁺ + C₆H₄(OH)₂ + OH⁻ → Ag + C₆H₄O₂ + H₂O
• 2 Ag⁺ + C₆H₄(OH)₂ + 2 OH⁻ → 2 Ag + C₆H₄O₂ + 2 H₂O
• 2 Ag⁺ + 2 C₆H₄(OH)₂ + 2 OH⁻ → 2 Ag + 2 C₆H₄O₂ + 2 H₂O
• Ag⁺ + C₆H₄(OH)₂ + 2 OH⁻ → Ag + C₆H₄O₂ + 2 H₂O + e⁻

Show the answer and why

Answer: 2 Ag⁺ + C₆H₄(OH)₂ + 2 OH⁻ → 2 Ag + C₆H₄O₂ + 2 H₂O

The oxidation releases two electrons and each silver ion takes one, so the silver half-equation is multiplied by two before adding. The electrons then cancel, which is the check that the multiplication was right. Charge: +2 from two silver ions and −2 from two hydroxides gives zero on the left, and zero on the right. The version that doubles the developing agent as well leaves four electrons unaccounted for, and the version that keeps a spare electron on the right is always a sign the counts have not been matched.

Question 3. Two developing agents are compared. One has a higher standard potential for its oxidised/reduced couple. Under what circumstance does the table let you predict which develops a film faster?

• Always: a higher potential means a faster developer
• Never: potentials say nothing about developers
• Only for paper developers, not film developers
• Under no ordinary circumstance, because the tabulated value applies at 1 mol/L and 25 °C, ignores pH, adsorption on the grain surface and the rate at which the developer reaches the grain — the table ranks tendency, not speed

Show the answer and why

Answer: Under no ordinary circumstance, because the tabulated value applies at 1 mol/L and 25 °C, ignores pH, adsorption on the grain surface and the rate at which the developer reaches the grain — the table ranks tendency, not speed

Kodak's 1928 primer makes exactly this point about what it calls reduction potential: it "alone does not determine the speed with which the developer develops the image, because this depends chiefly upon the rate at which the developer diffuses into the film and on the quantity of developing agent and other substances in the developer." What a higher potential buys is the ability to keep working under adverse conditions — cold, or a lot of bromide. Tendency and rate are different subjects, and the kinetics page is where the second one lives.

Question 4. A litre of developer is to carry enough hydroquinone (molar mass 110.11) to supply the electrons for 0.06 mol of silver. Ignoring everything else, what mass of hydroquinone is the bare minimum?

• About 3.3 g, because each molecule supplies two electrons and so reduces two silver ions
• About 1.7 g
• About 6.6 g, because each silver ion needs a molecule of its own
• About 13.2 g, because each silver ion needs two molecules

Show the answer and why

Answer: About 3.3 g, because each molecule supplies two electrons and so reduces two silver ions

Three steps, kept separate. Moles of silver to be reduced: 0.06. Electrons needed: 0.06 mol, one per silver ion. Moles of hydroquinone: each molecule gives two electrons, so 0.06 ÷ 2 = 0.03 mol. Mass: 0.03 × 110.11 = 3.30 g. The 1.7 g answer comes from halving twice; the 6.6 g answer from forgetting that the agent is a two-electron donor. Note what this figure is and is not: it is the stoichiometric floor, and a real formula carries several times as much, because the last grains still have to be reached at a useful rate by an agent that has been partly spent and partly oxidised by air.

Question 5. A print developer that gave good blacks last week gives muddy greys today, and the solution has gone brown. What has been oxidised, by what, and what does the brown tell you?

• The silver has been oxidised by the developing agent; the brown is dissolved silver
• The sulfite has been oxidised by the developing agent, and the brown is sulfate
• The developing agent has been oxidised by atmospheric oxygen, and the brown is its oxidation product accumulating; there are fewer electrons available, so the deepest tones cannot be reached
• The bromide has been oxidised to bromine, and the brown is bromine

Show the answer and why

Answer: The developing agent has been oxidised by atmospheric oxygen, and the brown is its oxidation product accumulating; there are fewer electrons available, so the deepest tones cannot be reached

Oxygen is the oxidising agent that is always present, and the developing agent is the reducing agent most exposed to it. The coloured products are oxidation products of the agent itself — quinone and what follows it. Sulfate is colourless, so the answer blaming the sulfite fails on the evidence even though the sulfite has indeed been spent along the way. The photographic consequence follows from the electron count: a developer that has given away much of its reducing power cannot deliver enough electrons to bring the shadows to maximum density in the usual time, and the print reads as flat and grey.

Question 6. You are handed an unfamiliar developer formula listing, per litre: 2 g of an organic compound, 100 g of sodium sulfite, 5 g of borax and 0.5 g of potassium bromide. Which ingredient is doing the reducing, and what is the 100 g doing?

• The sulfite is the developing agent, and the organic compound is a preservative
• The organic compound is the developing agent; the sulfite is the preservative, present in bulk so that it is oxidised in place of the agent and regenerates it
• The borax is the developing agent and the sulfite is the restrainer
• The bromide is the developing agent and the borax is the preservative

Show the answer and why

Answer: The organic compound is the developing agent; the sulfite is the preservative, present in bulk so that it is oxidised in place of the agent and regenerates it

Read the quantities. A developing agent works at a gram or a few grams per litre; a preservative is present in bulk because it is meant to be consumed. Borax at 5 g is the alkali, setting the pH at which the agent is active and holding it there, and potassium bromide at half a gram is the restrainer, suppressing the free silver ion concentration by the common-ion effect. Only two of the four are redox players. That is the reading this page exists to give you, and the formula described is close to the shape of Kodak D-76.

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