Anyone with an in-depth knowledge of the chemical processes in film development?

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I've been trying to find something good on the interwebz, but I'm having trouble. I came across this site

Chemical Engineering Resources - Cheresources.com - Indexut I thought there might be someone here able to answer questions. I'm a chemistry student (second year), and I find this interesting.....and that I owe it to my education to understand it :lol:

Thanks :)
 
It depends on what you call 'in-depth knowledge'. The more you know, the more you know you don't know. I'll certainly try to answer any questions you have.
 
LISTEN TO HELEN, she has forgotten more than most know.
 
Agreed ^^^
I love reading Helen's posts, it's incredible how much she knows :)

Another thought as well - are you interested in the ingredients (not sure if that's the right word?) of the processing chemicals? I've seen books that describe how to make your own developer, etc. -a friend of mine has a few older ones. If you want, I can find out the details on them and let you know.
 
I'll write when I get home. Waiting at the airport now, writing from my iPhone.

Thanks for wanting to help!

Ohh, that's the article I've read, but I have questions, obviously.
 
There is also Steve Anchell's :"Darkroom Cookbook"
 
Okay, so.....

As I've understood it, the silver grains on the film are salts of silver halids (either -Cl, -Br or -I). As light hits it (energy) the salt it split into ions, Ag+ and the negative halogen ion - basically of the same principle as UV-radiation initiated ozone depletion up in the higher troposphere/lower stratosphere. The bromide ion (I use Br for the purpose of explaining) loses the surplus electron and creates a bromine atom. Initially, we get a bromide ion - which loses its electron... I'm not quite getting why that is. It may be basic chemistry, but I'm confused. The silver ion then reacts with the electron to form a free silver atom.

These silver halide grains are "clusters" of silver salts. When being exposed to light, some of these salts are split, as mentioned above. The more light that hits these "grains", the more free silver is formed. What happens to the free bromine atoms? Alsothe latent image, which I've understood is the invisible image that becomes visible when exposed to the "developer", are these free silver atoms sitting on the rest of the grain.

From http://blog.lib.umn.edu/amattern/1701-2012-spring/Chemistry%20of%20Photography%201.pdf

"
Association within the grains produces species such as Ag2+, Ag20, Ag3+, Ag30, Ag4+, and Ag40."
(first numbers being index numbers, last numbers indicating the charge)

I'm not quite getting this. Is this the free silver associating within the grain? Or the silver salts associating? Are they bonded together by covalent bonds? Or merely interacting? Anyway, not getting that. And, I think I read, these different species are reduced at different rates, and Ag40 being readily reduced by a reducing agent resulting in large amounts of free silver. What of the other species? Any help on this would much appreciated!


Here I've talked about what happens to the film when exposed to light. Thanks for taking the time to help me, or trying! :)
 
The basics are simple. Silver halides crystallise in the cubic system with all bonds ionic. Bond energies are such that incoming photons can break those bonds and liberate neutral silver atoms (which stay in situ) and neutral halogen atoms which can wander out of the crystal and be captured by a substrate, gelatin for example.

A disrupted silver halide crystal (the latent image) has internal stresses which make it vulnerable to chemical reduction yielding silver and free halogen. A typical mild reducing agent is photographic developer.

Fixer is anything that dissolves silver halide but not silver.

The rest is easy; just 150 years of research and a million words to fill in the fine details.
 
Maris, thanks for the nice summary. So those species I read about, those are crystallized silver that sits on the grains (silver halide). Are all the crystallized silver species reduced by the reducing agent.

And about this latent image... When the film has been exposed to light, there are freed silver atoms. These sit on the silver halide grains, and act as a "centre for reduction" (in lack of a better expression). These centres, which may not be much bigger than a few silver atoms, are the latent image. When the developer is applied, electrons are freed to the silver halide grains (crystals) and reduce these to pure silver. The point is that the developer is strong enough to reduce silver halids close to the centre, but not the entire crystal...(I can image that producing a lower contrast image). The reduction centres act as catalysts, am I right? If there were no latent image, the developer would act on the crystals evenly, apprx, leaving a completely flat image (if allowed enough time to react)?

Or am I completely off here?
 
Yes, the small clump of silver atoms that makes up the latent image speck acts as a catalyst, and the silver grain grows from there, usually dendritically*, and yes, the whole film area would be fogged evenly by the developer eventually - some fogging usually occurs during normal development.

* The developed grain does not look anything like the silver halide microcystal - the developed silver image is usually made of tangled clumps of spidery dendrites. This is different from an image that is 'printed out' - ie made solely by the action of light.

To go back a step: there are two sources of electrons - bromide (or other halide, of course) ions and dye molecules. The quantum of light frees an electron from the bromide ion to create a temporary bromine atom (or iodide - iodine etc) which happens to be immobile in the lattice of the microcrystal. The electron is highly mobile through the lattice, of course. The electrons are drawn to a sensitivity speck (can be a variety of things, for example crystal irregularities that create a local charge imbalance, according to different theories) on the surface of the microcrystal and silver ions, which are much smaller and more mobile than the halide or halogen, then travel towards the charged sensitivty speck. It's a race. If too few silver ions get there compared to the rate of arrival of the electrons the speck becomes too negative and further electrons are repelled (high intensity reciprocity failure). If the electrons arrive too slowly, the low attracting force means that silver ions don't make it to the speck fast enough - they have time to recombine into the lattice (low intensity reciprocity failure). There needs to be at least 4 or 5 silver atoms in the clump to make the speck developable (ie a latent image). Two silver atoms may not be developable (sub-latent). A single microcrystal will have between one million and ten billion ion pairs (projected area between 0.1 square microns and 1 square micron, to give an idea of the size of a silver halide 'grain').Caveat: it's very difficult to study this process, because all evidence is usually destroyed as it becomes observable - there is more than one theory, and likely more than one process. I'm giving a simple version.

Meanwhile the 'positive hole' that is the halogen atom swaps an electron with its neighbour and the positive hole moves in that way to the edge of the microcrystal, where the halogen will be received by the gelatin and, if present, a secondary receptor.

Electrons can also be donated by excitons coming from a sensitizing dye molecule that has absorbed a quantum of light. In this case a positively charged dye molecule remains in the lattice in place of the silver ion.
 
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An exposed silver halide crystal develops to completion. It's a bit like a knitted jumper unravells by continually pulling at a loose thread at the hem. Contrast control formulations depend on a silver halide crystals mix with a wide range of particle sizes. The small crystals have a smaller optical cross-section and are less likely to be hit by enough photons. Big crystals are easily hit and most become developable. Big crystals record shadows, small ones highlights.

Developer is a slow reducing agent with an accelerator, usually alkali, to speed things up. Plus a restrainer to prevent everything, exposed and unexposed, from developing. Plus an antioxidant to stop atmospheric oxygen killing the reducing agent before it gets to the silver halide crystals.

There's more than one persons lifetime to know it all!
 
Thanks a lot!

"If too few silver ions get there compared to the rate of arrival of the electrons the speck becomes too negative and further electrons are repelled"

We're now in the process of exposing the film, yes? The silver that's wandering toward this speck.. why is it wandering? Electrons wander toward positive charges (or less negative charges than they are around). What does the silver achieve by wandering? A lower energy state (less entropy) I would guess, but how?

Also, these silver atoms that are "racing" with the electrons...when these have "arrived", this is what makes up for the latent image, yes? Does the electron react with the silver? What happens to the electron? Or is it the silver ion that's racing with the electron, and to have

Sub latent meaning that the catalytic site isn't strong enough to carry out the reactions specifically on the silver atoms?

Just for the record: the making of the silver atom and halogen atom (referred to as X, that's very common) would be something like this?


AgX + hv --> Ag+ + X-
X- + hv --> X + e-
Ag+ + e- --> Ag
 
"If too few silver ions get there compared to the rate of arrival of the electrons the speck becomes too negative and further electrons are repelled"

We're now in the process of exposing the film, yes?


Yes.

The silver that's wandering toward this speck.. why is it wandering? Electrons wander toward positive charges (or less negative charges than they are around). What does the silver achieve by wandering? A lower energy state (less entropy) I would guess, but how?

The silver is mobile within the lattice, but not as mobile as the electrons. Where a sensitivity speck exists, the electrons are more attracted to it than to individual silver ions in the charge-balanced lattice. (the sensitivity speck may be a discontinuity in the lattice, which unbalances the charge at the surface of the microcrystal) Once there becomes a stable negatively-charged speck, local silver ions can migrate towards it. If the formation of the silver atom and the halogen atom occurrs in the same place, recombination would be too easy.

Also, these silver atoms that are "racing" with the electrons...when these have "arrived", this is what makes up for the latent image, yes?

Yes.

Does the electron react with the silver? What happens to the electron? Or is it the silver ion that's racing with the electron, and to have

Yes, The electron takes part in the reduction of the silver ion to silver by supplying the missing electron.


Sub latent meaning that the catalytic site isn't strong enough to carry out the reactions specifically on the silver atoms?

I think you mean 'on the silver ions', but yes, a sub-latent site does not have sufficient silver atoms to catalyse the selective local reduction of silver ions by the developer.

Just for the record: the making of the silver atom and halogen atom (referred to as X, that's very common) would be something like this?


AgX + hv --> Ag+ + X-
X- + hv --> X + e-
Ag+ + e- --> Ag

The first line isn't necessary. The silver halide is present as an ionic crystalline lattice, not as AgX molecules. You should also remember that the spectral sensitizing dyes play a big role at the longer end of the visible spectrum, where the light quanta aren't energetic enough to dislodge an electron from the chloride / bromide / iodide ion (fluoride isn't used). The colour of the silver salt gives a clue about what light it absorbs: silver chloride is white, the other two are pale yellow or yellow-brown.
 

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