question on plating silver: how can I tell if the silver layer is too thin?

[RonF]

I ordered some polished copper plates and I am going to try electroplating with silver to make dag plates. I don’t have a scale that is sensitive enough to measure a 10 micron layer of silver on a 4×5 plate. How will I know if my silver is not on thickly enough? Will I simply not get an image ? Or are there other issues that will arise?

I am aware that I am entering an area of technique that is tricky. But the guy who previously sold me plates with destroyed images is no longer going to sell me any, he now plans to sell to the Eastman house and one other buyer. I don’t want to spend $30+ per plate, and besides, plating seems like a good skill to have. I already have plans to make some jewelery too. Perhaps if I get it down I will be able to sell plates eventually.

I am going to use a solution by Krohn. It contains small amounts of cyanide. Of course I am proceeding cautiously.

[photolytic]

http://en.wikipedia.org/wiki/Faradays_laws_of_electrolysis

Faraday’s 2nd Law of Electrolysis – For a given quantity of electricity (electric charge), the mass of an elemental material altered at an electrode is directly proportional to the element’s equivalent weight. The equivalent weight of a substance is its molar mass divided by an integer that depends on the reaction undergone by the material.

Mathematical form

Faraday’s laws can be summarized by

m = (Q)/F) x (M/z)

where

m is the mass of the metal deposited at an electrode

Q is the total electric charge passed through the metal solution.

F = 96,485 C mol-1 is the Faraday constant

M is the molar mass of the metal which is 107.88 gm/mol for silver

z is the valency number of ions of the metal ( 1 in the case of silver)

Note that M / z is the same as the equivalent weight of the metal deposited.

For Faraday’s second law, Q, F, and z are constants, so that the larger the value of M / z (equivalent weight) the larger m will be.

In the simple case of constant-current electrolysis, Q (coulombs) = electric current in amps (coulombs/second) x time t (seconds)

For a 4 x5 inch plate (area 129.032 cm2), a 10 micron (.001cm) layer of silver (10.49 gm/cc) weighs .000134535 grams.

.000134535 grams of silver contains m / 107.88 or .00001254677 moles of silver

which when multiplied by Faraday’s constant (96485 Coulombs/mole) will require 1.210575 coulombs of electircal charge.

Assuming a reasonable plating time of 5 minutes or 300 seconds, a current of .00403525 amps or 4 milliamps will be required.

Therefore you need to get yourself an accurate milliamp meter to monitor this current plus a DC rectifier or battery current source.

[RonF]

Thanks a lot, photolytic! My dad (an electrical engineer) had the same general idea. Now that you have done the math for me, this will be a lot easier.

I am pleasantly surprised by how much effort modern daguerreotypists will put into helping one another. What a wonderful thing this forum is.

I can’t wait until my plates arrive. I will be sure to keep you all in the loop. Hopefully my update will be on the topic of “my latest dag”.

[Author]

Posts