0
$\begingroup$

I understand the SI meaning of a Coulomb but how were the individual electrons counted in the measurement to know how many it would take to constitute a Coulomb?

$\endgroup$
1
$\begingroup$

Please check the following references at the bottom (from the earliest known usages of Coulomb in unabridged Oxford English Dictionary). Also, check the early definition of ampere which was also defined on the basis of silver nitrate electrolysis.

from the wt of Ag+ deposited, by Faraday's law, we have one to one correspondence:

Ag+ (solution) + electron ---> Ag (metal), thus from the weight of Ag deposited during electrolysis, you can count electrons involved in the process.

You can independently measure charge passed through a solution via time (t) and current (I). Q= Ixt

from OED:

1881 Nature 29 Sept. 512/2 The name Coulomb to be given to the quantity of electricity defined by the condition that an ampère gives one coulomb per second.

1892 Lightning 3 Mar. (Spec. Suppl.) Gloss. Electrical Terms s.v. The Coulomb is the quantity of electricity, which, when passed through a solution of nitrate of silver, deposits ·001118 of a gramme of silver on the plate by which it leaves the liquid.

$\endgroup$
4
  • $\begingroup$ wow ...I assume from this one number of .001118 gramme we can then calculate the number of electrical charges ? It would be interesting to see the calculation that actually gives the number of electrons derived by this observation. 10 to the minus 18 is a very small number and I only see 6 places from the .001118. Natural for me to wonder how the rest of the calculation was done. $\endgroup$
    – Sedumjoy
    Dec 21 '18 at 2:30
  • 1
    $\begingroup$ I would be interested to see the original calculations too. However, from a modern perspective, convert 0.001118 g Ag to mol of Ag. Let this be m. From the m mol Ag, we know that m mol of electrons must have been used to reduce it. Convert the m mol electron to the number of electrons by Avogadro's number. You roughly get the Coulombs. $\endgroup$
    – M. Farooq
    Dec 21 '18 at 5:15
  • $\begingroup$ ah yes...would you be tempted to say if it was not for Avogadro then the coulomb would not have been able to be calculated ? $\endgroup$
    – Sedumjoy
    Dec 21 '18 at 17:24
  • 1
    $\begingroup$ Avogadro as well as Faraday's laws. In earlier works, you will see the term equivalents instead of moles. The arguments still remain valid. $\endgroup$
    – M. Farooq
    Dec 21 '18 at 17:28
1
$\begingroup$

This was discovered in the experiments with electrolysis. The idea is that when electric current passes through a solution, positive ions accumulate on one side, and negative on another. We know the charge in terms of the number of electrons of one ion (+n if n electrons are missing, -n if there are n extra electrons). The number of ions that accumulated can be found when you know their atomic weight and the weight of the accumulated material. On the other hand, by measuring time and current passing through the solution we can find out how many Coulombs correspond to this charge.

$\endgroup$
2
  • 1
    $\begingroup$ In describing this chain of reasoning, you refer to electrons, but I think all of this probably predated the discovery of the electron or Millikan's determination of $e$ by quite some time. $\endgroup$
    – user466
    Dec 20 '18 at 19:11
  • 1
    $\begingroup$ You are right. But the question is about electrons:-) In reality, historically it was discovered that there is an "elementary charge" (in the same experiments with electrolysis) which was later identified with electron (or proton). $\endgroup$ Dec 20 '18 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.