I know that the OP said he tested them for silver(i think he said that) but it seems reasonable that these may have been part of a fishing associated find. Similar to the old split shot weights we crimp onto our line, one weight then two...whatever it took. Could this be a fishing line with crimp on weights. I don't see anything wrong with this except the silver part of it. JMO....vn
2 Ag+ (aq) + 2 e- ------> 2 Ag (s)
Here, two silver ions (silver with a positive charge) are being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid, respectively. We can now combine the two (2) half-reactions to form a redox equation:
We can also discuss the individual components of these reactions as follows. If a chemical causes another substance to be oxidized, we call it the oxidizing agent. In the equation above, Ag+ is the oxidizing agent, because it causes Cu(s) to lose electrons. Oxidants get reduced in the process by a reducing agent. Cu(s) is, naturally, the reducing agent in this case, as it causes Ag+ to gain electrons.
As a summary, here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium):
Divide the equation into an oxidation half-reaction and a reduction half-reaction
Balance these
Balance the elements other than H and O
Balance the O by adding H2O
Balance the H by adding H+
Balance the charge by adding e-
Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other
Combine the half-reactions and cancel
**Add OH- to each side until all H+ is gone and then cancel again**
In considering redox reactions, you must have some sense of the oxidation number (ON) of the compound. The oxidation number is defined as the effective charge on an atom in a compound, calculated according to a prescribed set of rules. An increase in oxidation number corresponds to oxidation, and a decrease to reduction. The oxidation number of a compound has some analogy to the pH and pK measurements found in acids and bases -- the oxidation number suggests the strength or tendency of the compound to be oxidized or reduced, to serve as an oxidizing agent or reducing agent. The rules are shown below. Go through them in the order given until you have an oxidation number assigned.
For atoms in their elemental form, the oxidation number is 0
For ions, the oxidation number is equal to their charge
For single hydrogen, the number is usually +1 but in some cases it is -1
For oxygen, the number is usually -2
The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.
As a side note, the term "oxidation", with its obvious root from the word "oxygen", assumes that oxygen has an oxidation number of -2. Using this as a benchmark, oxidation numbers were assigned to all other elements. For example, if we look at H2O, and assign the value of -2 to the oxygen atom, the hydrogens must each have an oxidation number of +1 by default, since water is a neutral molecule. As an example, what is the oxidation number of sulfur in sulfur dioxide (SO2)? Given that each oxygen atom has a -2 charge, and knowing that the molecule is neutral, the oxidation number for sulfur must be +4. What about for a sulfate ion (SO4 with a total charge of -2)? Again, the charge of all the oxygen atoms is 4 x -2 = -8. Sulfur must then have an oxidation number of +6, since +6 + (-8) = -2, the total charge on the ion. Since the sulfur in sulfate has a higher oxidation number than in sulfur dioxide, it is said to be more highly oxidized.
Working with redox reactions is fundamentally a bookkeeping issue. You need to be able to account for all of the electrons as they transfer from one species to another. There are a number of rules and tricks for balancing redox reactions, but basically they all boil down to dealing with each of the two half-reactions individually. Consider for example the reaction of aluminum metal to form alumina (Al2O3). The unbalanced reaction is as follows:
Looking at each half reaction separately:
This reaction shows aluminum metal being oxidized to form an aluminum ion with a +3 charge. The half-reaction below shows oxygen being reduced to form two (2) oxygen ions, each with a charge of -2.
If we combine those two (2) half-reactions, we must make the number of electrons equal on both sides. The number 12 is a common multiple of three (3) and four (4), so we multiply the aluminum reaction by four (4) and the oxygen reaction by three (3) to get 12 electrons on both sides. Now, simply combine the reactions. Notice that we have 12 electrons on both sides, which cancel out. The final step is to combine the aluminum and oxygen ions on the right side using a cross multiply technique:
Taking care of the number of atoms, you should end up with:
One of the more useful calculations in redox reactions is the Nernst Equation. This equation allows us to calculate the electric potential of a redox reaction in "non-standard" situations. There exist tables of how much voltage, or potential, a reaction is capable of producing or consuming. These tables, known as standard potential tables, are created by measuring potential at "standard" conditions, with a pressure of 1 bar (≅1 atm), a temperature of 298° K (or 25° C, or room temperature) and with a concentration of 1.0 M for each of the products. This standard potential, or E°, can be corrected by a factor that includes the actual temperature of the reaction, the number of moles of electrons being transferred, and the concentrations of the redox reactants and products. The equation is: