In this video we’ll use the Periodic table and a few simple rules to find the protons, electrons, and neutrons for the element Silver (Age). From the Periodi. Silver is similar in its physical and chemical properties to its two vertical neighbours in group 11 of the periodic table, copper and gold.Its 47 electrons are arranged in the configuration Kr4d 10 5s 1, similarly to copper (Ar3d 10 4s 1) and gold (Xe4f 14 5d 10 6s 1); group 11 is one of the few groups in the d-block which has a completely consistent set of electron configurations.

Electrolysis

Electrolysis involves passing an electric current through either a molten salt or an ionic solution. The ions are 'forced' to undergo either oxidation (at the anode) or reduction (at the cathode). Most electrolysis problems are really stoichiometry problems with the addition of an amount of electric current. The quantities of substances produced or consumed by the electrolysis process is dependent upon the following:

• electric current measured in amperes or amps
• time measured in seconds
• the number of electrons required to produce or consume 1 mole of the substance

Amps, Time, Coulombs, Faradays, and Moles of Electrons

Three equations relate these quantities:

• amperes x time = Coulombs
• 96,485 coulombs = 1 Faraday
• 1 Faraday = 1 mole of electrons

The thought process for interconverting between amperes and moles of electrons is:

amps & time ' no save> Coulombs ' no save> Faradays ' no save> moles of electrons

Use of these equations are illustrated in the following sections.

Calculating the Quantity of Substance Produced or Consumed

To determine the quantity of substance either produced or consumed during electrolysis given the time a known current flowed::

• Write the balanced half-reactions involved.
• Calculate the number of moles of electrons that were transferred.
• Calculate the number of moles of substance that was produced/consumed at the electrode.
• Convert the moles of substance to desired units of measure.

Silver Number Of Electrons

Example: A 40.0 amp current flowed through molten iron(III) chloride for 10.0 hours (36,000 s). Determine the mass of iron and the volume of chlorine gas (measured at 25^oC and 1 atm) that is produced during this time.

• Write the half-reactions that take place at the anode and at the cathode.

anode (oxidation): 2 Cl^- ' nosave> Cl₂(g) + 2 e^-

cathode (reduction) Fe³⁺ + 3 e^- ' nosave> Fe(s)

• Calculate the number of moles of electrons.
• Calculate the moles of iron and of chlorine produced using the number of moles of electrons calculated and the stoichiometries from the balanced half-reactions. According to the equations, three moles of electrons produce one mole of iron and 2 moles of electrons produce 1 mole of chlorine gas.

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• Calculate the mass of iron using the molar mass and calculate the volume of chlorine gas using the ideal gas law (PV = nRT).

2' NOSAVE height=115 width=502>

Calculating the Time Required

To determine the quantity of time required to produce a known quantity of a substance given the amount of current that flowed:

Silver Element Number Of Electrons

• Find the quantity of substance produced/consumed in moles.
• Write the balanced half-reaction involved.
• Calculate the number of moles of electrons required.
• Convert the moles of electrons into coulombs.
• Calculate the time required.

Example: How long must a 20.0 amp current flow through a solution of ZnSO₄ in order to produce 25.00 g of Zn metal.

• Convert the mass of Zn produced into moles using the molar mass of Zn.
• Write the half-reaction for the production of Zn at the cathode.

Zn²⁺(aq) + 2 e^- ' nosave> Zn(s)

• Calculate the moles of e^- required to produce the moles of Zn using the stoichiometry of the balanced half-reaction. According to the equation 2 moles of electrons will produce one mole of zinc.
• Convert the moles of electrons into coulombs of charge using Faraday's constant.
• Calculate the time using the current and the coulombs of charge.

Calculating the Current Required

To determine the amount of current necessary to produce a known quantity of substance in a given amount of time:

• Find the quantity of substance produced/or consumed in moles.
• Write the equation for the half-reaction taking place.
• Calculate the number of moles of electrons required.
• Convert the moles of electrons into coulombs of charge.
• Calculate the current required.

Example: What current is required to produce 400.0 L of hydrogen gas, measured at STP, from the electrolysis of water in 1 hour (3600 s)?

• Calculate the number of moles of H₂. (Remember, at STP, 1 mole of any gas occupies 22.4 L.)
• Write the equation for the half-reaction that takes place.

Hydrogen is produced during the reduction of water at the cathode. The equation for this half-reaction is:

4 e^- + 4 H₂O(l) ' nosave> 2 H₂(g) + 4 OH^-(aq)

• Calculate the number of moles of electrons. According to the stoichiometry of the equation, 4 mole of e^- are required to produce 2 moles of hydrogen gas, or 2 moles of e^-'s for every one mole of hydrogen gas.
• Convert the moles of electrons into coulombs of charge.
• Calculate the current required.

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What is the electron configuration of Ag?

2 Answers

The electron configuration for silver (Ag) is based upon the place meant of silver in the fifth row of the periodic table in the 11th column of the periodic table or the 9th column of the transition metal or d block. Therefore th electron configuration for silver must end as #4d^9#,

Mobile empire. #1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^9#

This notation can be written in core notation or noble gas notation by replacing the #1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6# with the noble gas [Kr].

#[Kr] 5s^2 4d^9#

For some of the transition metals they will actually transfer an s electron to complete the d orbital, making silver,

#[Kr] 5s^1 4d^10#

I hope this was helpful.
SMARTERTEACHER

Explanation:

Silver, #'Ag'#, is located in period 5, group 11 of the periodic table, and has an atomic number equal to #47#.

This tells you that a neutral silver atom will have a total of #47# electrons surrounding its nucleus.

Now, you have to be a little careful with silver because it is a transition metal, which implies that the occupied d-orbitals are actually lower in energy than the s-orbitals that belong to the highest energy level.

So, here's how silver's electron configuration would look if it followed the Aufbau principle to the letter

#'Ag: ' 1s^2 2s^2 2sp^6 3s^2 3p^6 color(blue)(4s^2 4p^6) color(red)(3d^10) 5s^2 4d^9#

Now, for the energy level #n#, the d-orbitals that belong to the #(n-1)# energy level are lower in energy than the s and p orbitals that belong to the #n# energy level.

This means that you will have to switch the 3d orbitals on one hand, and the 4s and 4p orbitals on the other.

This will get you

#'Ag: ' 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 5s^2 4d^9#

Now do the same for the 4d and 5s orbitals

#'Ag: ' 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 4d^9 5s^2#

The thing to remember here is that in silver's case, the 4d orbitals will be completely filled. That implies that you won't have two electrons in the 5s orbital, since one will be kept in the lower 4d orbitals.

This means that the electron configuration of silver will be

#'Ag: ' 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 4d^10 5s^1#

Using the noble gas shorthand notation will get you

#'Ag: ' overbrace(1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6))^(color(green)(['Kr'])) 4d^10 5s^1#

#'Ag: ' ['Kr'] 4d^10 5s^1#

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