AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Since the barium ion will be isoelectronic to the nearest noble gas, which is xenon, the electronic configuration for Ba2+ is: [Xe]
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
The hydrogen and oxygen atoms that combine to form water molecules are bound together by covalent bonds. The electron from the hydrogen splits its time between the incomplete outer shell of the hydrogen atom and the incomplete outer shell of the oxygen atom.
