Electrolysis/Electrochemistry

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Electrolysis/Electrochemistry
MyElimu Offline
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Post Icon Electrolysis/Electrochemistry

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 Electrolysis - decomposition of compound using electricity 

Electrolyte - 
an ionic compound which conducts electric current in molten or aqueous solution, being decomposed in the process. 

Electrode - 
a rod or plate where electricity enters or leaves electrolyte during electrolysis. 
Reactions occur at electrodes. 

Discharge -
 the removal of electrons from negative ions to form atoms or the gain of electrons of positive ions to become atoms. 

Anode
  • positive electrode connected to positive terminal of d.c. source. 
  • Oxidation occurs here. 
  • Anode loses negative charge as electrons flow towards the battery, leaving anode positively charged.
  •  This causes anion to discharge its electrons here to replace lost electrons and also, negative charge are attracted to positive charge.
 Cathode
  •  negative electrode connected to negative terminal of d.c. source. 
  • Reduction occurs here. 
  • Cathode gains negative charge as electrons flow from the battery towards the cathode, making cathode negatively charged. 
  • This causes cation to be attracted and gains electrons to be an atom.
 Anion
  •  negative ion
  • attracted to anode.
 Cation
  •  positive ion
  • attracted to cathode.
  •  
 Electrolysis of Molten Compounds
  • Molten/aqueous ionic compounds conduct electricity because ions free to move. 
  • In solid state, these ions are held in fixed position within the crystal lattice. 
  • Hence solid ionic compounds do not conduct electricity.
  • When molten binary compound is electrolysed, metal is formed on cathode while non-metal is formed on anode.
Example
Electrolysis of molten PbBr2

To make molten lead(II) bromide, PbBr2, we strongly heat the solid until it melts. To electrolyse it, pass current through the molten PbBr2.

What happens:

Ions present: Pb2+ and Br-

Reaction at Anode

Br- loses electrons at anode to become Br atoms. Br atoms created form bond together to make Br2 gas.

2Br-(aq) --> Br2(g)+ 2e-

Reaction at Cathode

Pb2+ gains electrons at cathode to become Pb atoms becoming liquid lead (II).
Pb2+(aq) + 2e- --> Pb(l)

Overall equation

PbBr2(l) --> Pb(l) + Br2(g)Electrolysis of Aqueous Solution
  • Aqueous solutions contain additional H+ and OH- ions of water, totaling 4 ions in the solution : 
    • 2 from electrolyte, 2 from water. 
    • Only 2 of these are discharged.
  • Electrolysis of aqueous solutions use the theory of selective discharge.
[Image: Screen%20shot%202011-07-03%20at%20PM%2002.48.16.png]
At cathode
  • In CONCENTRATED solutions of nickel/lead compound, nickel/lead will be discharged instead of hydrogen ions of water which is less reactive than nickel/lead.
  • In VERY DILUTE solutions, hydrogen, copper and silver ions are preferable to be discharged, according to its ease to be discharged.
  • Reactive ions (potassium, sodium, calcium, magnesium, aluminium) will NEVER BE DISCHARGED in either concentrated or dilute condition. Instead, hydrogen ions from water will be discharged at cathode.
At anode
  • In CONCENTRATED solutions, iodine/chlorine/bromine ions are preferable to be discharged, although it’s harder to discharged compared to hydroxide ions.
  • In VERY DILUTE solutions containing iodide/chloride/bromide ions, hydroxide ions of water will be discharged instead of iodide/chloride/bromide, according to ease of discharge.
  • Sulphate and nitrate are NEVER DISCHARGED in concentrated/dilute solutions.
ExamplesA. Concentration Solutions

Electrolysis of Concentrated NaCl

What happens:

Ions Present: Na+, H+, OH- and Cl-

Reaction at Anode
  • Cl- loses electrons at anode to become Cl atoms, although OH- is easier to discharge. 
  • Cl atoms created form covalent bond together to make Clgas.
  • 2Cl(aq) --> Cl(g) + 2e-
Reaction at Cathode
  • H+ gains electrons at cathode to become H atoms becoming hydrogen gas
  • 2H(aq) + 2e- --> H2 (l)
Overall Equation

2HCl(l) --> H2(l) + Cl2(g)

Note: any cation and anion left undischarged in solution forms new bonds between them. 
E.g. in above, leftovers Na+ and OH- combine to form NaOH.

B. Very Dilute Solutions

Electrolysis of Dilute H2SO4

What happens:

Ions Present: H+, OH- and SO42-

Reaction at Anode
  • OH- loses electrons at anode to become O2 and H2O.
  • 4OH- (aq) --> O2 (g) + 2H2O (l) + 4e-
Reaction at Cathode
  • H+ gains electrons at cathode to become H atoms becoming hydrogen gas.
  • 2H+(aq) + 2e- --> H2 (g)
Overall Equation
  • Both equations must be balanced first. 
  • The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.
    • (2H+(aq) + 2e- --> H2 (g)) x 2 = 4H +(aq) + 4e--> 2H(g)
  • Now we can combine the equations, forming: 
    • 4H+ (aq) + 4OH(aq) --> 2H(g) + O(g) + 2H2O (l)
  • 4H+ and 4OH+ ions, however, combine to form 4H2O molecules. 
    • Hence: 4H2O (l) --> 2H2 (g) + O(g)+ 2H2O (l)
  • H2O molecules are formed on both sides. 
    • Therefore, they cancel the coefficients: 2H2O (l) --> 2H(g) + O(g)
  • Since only water is electrolysed, the sulfuric acid now only becomes concentrated.
Electrolysis using different types of electrodes
  • Inert Electrodes are electrodes which do not react with electrolyte or products during electrolysis.
    • Eg. platinum and graphite.
  • Active Electrodes are electrodes which react with products of electrolysis, affecting the course of electrolysis. 
    • Eg. copper.
A. Electrolysis of CuSO4 Using Inert Electrodes (e.g. carbon)

What happens:

Ions Present: Cu2+, H+, OH- and SO42-

Reaction at Anode
  • OH- loses electrons at anode to become O2 and H2O.
  • 4OH- (aq) --> O(g) + 2H2O (l) +4e-
Reaction at Cathode
  • Cu2+ gains electrons at cathode to become Cu atoms becoming liquid copper. 
  • Hydrogen ions are not discharged because copper is easier to discharge.
  • Cu2+ (aq) + 2e- --> Cu (s)
Overall Equation
  • Both equations must be balanced first. 
  • The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.
    • (Cu2+ (aq) + 2e- --> Cu (s)) x 2 = 2Cu2+ (aq) + 4e- --> 2Cu (s)
  • Now we can combine the equations, forming:
    • 2Cu(OH)(aq) --> 2Cu (s) + O(g) + 2H2O (l)
  • Since copper ions in solution are used up, the blue colour fades. 
  • Hydrogen and sulphate ions left forms sulphuric acid.
B. Electrolysis of CuSO4 Using Active Electrodes (e.g. copper)

Ions Present: Cu2+, H+, OH- and SO42-

Reaction at Anode
  • Both SO42- and OH- gets attracted here but not discharged. Instead, the copper anode discharged by losing electrons to form Cu2+. So, the electrode size decreases.
  • Cu (s) --> Cu2+ (aq) + 2e-
Reaction at Cathode
  • Cu2+ produced from anode gains electrons at cathode to become Cu atoms becoming copper. Hence, the copper is deposited here and the electrode grows.
  • Cu2+ (aq) + 2e- --> Cu (s)
Overall Change
  • There is no change in solution contents as for every lost of Cu2+ ions at cathode is replaced by Cu2+ions released by dissolving anode. 
  • Only the cathode increases size by gaining copper and anode decreases size by losing copper. 
  • We can use this method to create pure copper on cathode by using pure copper on cathode and impure copper on anode. 
  • Impurities of anode falls under it.

 

 
(This post was last modified: 02-26-2014 08:54 AM by MyElimu.)
02-26-2014 08:23 AM
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