Electrolysis - A-level chem

• hoay

  11 Feb 15, 22:49

  The by-product of electrolysis of brine is sodium hydroxide. The reaction that is quoted is 2H2O + 2e - H2 + 2OH- .....which occur at cathode. The proble is that at cathode cations are discharged such as here H+ ions are discharged. No idea how water being a neutral specie migrates to cathode ?? Please explain.

   

• UltimaOnline

  12 Feb 15, 01:14

  Originally posted by hoay:

  The by-product of electrolysis of brine is sodium hydroxide. The reaction that is quoted is 2H2O + 2e - H2 + 2OH- .....which occur at cathode. The proble is that at cathode cations are discharged such as here H+ ions are discharged. No idea how water being a neutral specie migrates to cathode ?? Please explain.

   

  
  
  
  Within the H2O molecule, because O atoms are more electronegative than H atoms, hence the H atoms, having a partial positive charge, are electrostatically attracted to, and accept electrons from, the cathode. For every H2O molecule, 1 H atom accepts 1 electron, generating an OH- ion and a H atom, which readily combines with another H atom from the reduction of the next H2O molecule, thus generating H2 gas (with positive, thermodynamically favorable entropy change).
  
  If you prefer, an alternative acceptable way to look at it, is that H+ ions (or H3O+ ions) from the auto-dissociation of H2O molecules, are electrostatically attracted to, and accept electrons from, the cathode, to be reduced to H atoms, which readily combines with the next H atom to generate H2 gas (again with positive, thermodynamically favorable entropy change), thereby causing the position of equilibrium of the auto-dissociation of water to shift to the RHS as predicted by Le Chatelier's principle, thus generating more OH- ions as a byproduct.
  
  Both ways of looking at this are correct, use whichever you prefer, whichever helps you understand better, and Cambridge certainly won't ask for such explanatory details at A levels in any case.
• hoay

  12 Feb 15, 02:06

  Oh ! I missed this point. Thank you.

  One more thing.........

  For ions to discharge the E should be atleast +0.30 V such as Cu+2 are discharged in preference to H+ in the electrrolysis of CuSO4 (aq). If Cl- ions and Br- ions are present in the same solution with equal concentration then which is oxidized preferentially Cl- or Br- ?? E for Cl- is more postive (+1.36V) than Br- (+1.07V) then Cl- should discharged at anode but Br- is discharged in actual practice....any expalnation??

   

• UltimaOnline

  12 Feb 15, 03:42

  Originally posted by hoay:

  Oh ! I missed this point. Thank you.

  One more thing.........

  For ions to discharge the E should be atleast +0.30 V such as Cu+2 are discharged in preference to H+ in the electrrolysis of CuSO4 (aq). If Cl- ions and Br- ions are present in the same solution with equal concentration then which is oxidized preferentially Cl- or Br- ?? E for Cl- is more postive (+1.36V) than Br- (+1.07V) then Cl- should discharged at anode but Br- is discharged in actual practice....any expalnation??

   

  
  It is advisable to specify either "reduction potential" or "oxidation potential" when discussing feasibility of redox reactions.
  
  The standard oxidation potential of 2Br- to Br2 is -1.07 V, which is more positive compared to the standard oxidation potential of 2Cl- to Cl2, which is -1.36 V.
  
  Hence, assuming standard or equal molarities, the oxidation of Br- ions to Br2 is more feasible compared to the oxidation of Cl- ions to Cl2 at the anode.
• hoay

  13 Feb 15, 02:35

  Nitrate salts are not used as electrolyte insted of sulfates such as copper(II) nitrate. Is it due to the NO3- (+0.94V) having lower reduction potential than sulfate ion (+2.01V) so they may be discharged at anode eleasing toxic NO2 gas...? 

• UltimaOnline

  13 Feb 15, 02:52

  Originally posted by hoay:

  Nitrate salts are not used as electrolyte insted of sulfates such as copper(II) nitrate. Is it due to the NO3- (+0.94V) having lower reduction potential than sulfate ion (+2.01V) so they may be discharged at anode eleasing toxic NO2 gas...? 

  
  At the anode, oxidation occurs. Both the nitrate(V) ion NO3- and the sulfate(VI) ion SO42- cannot be oxidized at the anode because both the Group V Nitrogen and Group VI Sulfur are already at their maximum oxidation states of +5 and +6 respectively. It is possible for the oxygens to be oxidized (instead of the nitrogens and the sulfurs), but under aqueous conditions, it is more thermodynamically feasible to oxidize the oxygens from the water solvent, or if you prefer to think of it as, from the hydroxide OH- ions from the auto-dissociation of the water solvent, hence generating O2 gas at the anode, rather than oxidizing the oxygens in the nitrate(V) or sulfate(VI) ions. Therefore for A level purposes, the nitrate(V) and sulfate(VI) ions are to be considered inert and unreactive when carrying out electrolysis of aqueous salts (electrolysis of molten nitrates and sulfates are beyond the scope of the A level syllabus).
• hoay

  13 Feb 15, 03:56

  What about the following reaction  

  S2O82- + 2e- 2SO42-  +2.01 V 

  Sulfate ions can be oxidized as the reaction can be reversed. If we increase the [SO42-] ions the E becomes less positive thereby making it feasible to be oxidized.

  Secondly, can u through some light on Overpotential. 

   

• UltimaOnline

  13 Feb 15, 05:36

  Originally posted by hoay:

  What about the following reaction  

  S2O82- + 2e- 2SO42-  +2.01 V 

  Sulfate ions can be oxidized as the reaction can be reversed. If we increase the [SO42-] ions the E becomes less positive thereby making it feasible to be oxidized.

  Secondly, can u through some light on Overpotential. 

   

  
  Oxidation of sulfate(VI) to peroxodisulfate(VI) S2O82- with the 2 peroxo O atoms having an OS of -1, is indeed what I was referring to in my previous post, that is, oxidizing the oxygens rather than the sulfurs. However, as also mentioned in my previous post, the oxidation of the oxygens in the water solvent (and by definition of solvent, it's present in much higher molarity than the sulfate(VI) ions) to generate O2 gas, is more thermodynamically feasible.
  
  For A level purposes, unless otherwise indicated in the question, electrolysis of aqueous sulfates will not oxidize the sulfate(VI) ion into the peroxodisulfate(VI) ion.
  
  Overpotential is a complex matter (as this is a broad term that includes different types of overpotential with competely different and even unrelated underlying chemistry and physics reasons), and is safely beyond the scope of the A level syllabus, and as such I suggest you do not confuse your students with this. On rare occasions when the Cambridge A level exam question includes beyond-the-syllabus concepts such as overpotential, sufficient data (eg. relevant formulae) will certainly be provided in the question to allow the A level student to cope.
• hoay

  24 Feb 15, 18:30

  In the elctrolysis of aq CuSO4 using carbon electrode OH- ions discharge at the anode producing oxygen because OH- comes from the dissociation of water.

  The electrolytic refinning of impure copper (CuSO4 being the elcterolyte again) using copper elcterode does not give oxygen at anode since anode dissolves. Why the OH- which comes from the dissociation of water are dishagred at anode?.....Is anode dissolve so rapidly..?   

• UltimaOnline

  24 Feb 15, 20:53

  Originally posted by hoay:

  In the elctrolysis of aq CuSO4 using carbon electrode OH- ions discharge at the anode producing oxygen because OH- comes from the dissociation of water.

  The electrolytic refinning of impure copper (CuSO4 being the elcterolyte again) using copper elcterode does not give oxygen at anode since anode dissolves. Why the OH- which comes from the dissociation of water are dishagred at anode?.....Is anode dissolve so rapidly..?   

  
  At the anode, the copper metal of the anode electrode is itself oxidized, thus generating soluble Cu2+(aq) and thus 'dissolving'.
  
  The oxidation potential for Cu(s) to Cu2+(aq) is more positive than the oxidation potential of either OH- or H2O, to be oxidized to O2. You cannot use Data Booklet values for this, as Data Booklet are for standard conditions (including molarities) only.
  
  The copper electrode is solid, and by definition of solids, thus have a much higher molarity compared aqueous ions (so much so that we don't even give any value for the molarity of solids).
  
  Therefore, for A & O level Chemistry purposes, consider the reactivity of the electrodes. Generally, copper and any metal more reactive than copper, would be preferentially oxidized if used as the anode electrode, over any other species present in the aqueous state.
• hoay

  4 Mar 15, 16:16

  What is always observed when aqueous CuSO4 is electrolyzed using carbon electrodes.

  A Copper (II) ions are reduced at cathode

  B Sulfate ions migrate to anode

  C The blue solution fades

  D oxygen is produced at anode

   

  Is D the correct answer?? since if A is a correct answer then C would have been correct also. This is because OH- will migrate towards anode in preference to sulfate ions as the E is feasibly positive than sulfate ions. Am i correct??

   

• UltimaOnline

  4 Mar 15, 23:27

  Originally posted by hoay:

  What is always observed when aqueous CuSO4 is electrolyzed using carbon electrodes.

  A Copper (II) ions are reduced at cathode

  B Sulfate ions migrate to anode

  C The blue solution fades

  D oxygen is produced at anode

   

  Is D the correct answer?? since if A is a correct answer then C would have been correct also. This is because OH- will migrate towards anode in preference to sulfate ions as the E is feasibly positive than sulfate ions. Am i correct??

   

  
  
  
  This is a trick question. The only option in which you can visually observe is option C : blue colour fading.