Standard electrode Potential.....

• Kahynickel

  2 Sep 15, 02:59

  considering the following two pieces of information:

  1- For a reaction to be feasible and spontaneous Ecell should equal to or greater than +0.34V. Can someone tell the origin of this value? I mean any book?

  2- Applying E to the discharge of ions during the electrolysis, ions with the more positive value will be discharged in preference to the less positive one. For example at cathode Cu2+ ions [+0.34V] are discharged in comparison to H+ [0.00V] during the electrolysis of copper sulfate. But F- ions [+2.87V] do not discharge in preference to OH- [+0.40V] during the electrolysis of aqueous sodium chloride and O2 is formed not F2 at anode. Please explain.  

• UltimaOnline

  2 Sep 15, 04:45

  If these qns are part of an exam question, it's better to type out or post a screenshot or link to the exam paper (eg. from CIE website).
  
  Cell potential needs to be greater than zero volts (not +0.34V), for the reaction to be thermodynamically feasible.
  
  The species with the more positive reduction potential is reduced at the cathode.
  The species with the more positive oxidation potential is oxidized at the anode.
  
  F- and OH- are both anions (ie. electron-rich), hence both migrate to the anode for the possibility of being oxidized. However, the oxidation potential of OH- to O2 is -0.40V, which is more positive than the oxidation potential of F- to F2, which is -2.87V. Hence OH- is preferentially oxidized to O2 at the anode.
• Kahynickel

  2 Sep 15, 23:49

  What if we do not reverse the values as it is not necessary to reverse the value??The Ecell can be calculated using ER - EL formula.  

• UltimaOnline

  2 Sep 15, 23:58

  Originally posted by Kahynickel:

  What if we do not reverse the values as it is not necessary to reverse the value??The Ecell can be calculated using ER - EL formula.  

  
  Yes you'll get the same answer, but chemically speaking (ie. in terms of more deeply understanding and applying chemistry concepts rather than just blindly memorizing formulae without understanding), you should advise your students that it makes way more sense to ADD reduction potential @ cathode, to the oxidation potential @ anode to obtain cell potential; instead of SUBTRACTING reduction potential @ anode from reduction potential @ cathode, because after all, oxidation (not reduction) occurs at the anode.
• Kahynickel

  8 Sep 15, 04:00

  Cu2+ + 2e = Cu   +0.34V (for 1M). 

  If the [Cu2+] is increased right-hand is favoured i.e more Cu will be formed. The value of this half -cell will be more positive (  > +0.34V)

  following the same line.....

  Zn2+   +   2e = Zn    -0.76V. { for 1 M}

  If the [Zn2+] is increased then the forward reaction is favoured i.e the product Zn is formed more and more. The E (-0.76V) will become more negative. Is it so??

   

• UltimaOnline

  8 Sep 15, 04:37

  Originally posted by Kahynickel:

  Cu2+ + 2e = Cu   +0.34V (for 1M). 

  If the [Cu2+] is increased right-hand is favoured i.e more Cu will be formed. The value of this half -cell will be more positive (  > +0.34V)

  following the same line.....

  Zn2+   +   2e = Zn    -0.76V. { for 1 M}

  If the [Zn2+] is increased then the forward reaction is favoured i.e the product Zn is formed more and more. The E (-0.76V) will become more negative. Is it so??

   

  
  If molarity of Zn2+ is increased, the position of equilibrium will shift to the RHS, ie. reduction potential of Zn2+ to Zn will become more positive (alternatively, oxidation potential of Zn to Zn2+ will become more negative).
• Kahynickel

  9 Sep 15, 00:10

  In both these non-standard situations the E will become more +ve since the reaction is on RHS. So it does not matter whether the half-cell is +ve or -ve.

• UltimaOnline

  9 Sep 15, 01:20

  Yes, correct.

  Originally posted by Kahynickel:

  In both these non-standard situations the E will become more +ve since the reaction is on RHS. So it does not matter whether the half-cell is +ve or -ve.