Constructing redox reactions

• anpanman

  5 Mar 10, 18:31

  I need some help in constructing component half equations from overall equations regarding redox reactions

  Say we have Iodate(V)Ion, IO3 -

  the overall equation being: IO3-  +  5I-   + 6H+ -> 3I2 + 3H2O

  How can I derive the reduction and oxidation equations from here? Do give me a step-by-step solution... because I'm clueless as to where to start. Thanks all!

• UltimaOnline

  5 Mar 10, 19:48

  Originally posted by anpanman:

  I need some help in constructing component half equations from overall equations regarding redox reactions

  Say we have Iodate(V)Ion, IO3 -

  the overall equation being: IO3-  +  5I-   + 6H+ -> 3I2 + 3H2O

  How can I derive the reduction and oxidation equations from here? Do give me a step-by-step solution... because I'm clueless as to where to start. Thanks all!

  
  Based on OS (oxidation state aka oxidation number), determine which species on the LHS is oxidized, and which is reduced.

   

  For this equation, iodate(V) ion is reduced to molecular iodine (OS from +5 to 0), and the iodide ion is oxidized to molecular iodine (OS from -1 to 0).

   

  So there you have your half-equations. Balance them as per standard rules of balancing redox equations.

   

  Step 1

  Balance the hetero-element atoms. (ie. atoms of any other element other than O and H).

   

  Step 2

  Balance the oxygen atoms by adding water.

   

  Step 3

  Balance the hydrogen atoms by adding protons (ie. H+).

   

  Step 4

  Balance the charges by adding electrons (ie. e-).

   

  Step 5

  For alkaline conditions, neutralize away any H+ by adding enough OH- ions on both sides.

• anpanman

  5 Mar 10, 22:25

  Originally posted by UltimaOnline:

  
  Based on OS (oxidation state aka oxidation number), determine which species on the LHS is oxidized, and which is reduced.

   

  For this equation, iodate(V) ion is reduced to molecular iodine (OS from +5 to 0), and the iodide ion is oxidized to molecular iodine (OS from -1 to 0).

   

  So there you have your half-equations. Balance them as per standard rules of balancing redox equations.

   

  Step 1

  Balance the hetero-element atoms. (ie. atoms of any other element other than O and H).

   

  Step 2

  Balance the oxygen atoms by adding water.

   

  Step 3

  Balance the hydrogen atoms by adding protons (ie. H+).

   

  Step 4

  Balance the charges by adding electrons (ie. e-).

   

  Step 5

  For alkaline conditions, neutralize away any H+ by adding enough OH- ions on both sides.

  Thanks for helping!

  And may I ask anyone over here, is nitric (V) acid is the same as nitric acid? hahah, saw a question with nitric(V) acid and i tried to look up using google but didnt churn up useful results. but i guess it's the same anyway.

• UltimaOnline

  5 Mar 10, 22:47

  Originally posted by anpanman:

  Thanks for helping!

  And may I ask anyone over here, is nitric (V) acid is the same as nitric acid? hahah, saw a question with nitric(V) acid and i tried to look up using google but didnt churn up useful results. but i guess it's the same anyway.

  
  Latin name vs Stock names.

   

  HNO2

  Latin name : Nitrous acid

  Stock name : Nitric(III) acid

   

  HNO3

  Latin name : Nitric acid

  Stock name : Nitric(V) acid

   

  NO2-

  Latin name : nitrite ion

  Stock name : nitrate(III) ion

   

  NO3-

  Latin name : nitrate ion

  Stock name : nitrate(V) ion

   

   

  FYI, in addition to latin and stock names, there are also prefix names. And for organic compounds, there are systematic IUPAC names versus common names.

   

  Eg. PCl5 is known as phosphorous pentachloride (prefix name) and phosphorous(V) chloride.

   

  Eg. H3COH3 is known as methanal (systematic IUPAC name) and formaldehyde (common name).

   

  And so on and so forth.

   

   

  Chemistry is soooo fun!