[mol]

write the balanced redox reaction, determine E* cell (*= zero) for
1. the reduction of dichromate ions to chromium (3) ions by Mn 2+, forming MnO2.
2. the oxidation of As O3 (3-) to H2AsO4 by the reduction of iodine to iodide ions.
 I could not find the std redctn potential values for these polyatomic iosn as I found for the individual ions from the table. anyone pls help me with this problem?

[Borek]

Here perhaps:

http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

[mol]

Thank you Borek. That link gave me the direct vaules for the half reactions. I need to know one more thing because in my book the table does not have all the values like this one.... but still if I am asked to solve the problem is there any other alternative way that I should take into acccount. say for eg... 

the answer given to those 2 eqtns were:
1.     2H^+  +Cr207^2- +3Mn^2+  ----->3MnO2 + 2Cr^3+  +H20
        E^0cell =  +1.33V - (+1.23v) = +0.10v

2.     H20  + As03^3-  +I2  ------->H2AsO4^- +2I^-
        E^0Ccell   =+o.54V-  (+0.58V) = -0.04V
I have the values for the H+ reaction and 02 reactions and Mn and for Cr3+. Not for the Cr207 or H2AsO4.does this mean I should combine any 2 reaction(like H2+ and Mn and O2 values for the first one and combine H+ and AsO3 for the 2nd one) to get the correct value of ONE half reaction or is that totally a wrong approach?

[Borek]

Determine HALF reactions first. And write them here, so that we both know what you are talking about.

[mol]

The question says:
write the balanced redox reaction,and  determine E* cell (*= zero) for
1. the reduction of dichromate ions to chromium (3) ions by Mn 2+, forming MnO2.

2. the oxidation of As O3 (3-) to H2AsO4 by the reduction of iodine to iodide ions.


the answer given to those 2 eqtns were:
1.     2H^+  +Cr207^2- +3Mn^2+  ----->3MnO2 + 2Cr^3+  +H20
        E^0cell =  +1.33V - (+1.23v) = +0.10v

2.     H20  + AsO3^3-  +I2  ------->H2AsO4^- +2I^-
        E^0Ccell   =+o.54V-  (+0.58V) = -0.04V
Only the qstn and answer like shown above was all given in the book.

This is how I tried to solve:

1. 14H^+   + Cr207^2-  ---> 2Cr^3+ + 7H20 +4e-
Cr is oxidised from 2- to 3+
so this is the  oxidation half-reaction.
   
3Mn^2+  6H20 +6e- ----> 3MnO2 +  12H^+
Mn is reduced from 2+ to 0
So this is the redctn half-reaction.

E* cell =E cathode- E anode


2.H20 + AsO3^3-   ----->H2AsO4^-  +2e-
 As changes from -3 to -1 charge. so oxidation half-rctn.

I2 +2e-  -----> 2I^-
Iodine changes from 0 to -2 .so reduction half-reaction.
After this step im stuck in finding the E cathode and E anode values from the table
because I dont see any values for these half reactions specifically.
can u pls help me where I am going wrong?
thanks

[Borek]

2.H20 + AsO3^3-   ----->H2AsO4^-  +2e-
 As changes from -3 to -1 charge. so oxidation half-rctn.

This is not different from the reaction of oxidation of H₃AsO₃ to H₃AsO₄. you may need to account for pH to calculate exact potential (after all that's what the Nernst equation is for), but data from the wikipedia page is enough.

[mol]

Thanx  a lot Borek for heling me with the infos . I will get them from the wiki table.