[mooliak]

Dissolving Cu with Cr2O72-

I am working on a project to dissolve copper with chromic acid, and I want to fully understand it. I want to generate a Nernst equation that gives the galvanic cell potential for the dissolution, and then learn how to produce equations for the reversal with electrolysis. For me, this is very complicated, and I would really appreciate some help. These are my efforts so far.
 
First, the dissolution.

The Cr6+ comes in the form of CrO3 granules which are dissolved in water, acidified with 200 g/L H2SO4 .  This I believe forms chromic acid, H2CrO4 which disociates and dimerises to form the Cr2O72- dichromate ion, explaining why it goes orange. When metallic Cu is placed in this, it gives a strongly exothermic reaction, and dissolves, the half cell reactions are as follows:

     Cr2O72-  +14 H+    +6e-                2 Cr3+  +7 H2O      E0   = 1.33 V     (1)
      
           Cu2+  +  2e-                Cu            E0   = 0.34 V   (2)

   Clearly the copper will be oxidised so it can be re written:

               Cu          Cu2+  +  2e-             E0   = - 0.34 V   (3)
And balanced  3 x (Cu       Cu2+  +  2e-   ) 

 Adding 1 & 3 x 3, we get:

  Cr2O72- +14 H+ +3Cu +6e-          2 Cr3+ +3Cu2+ +6e- +7 H2O   E0  = 0.99 V  (4)

My concentrations are Cu2+  50 g/L, H2SO4 200g/L. In order to get the copper conc., the amount of Cr6+ required is : Cu2+ 50 g/L = 50/63.5 M = 0.79M and (4) tells us the mole ratio of Cr6+  to  Cu2+   is 2:3 so we need 2/3 x 0.79 M  Cr6+, this is 0.53 M or 0.53 x 52 g/L   which is 27.4g/L Cr6+ (reckon this will be present as dichromate, see shortly)

Then we get to the bit I’m not sure of (assuming I haven’t cocked the above up !). The electrode potentials are correct in standard conditions, but I’m not 100% on how to develop the Nernst equation.

My concentrations are as follows:
 
Reactants

No matter what form it’s in, the Cr6+ is 27.4 g/L so if it is dichromate, it is
  (2x52 + 7x16) / (2x52)  x 27.4 g/L which is 56.9 g/L Cr2O72-  assuming the chromic acid is fully disociated. This means that the molarity of dichromate is 56.9 / 216 or 0.26 M.

re H+ , H2SO4 200g/L, molarity is 200 / (2+32+4x16) or 2.04 M.

Question: Is it safe to assume that the sulphuric only provides one proton according to H2SO4     H+  +  HSO4-   and that the molarity of  H+ is therefore 2?  Assuming it is, we get :
(Cu doesn’t count because it’s in the bulk metallic state.)

Products:
Cr3+ : this is still 27.4 g/L, because by titration, none of the 6+ is evident.
So 27.4 / 52  gives  Cr6+ = 0.52M   

  Cu2+ is at 50 g/L which is 50 / 63.5  so Cu2+ = 0.79 M

H2O  doesn’t count  because it’s not ionic.

Assuming that you haven’t all gone to sleep yet, finally we get Nernsty out !          E =   E0 – RT/nF. ln Q
      
 n = 6,  E0 = 0.99  V,  RT/F = 8.314 x 298 / 96485 = 0.02568

        Q = [Cr3+]^2 [Cu2+]^3  / { [Cr2O72-] [H+]^14 } 
    = 0.522 x 0.793 / (0.26 x 2 ^14)
Q = 0.000031

    E =   E0 - 0.02568 x ln (0.000031)  = 0.99 - 0.0257 x (-10.382)  =  1.26

   E   =  1.26 V

  Do you reckon the above is anywhere near accurate ?
  To see it practically, I thought of setting it up against one of the known reference cells  like Calomel, and measuring it with a copper electrode in a  Cr2O72- , Cr3+ , Cu2+ & H2SO4 solution of appropriate concentration. Then I thought why not use an easy cell as ref. Eg.  Cu in  CuSO4, would this work,  (obviously with the correct adjustment), or have I got the wrong idea?

Also, to set up a cell with a Pt wire, it seems to cost a fortune to buy the wire, way above the metal content.  a cheap way to do this seems to be to buy and dismantle a Pt100 thermocouple. Presumably the wires are pretty thin if they only cost £20. Does the Pt need any special treatment, or would this be ok for a decent result ?

Hope all this isn’t asking too much.

[Borek]

dissolve copper with chromic acid

Interesting, similar question was already asked today.

You may have to take copper chromate/dichromate solubility into account. But perhaps it won't be necessary, it is possible that pH is low enough to protonate Cr₂O₇^2-, making precipitation impossible.

200g/L of H₂SO₄ is about 2M, ionic strength is high enough to make precise calculations impossible.

Yes, at 2M you can safely assume it is HSO₄^- only. This is an approximation, but good enough.