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.