Post a New Question

chemistry

posted by .

The electrochemical cell described by the cell notation has a standard emf (electromotive force) of -0.68 V. Calculate the value (kJ) for the standard free energy change of the cell. Round your answer to 3 significant figures.

Pt(s) l MnO2(s) l MnO4-(aq), H+(aq) ll H+(aq), VO2+(aq), VO2+(aq) l Pt(s)

  • chemistry -

    i know how to do the problem but im not sure how the H+ effects the charge

  • chemistry -

    I think it's just delta G = -nFE and H^+ doesn't fit in unless you are calculating K.

  • chemistry -

    but don't you need to incorporate the charge of H+ when finding n

  • chemistry -

    The charge must balance, yes, but the n is determined from MnO4- to MnO2 and VO2+ to VO2+ (why do you have two VO2+?--one of them should be something else.

  • chemistry -

    oh, im sorry. i wrote it wrong. But i also had one more question. Do the coeffiecents have any effect on our calculation of "n"?

  • chemistry -

    for example this problem. the n is 6. how do we derive that?

    The galvanic cell described by the balanced chemical equation has a standard emf of 0.10 V. Calculate the maximum electrical work (kJ) the cell has done if 1987.2 g of Cr2O72-(aq) (Molar Mass - 216.00 g/mol) reacts. Round your answer to 3 significant figures.

    2Cr2O7^2-(aq) + 16H^+(aq) → 3O2(g) + 4Cr^3+(aq) + 8H2O(l)

  • chemistry -

    Yes, the coefficients matter. When calculating the n for a reaction, it is the TOTAL n for the TOTAL atoms.
    Cr in Cr2O7^-2 on the left is +5 for each Cr atom 6. That is 24 for 4 Cr atoms. On the right we have 4 Cr^+3 atoms at 12. Change is 12-6 = 6

  • chemistry -

    Im sorry but i don't get it..... wouldnt the Cr in C2O72- have a 6+ charge? and Cr3+ has 3+ charge. So the left Cr would be 6*4 =24 and the right Cr is 3*4= 12 so wouldnt that be 24-12= 12? what was your logic behind reducing it to 6? im sorry but this really confuses me

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question