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chemistry

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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"?

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    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)

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    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

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