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

chemistry 
DrBob222,
I think it's just delta G = nFE and H^+ doesn't fit in unless you are calculating K.

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

chemistry 
DrBob222,
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 
me,
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 
me,
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 
DrBob222,
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 126 = 6 
chemistry 
me,
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 2412= 12? what was your logic behind reducing it to 6? im sorry but this really confuses me