Why does ΔG for a reaction scale with reaction quantity but E does not? For example, ΔG0rxn for the combustion of 1 mol of hydrogen is 1 × –237 kJ∕mol = –237 kJ,while ΔG0rxn for the combustion of 2 mol of hydrogen is 2 × –237 kJ∕mol =–474 kJ. In both case

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Feb 9, 2018

#E_(cell)^@# scales with the mols of electrons, as well as the mols of reactants... But they cancel out.

#DeltaG_(rxn)# is related to #RTlnQ#, and we know already that #Q -> Q^c# if the reaction is scaled, so #DeltaG_(rxn)# is scaled by #c# since #RTlnQ^c = cRTlnQ#.


In the Nernst equation we have

#E_(cell) = E_(cell)^@ - (RT)/(nF) lnQ#,

and the units work out as

#(("V" cdot "C/mol reactant" cdot "K")("K"))/(("mol e"^(-)"/mol reactant")("C/mol e"^(-)))#

to give units of #"V"#. It must, or else #E_(cell)^@# can't add to #-(RT)/(nF)lnQ#...

If you scale the reaction by a factor of #c#, you raise #Q# to the #c# and multiply the number of electrons #n# by #c#.

For some general redox reaction, we would get something like:

#M^(+)(aq) + A(s) rightleftharpoons A^(+)(aq) + M(s)#

#" "" "" "darr#

#cM^(+)(aq) + cA(s) rightleftharpoons cA^(+)(aq) + cM(s)#

The charge contribution on each side clearly changes. Thus, you would get that the number of electrons changes as #n -> cn# and #Q -> Q^c#. As a result:

#-(RT)/(nF) lnQ -> [-(RT)/(cnF) lnQ^c = -(RT)/(cancel(c)nF)cdotcancel(c)lnQ]#

#= -(RT)/(nF) lnQ#

which is the same expression as before. If we do not affect #-(RT)/(nF) lnQ#, then by inference, #E_(cell)# and #E_(cell)^@# do not change due to scaling the reaction by a constant.

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