My lecturer told us that the Q term in the Nernst equation is [Reduced state]/[Oxidized state] which feels really wrong to me. Shouldn't it be [Product]/[Reactant]?
And in his example question it stated:
Given Cu
2++2e
Cu E°=0.337V
Zn
2++2e
Zn E°=-0.763V
Cell notation: Zn
2+/Zn//Cu
2+/Cu
Calculate the cell potential if the concentrations of CuSO4 is 0.1M but ZnSO4 is 0.01M.
So the first problem I have with the question is that the cell notation should have been Zn/Zn
2+//Cu
2+/Cu instead to show that Zn was oxidized and hence its the anode. But that's just a minor point.
The solution he provided was this:
Ecell°=0.337-(-0.763)=1.1V
E=E°-RT/nFln[Cu
2+]/[Zn
2+] and he solve for E from here.
But shouldn't it be E=E°-RT/nFln[Zn2+]/[Cu2+] because the net cell reaction should have been Zn+Cu
2+ Zn
2+ +Cu? I have no idea about the rationale on why he uses Q=[reduced state]/[oxidized state] and after asking him he told me to re-read the notes. This completely contradicts what I have learned in my previous modules.