[confusedstud]

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 Cu²⁺+2e Cu E°=0.337V
Zn²⁺+2e Zn E°=-0.763V
Cell notation: Zn²⁺/Zn//Cu²⁺/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/Zn²⁺//Cu²⁺/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[Cu²⁺]/[Zn²⁺] 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+Cu²⁺ Zn²⁺ +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.

[mjc123]

I think you're right. I don't know what he means by "Q=[reduced state]/[oxidized state]" because Zn²⁺ and Cu²⁺ are both oxidised states. Perhaps he meant E_M2+/M = E° - RT/nF ln[M]/[M²⁺], in which case he should have got E_cell = E_cell° - RT/nF ln{[Cu]/[Cu²⁺]*[Zn²⁺]/[Zn]}; and [Zn] = [Cu] = 1.
This can be tricky to remember - I just made a mistake in another thread - and I always like to go by Le Chatelier - what would be the effect of increasing, say, [Cu²⁺]? Which way would the equilibrium go? Would that make E more positive or more negative?

[Hunter2]

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 Cu²⁺+2e Cu E°=0.337V
Zn²⁺+2e Zn E°=-0.763V
Cell notation: Zn²⁺/Zn//Cu²⁺/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/Zn²⁺//Cu²⁺/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[Cu²⁺]/[Zn²⁺] 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+Cu²⁺ Zn²⁺ +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.

All calculations are correct.

But mostly the Nernst equation is founf in this way: E = E₀ + RT/nF* ln Eox/Ered.


He used a Minus instead lnx = - 1/lnx

[confusedstud]

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 Cu²⁺+2e Cu E°=0.337V
Zn²⁺+2e Zn E°=-0.763V
Cell notation: Zn²⁺/Zn//Cu²⁺/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/Zn²⁺//Cu²⁺/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[Cu²⁺]/[Zn²⁺] 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+Cu²⁺ Zn²⁺ +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.

All calculations are correct.

But mostly the Nernst equation is founf in this way: E = E₀ + RT/nF* ln Eox/Ered.


He used a Minus instead lnx = - 1/lnx

Hmm isn't the derivation like this
ΔG=ΔG°+RTlnQ so -nFE==nFE°+RTlnQ and dividing -nF by both sides gives us E=E°-RT/nFlnQ? And for the equation isn't Q=[Zn2+]/[Cu2+]?

[Hunter2]

Its only a matter of logarithm mathematics lnA/B = -lnB/A

[confusedstud]

Its only a matter of logarithm mathematics lnA/B = -lnB/A

Hmm then shouldn't my lecturer's equation be E=E°+RT/nFln[Cu2+]/[Zn2+] instead? But his equation still had a -RT/nFln[Cu2+]/[Zn2+] term.

[Hunter2]

You are right this must be an error.

a -RT/nFln[Cu2+]/[Zn2+]

Here a calculation from a german file

https://de.wikibooks.org/wiki/Anorganische_Chemie_f%C3%BCr_Sch%C3%BCler/_Elektrochemie_III_-_Konzentrationsabh%C3%A4ngigkeit_galvanischer_Elemente,_die_Nernst-Gleichung_und_ihre_Anwendung

[confusedstud]

You are right this must be an error.

a -RT/nFln[Cu2+]/[Zn2+]

Here a calculation from a german file

https://de.wikibooks.org/wiki/Anorganische_Chemie_f%C3%BCr_Sch%C3%BCler/_Elektrochemie_III_-_Konzentrationsabh%C3%A4ngigkeit_galvanischer_Elemente,_die_Nernst-Gleichung_und_ihre_Anwendung

How should I tell him about this problem? My lecturer simply says that I should re-read the notes whenever I ask him about this problem and he's quite senior so the other lecturers would tend to agree with him.

[Hunter2]

With facts. Send him in moderate words an Email and show your calculation. Or show him the link, maybe its also possible to find an english link for zinc copper Nernst equation.

[confusedstud]

With facts. Send him in moderate words an Email and show your calculation. Or show him the link, maybe its also possible to find an english link for zinc copper Nernst equation.

I will try that. But I will probably have to give his wrong answer in my examination papers. A similar incident happened in my last term when he told us that in the Haber Process, the pressure remains the same before and after reaction as the gases involved are considered ideal.. couldn't convince him that number of moles changes so P changes there so i doubt i can convince him about this too.

Have you experienced a similar scenario before?

[Old_Doc48]

You are right this must be an error.

a -RT/nFln[Cu2+]/[Zn2+]

Here a calculation from a german file

https://de.wikibooks.org/wiki/Anorganische_Chemie_f%C3%BCr_Sch%C3%BCler/_Elektrochemie_III_-_Konzentrationsabh%C3%A4ngigkeit_galvanischer_Elemente,_die_Nernst-Gleichung_und_ihre_Anwendung

How should I tell him about this problem? My lecturer simply says that I should re-read the notes whenever I ask him about this problem and he's quite senior so the other lecturers would tend to agree with him.

You sure your prof didn't say ['Reducing Agent']/['Oxidizing Agent']. This is a matter of semantics Q = [Concentration of Reducing Agent]/[Concentration of Oxidizing Agent]. Just a thought. 

[confusedstud]

You are right this must be an error.

a -RT/nFln[Cu2+]/[Zn2+]

Here a calculation from a german file

https://de.wikibooks.org/wiki/Anorganische_Chemie_f%C3%BCr_Sch%C3%BCler/_Elektrochemie_III_-_Konzentrationsabh%C3%A4ngigkeit_galvanischer_Elemente,_die_Nernst-Gleichung_und_ihre_Anwendung

How should I tell him about this problem? My lecturer simply says that I should re-read the notes whenever I ask him about this problem and he's quite senior so the other lecturers would tend to agree with him.

You sure your prof didn't say ['Reducing Agent']/['Oxidizing Agent']. This is a matter of semantics Q = [Concentration of Reducing Agent]/[Concentration of Oxidizing Agent]. Just a thought.

His notes stated Q=[Reduced state]/[oxidized state] though. But even so in the equation Zn+Cu2+ Zn2+ +Cu Zn is the reducing agent and Cu2+ is the oxidizing agent so wouldn't the equation here not work too? Because Q would be [Zn]/[Cu2+].

[Old_Doc48]

Yes, Zn(s) is the reducing agent and Cu⁺² is the oxidizing agent.  It was just a thought referencing semantics. Tough situation to be in, but I'd never recommend giving a 'wrong' answer on an exam if you are sure about the concept and calculation. I would guess there may be others in your class that have the same issue... There's always strength in numbers (be polite) but if 2 or three of you are presenting the same (correct) argument, it would be hard to defend and incorrect assumption... Anyways, Best of luck.