Question on chapter 10 BR electrochemistry

[dannybht]

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I don't get how the overall electric potential for number 32 is - .40...aren't you suppose use the Nernst equation when doing this not the E cell = Eoxidation + E reduction???

I got -0.76 using the Nernst equation

 

[ashtonjam]

Using the Nernst equation doesn't give you Estandard, but the problem specifically asks for Estandard (E with a dot in the right corner). The Nernst equation gives you Ecell using Estandard and concentrations at anode and cathode (both half cells together, which isn't what is being asked). In Reaction 1, it says Fe2+ -> Fe3+ in the presence of permanganate. So to get the right answer, you leave the second reaction (Fe2+ + 2e- -> Fe) in the same orientation and flip the first reaction to get (Fe -> Fe3+ + 3e-). Then you just add Eox + Ered = Estandard.

 

[dannybht]

View attachment 24911

How come in passage 13 of berkeley review it tells us to use the nernst equation delta g = -nFEstandard

 

[dannybht]

 

[ashtonjam]

That is actually really weird. The only thing I could find was a university website link here. It still doesn't entirely make sense because in both TBR problems you end up with a half reaction. Sparknotes also mentions that if there are leftover electrons, you can't just add the reaction EMFs together.

Example: Determine the standard potential of the Cu2+/Cu+ cell from two other standard cell potentials.
Eº(Cu2+/Cu) = +0.340V and Eº(Cu+/Cu) = +0.522 V.

Since Hess' Law allows us to add Gibbs energies for the reactions to arrive at the Gibbs energy of the desired reaction, we should go via Gibbs energies. Convert our standard potentials into Gibbs energies, perform the addition and then convert back to a standard potential.

The two reactions that will occur are:

(a) Cu(s) -> Cu2+(aq) + 2 e–. Eº = –0.340V ΔrG = –2(–0.340 V) F = 65.6 kJ/mol
(b) Cu+(aq) + e– -> Cu(s) Eº = +0.522V ΔrG = –(+0.522 V) F = -50.4 kJ/mol
(c) Cu2+(aq) + e– -> Cu+(aq) Eº = +0.182V ΔrG = ( 0.158 V) F = 15.2 kJ/mol

Now, since Eº = –ν ΔrG/F = –0.158V (ν = 1 in the final equation)

(note that the final equation in the text book example 7.4 is reversed, hence the sign change for Eº)

We cannot simply add the electrode potentials in this case because, the final reaction is still a half-reaction.

If the final reaction is a complete cell reaction, with no electrons remaining, then we can skip to just adding the cell potentials because the factor ν would be the same in all equations and hence, cancel out.

 

[dannybht]

Lol yeah right? I guess maybe the answer solution for the first BR problem I posted is wrong... if its true that half reactions need to use nernst and for full cell redox equations you can go straight using Ecell = Ered + E Oxidation?