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Question

It is my understanding that to apply Hess' Law to this equation the top equation sign would be reversed and the outcome would be +0.65 V. Often the system produces incorrect answers. I was looking through my text book to double check this, and there is an instance where it says E cell is E anode - E cathode, but I believe that rule does not fit this instance. Could somebody validate that this answer provided is indeed correct or not?

EDIT:

I would like to clarify the my answer was d) and the system provided that the correct answer was a).

EDIT 2:

I sent an email to my teacher to contest the question using an example from the text book. Given that Martin has provided me with an answer concurrent with the +0.45 V, then there must be a mistake in the text or I am misinterpreting something. I would really like to understand this.

At first my teacher told me that after flipping the equation you still need to subtract the anode from the cathode. So I responded like this:

I would like to reference the E Cell example from the text. In the text it explains that E cell = E cathode - E anode.

The example given in the book is identical to the one provided in the exam:

\begin{aligned} (1): &&\ce{3 Ag+(aq) + Fe (s) &-> 3 Ag (s) + Fe^{3+} (aq)}\\ (2): &&\ce{3 Ag+(aq) + 3 e- &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\ (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\ \end{aligned}

By using the same logic that I have applied to the exam question the E cell was found to be 0.84 V. This is because the sign on the E value of the Fe half reaction was reversed like I did to the top equation (which consequently is the cathode).

As the given equations line up with the primary equation and signs are flipped whenever the equations are flipped, then it is not important to decipher which is the anode and which is the cathode. This is also in accordance with Hess' Law.

That being said, I am confused whether the text book example was incorrect, the exam question was incorrect, or my understanding is incorrect.

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You have to take into account, that the given reactions are only reductions, while you need to have an oxidation, too. So one of the equations has to be reversed.

\begin{aligned} (A)E^\circ = -0.10~\mathrm{V}&&\ce{CO3^{2-} + 7H2O + 8e- &-> CH4 + 10 {}^{-}OH} \\ (B)E^\circ = +0.55~\mathrm{V}&&\ce{O2 + 2H2O + 4e- &-> 4 {}^{-}OH} \\\hline \color{\red}{-1}\cdot(A) + 2\cdot(B) &&\ce{CH4 + 2O2 + 10{}^{-}OH + 4H2O&-> CO3^{2-} + 7H2O + 4 {}^{-}OH}\\ \implies &&\ce{CH4 + 2O2 + 2{}^{-}OH &-> CO3^{2-} + 3H2O} \end{aligned}

Equation $(A)$ gives you the reduction of the carbonate ions, hence the oxidation of methane. Equation $(B)$ is the reduction of oxygen. You have a galvanic cell of the form $\ce{CH4|CO3^{2-}||O2|{}^{-}OH}$, hence the standard potential difference is \begin{aligned} \Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~ &= E^\circ(\ce{O2|{}^{-}OH}) -E^\circ(\ce{CH4|CO3^{2-}})\\ &= E^\circ(\ce{O2|{}^{-}OH}) -\left[\color{\red}{-1}\cdot E^\circ(\ce{CO3^{2-}|CH4})\right]\\ &= +0.55~\mathrm{V} -\left[\color{\red}{-1}\cdot (-0.10~\mathrm{V})\right]\\ \Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~ &= +0.45~\mathrm{V} \end{aligned}


Just to address the second edit of the OP and to clarify: \begin{aligned} (1): &&\ce{3 Ag+(aq) + Fe (s) &-> 3 Ag (s) + Fe^{3+} (aq)}\\ (2): &&\ce{3 Ag+(aq) + 3 e- &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\ (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\ \end{aligned}

The sign of the second equation has to be flipped and then subsequently subtracted from the first equation, hence the potential that should be found is $\Delta E^\circ = 0.74~\mathrm{V}$. Therefore, I believe your textbook is wrong.

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  • $\begingroup$ Martin, thank you for your answer. Since I have asked this question I have contested it with my teacher. I ended up being awarded the mark because I showed that the logic I applied was the same as the logic given in the text book. I will update my question, please have a look and give me your thoughts. Perhaps the text is misleading. In any case I would really like to know the answer (not that yours is wrong, but please have a look at my argument which I ended up winning the mark for the question). $\endgroup$ – Klik Jun 27 '14 at 23:20

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