Given the two reduction half-reactions for the following fuel cell, determine the standard cell potential $E^\circ_\mathrm{cell}$, $$\ce{CH4 (g) + 2 O2 (g) + 2 OH- (aq) -> CO3^2- (aq) + 3 H2O}.$$ Reduction half-reactions: \begin{align} \ce{CO3^2- (aq) + 7 H2O (l) + 8 e- &-> CH4 (g) + 10 OH- (aq)}& E^\circ &= \pu{-0.1 V}\\ \ce{O2 (g) + 2 H2O (l) + 4 e- &-> 4 OH- (aq)}& E^\circ &= \pu{+0.55 V} \end{align} Answers: (a) $\pu{+0.45 V}$ (b) $\pu{-0.45 V}$ (c) $\pu{-0.65 V}$ (d) $\pu{+0.65 V}$

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

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

I sent an email to my teacher to contest the question using an example from the textbook. Given that Martin has provided me with an answer concurrent with the $\pu{+0.45 V}$, 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 $$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode}.$$

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 $E_\mathrm{cell}$ was found to be $\pu{0.84 V}$. This is because the sign on the $E$ value of the $\ce{Fe}$ 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 textbook example was incorrect, the exam question was incorrect, or my understanding is incorrect.


2 Answers 2


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.


Your intuition is correct. The answer should be +0.65V as in accordance to Hess' law. However, your textbook contains an error on the Ecell formula as it should be $$E_\mathrm{cell} = E_\mathrm{cathode} - E_\mathrm{anode}$$ Note that the values used in this formula are the reduction potential values, hence the reason why Martin's answer leads to the incorrect +0.45V answer as it uses the oxidation potential for the anode instead. This formula also confirms our answer to be correct, with the reduction of O2 being the cathode and the other being the anode. $$E_\mathrm{cell} = +0.55~\mathrm V - (-0.1~\mathrm V)= +0.65~\mathrm V$$ The same applies to the textbook example. The answer derived from Hess' law is the same as the answer derived from the formula $$E_\mathrm{cell} = +0.80~\mathrm V - (-0.04~\mathrm V)= +0.84~\mathrm V$$


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