Calculating the cell potential of a half cell from two half cell equations [duplicate]

Question

I understand the premise of the question and how to solve it but I don't quite understand why I'm a factor of 2 out:

Construct a potential diagram (at $\mathrm{pH} = 0$) for the reduction of aqueous $\ce{HNO2}$ to $\ce{NO}$ and then to $\ce{N2O}$ given that $E^0$ for the $\ce{HNO2/NO}$ and $\ce{NO/N2O}$ couples are $\pu{+0.98 V}$ and $\pu{+1.59 V}$ respectively. Calculate $E^0$ for the following half-reaction: $$\ce{2HNO2 + 4H+ +4e- <=> N2O + 3H2O}$$

So by constructing the two half-equations:

$$\ce{HNO2 + H+ +e- <=> NO + H2O~~~~~~~~~~~~(E^\circ= +0.98V)}$$ $$\ce{2NO + 2H+ +2e- <=>N2O +H2O~~~~~~~(E^\circ= +1.59V)}$$

And so by adding 2 times the first equation to the second, you obtain the desired overall equation. And so you would add the individual reduction potentials to obtain the overall potential:

$$\pu{+0.98 V + 1.59 V} = \pu{+2.57 V}$$

The answer stated however is $\pu{+1.29 V}$, ie, half of my answer. Why do we need to divide by $2$ at the end here? The standard reduction potential doesn't depend of stoichiometry, and when doing similar questions combining two half cells into a full equation (with no electrons in the equation), simply subtraction or adding the equations is sufficient. Why the difference here?

Answer 1

Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.

So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving Latimer diagrams.

You would work on the portion starting from nitrous acid to nitrous oxide.

Answer 2

By definition : $\ce{\Delta $G = - zEF$}$, where $z$ is the number of electrons in the half-reaction, and $F$ is the Faraday (about $96'500$ Cb)

In the first half-reaction , $z = 1$, and $E = +~ 0.98$ V. So $\Delta G_1 = - ~0.98~ F$ (in Joules)

In the second half-reaction, $z = 2$, and $E = +~1.59$ V. So $\Delta G_2$ = - $2·1.59 F$ = - $3.18 F$ (in Joules)

For the final process, which is the sum of twice the $1$st and $2$nd equation, the free energies must be added. $\Delta G_f$ = $\ce{2 \Delta $G$_1} + {\Delta G_2} = 2(-~ 0.98~ F) + (- 3.18~ F) = -~ 5.14~ F$ (in Joules)

But the total number of electrons is $2+2·1 = 4$.

So the corresponding potential is : $\ce{E_f} = -\frac{\Delta G_t}{4~F} = \frac{5.14~ F}{4~ F} = + ~1.29$ V