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

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?

• The title seems rather confused to me. – Poutnik Sep 21 at 19:21
• Check how are related the potential of a half reaction and it's Gibbs energy charge. Then you are 1 step from recognizing the rules for deriving one halfreaction potential from other ones. – Poutnik Sep 21 at 19:31
• Conceptually related? chemistry.stackexchange.com/questions/139495/… – Safdar Sep 21 at 19:33
• @Maurice: might be an issue in calculations. I get $\pu{0.1285 V}$ though, which IIRC rounds to $\pu{0.128 V}$.. – Safdar Sep 21 at 19:34
• chemistry.stackexchange.com/questions/9454/… – Mithoron Sep 21 at 22:55

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.

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

• I thought of this method and got this answer but it's different from the textbook's so I assumed an error. Just a textbook typo then? – Josh Mitchell Sep 21 at 20:32
• hang on because surely if you just put these species in a Latimer diagram and worked out the potential for HNO2 to N2O you get (0.98+1.59)/2 = 1.29V, not 1.39. – Josh Mitchell Sep 21 at 21:02
• Yeah I think you've made a mistake because the Gibbs energy for the first reaction should be 2*-0.98F as you've multiplied that reaction by 2 to get the overall equation balanced. Following the rest through you get 1.29V. – Josh Mitchell Sep 21 at 21:07
• Thank you Josh Mitchell. I forgot the factor 2. I have edited and corrected my initial text. Thank you ! – Maurice Sep 22 at 8:08
• @Maurice No, $F$ is the Faraday constant (about $\pu{96500 C mol-1}).$ Wrong constant name, notation (prime?) and unit symbol. Also, none of the units you've spelled out should be capitalized. – andselisk Sep 22 at 8:25