[krishnaraj]

 in acid medium the standard reduction potential of NO converted to N2O is 1.59V .its standard potential in alkaline meduim at 298k would be?
My attempt-
a)2NO+2H ---> N2O +H20   E=1.59
b)2NO + H2O ---->N2O + 2OH-   E=x
c)2H2O---->2H+ + 2OH-   E=0.826 (calculated, correct and verified)
 Reversing b) we get E= -x
adding a and reversed b, we get reversed c. Therefore 1.59-x=-0.826, so x should be 1.59+0.826 but this is incorrect. Where am I wrong? any help would be appreciated.

[mjc123]

You've got the sign of C wrong. (Strictly it doesn't have an E because it is not an electrochemical reaction, but you can get the answer using free energies.) The correct equation is
H₂O + 2e  H₂ + 2OH^- E° = -0.826V
Combine that with 2H⁺ + 2e  H₂ E° = 0V to get equation C.
You can tell your sign's wrong because you know the water ionisation equilibrium is well to the left (K_w = 10^-14)

[krishnaraj]

I know RTlnK=nFE. R, n, F and T are positive while K is negative so E must be negative. Alright, I see your point. I will give it another attempt and get back if I have any trouble in a while, thank you

[krishnaraj]

Alright, I got my mistake and the answer. The way I obtained E for 2H+ + 2OH- = 2H2O was that the original equation (divided by 2) had 0 E and k=10^(-14)
while this one has k=(10^-28) therefore from nernst equation- E=0-(0.059/2)log(10^-28) which became negative 0.826

[mjc123]

0-(0.059/2)log(10^-28) = +0.826!
What is n for the reaction 2H⁺ + 2OH^- = 2H2O ? How many electrons are transferred?

[krishnaraj]

Ah!! such a silly mistake, I ignored that log(10^-28) is *minus* 28. 2 electrons are transferred. Its clear now, thank you very much

[mjc123]

No electrons are transferred. It is not an electrochemical reaction. Nothing is oxidised or reduced. That's why, properly speaking, it can't have an E.

[krishnaraj]

ohh, right- the oxidation number doesn't change. How were we able to use E to solve the problem correctly then?

[mjc123]

Consider the following equations:
(1) 2NO + 2H⁺ + 2e  N₂O + H₂O
(2) 2NO + H₂O + 2e  N₂O + 2OH^-
(3) 2H₂O  2H⁺ + 2OH^-
(4) 2H⁺ + 2e  H₂
(5) 2H₂O + 2e  H₂ + 2OH^-
You were given E°(1) and asked to find E°(2)
We notice that (2) = (1) + (3)
Thus ΔG°(2) = ΔG°(1) + ΔG°(3). If we know ΔG°(3) we can work out ΔG°(2) and hence derive E°(2)
Can we write nFE°(2) = nFE°(1) + nFE°(3)? What are n(3) and E°(3)? This, as written, is a chemical, not an electrochemical, reaction. Nothing is oxidised or reduced.
But wait. We can also write (3) = (5) - (4)
So ΔG°(3) = ΔG°(5) - ΔG°(4)
Then we can say (2) = (1) + (5) - (4)
So ΔG°(2) = ΔG°(1) + ΔG°(5) - ΔG°(4)
Then 2FE°(2) = 2FE°(1) + 2FE°(5) - 2FE°(4) or simply E°(2) = E°(1) + E°(5) - E°(4), or just  E°(1) + E°(5), as E°(4) = 0.

[krishnaraj]

I understand your reply to the Fe2+/3+ post even better now seeing its application here, thank you very much