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For this question why didn't you use the equation with the electrode potential +0.40 for oxygen ?We need to look at the bonding in the -CONH2 group.
Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond.
A pi bond is made by sideways overlap between p orbitals on the carbon and the oxygen.
In an amide, the lone pair on the nitrogen atom ends up almost parallel to these p orbitals, and overlaps with them as they form the pi bond.
image: http://www.chemguide.co.uk/organicprops/amides/amidedeloc.gif
The result of this is that the nitrogen lone pair becomes delocalised - in other words it is no longer found located on the nitrogen atom, but the electrons from it are spread out over the whole of that part of the molecule.
This has two effects which prevent the lone pair accepting hydrogen ions and acting as a base:
- Because the lone pair is no longer located on a single atom as an intensely negative region of space, it isn't anything like as attractive for a nearby hydrogen ion.
- Delocalisation makes molecules more stable. For the nitrogen to reclaim its lone pair and join to a hydrogen ion, the delocalisation would have to be broken, and that will cost energy.
Compare E values of the metal getting reduced at the cathode with that of H2 (which is zero). So if it's positive, the metal gets discharged. If negative, hydrogen gets discharged.
Ag+ + e– ⇌ Ag E = +0.80 so Ag discharged
Fe2+ + 2e– ⇌ Fe E = –0.44 so H2 discharged
Mg2+ + 2e– ⇌ Mg E = –2.38 so H2 discharged
Compare E values of the anion with that of
O2 + 4H+ + 4e– ⇌ 2H2O E = +1.23
F2 + 2e– ⇌ 2F– E = +2.87 more positive so O2 discharged
S2O82– + 2e– ⇌ 2SO42– E = +2.01 more positive so O2 discharged
Br2 + 2e– ⇌ 2Br– E = +1.07 less positive so Br2 discharged