Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, 
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
![[Cl^{-}{diluted}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bdiluted%7D%5D)
= 0.0222 M
![[Cl^{-}{concentrated}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D)
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
When there are pi bonds as well as the sigma bonds, and therefore a double or triple bond. When it can change where that double bond is you can draw a resonance structure. Ozone, O3, has two resonance structures because you can rearrange the bonds in it.
How am I supposed to help you if there's not picture of what the problem is
NH3 is a molecule with a nitrogen atom in the middle surrounded by 3 hydrogen atoms bonded covalently. Nitrogen has 5 valence electrons so, it will form sigma bonds (sp3 hybrid orbitals) with each of the hydrogen atom. The remaining lone pair electrons will be situated above the atom. You ca look at this image I've pulled from Google.
