Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: 
Reduction: 
---------------------------------------------------------------------------------
Overall: 
Nernst equation for this cell reaction at
-
![E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE_%7Bcell%7D%5E%7B0%7D-%5Cfrac%7B0.059%7D%7Bn%7Dlog%7B%5BAl%5E%7B3%2B%7D%5D%5E%7B2%7D%5BI%5E%7B-%7D%5D%5E%7B6%7D%7D)
where n is number of electrons exchanged during cell reaction,
is standard cell emf ,
is cell emf ,
is concentration of
and
is concentration of 
Plug in all the given values in the above equation -
![E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.20-%5Cfrac%7B0.059%7D%7B6%7Dlog%5B%284.5%5Ctimes%2010%5E%7B-3%7D%29%5E%7B2%7D%5Ctimes%20%280.15%29%5E%7B6%7D%5DV)
So, 
Answer:
The Net reaction is
-

-

-
Explanation:
From the Question we are told that the buffers are
and 
When NaOH is added the Net ionic reaction would be
-

-

-
A base and an Acid always react to form a salt and water
So, HCl + NaOH —> NaCl + HOH
Answer:
the ionic radius of the anion 
Explanation:
From the diagram shown below :
The anion
is located at the corners
The cation
is located at the body center
The Body diagonal length = 
∴ 
Given that :
(i.e the ratio of the ionic radius of the cation to the ionic radius of
the anion )

Also ; a = 664 pm
Then :

Therefore, the ionic radius of the anion 