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
Answer:
Removal of Third Electron
Explanation:
a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.
Ionization energy increases from bottom to top within a group, and increases from left to right within a period.
Answer:
Reaction I: Sodium + Aluminum chloride →Sodium chloride + Aluminum
Explanation:
Sodium being more reactive means that it will take the place of aluminium in whats called a displacement reaction and form.
Sodium chloride + Aluminum
Explanation:
When pH of the solution is 11.
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![11=-\log[H^+]](https://tex.z-dn.net/?f=11%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=1\times 10^{-11} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-11%7D%20M)
..(1)
At pH = 11, the concentration of 
ions is 
.
When the pH of the solution is 6.
![pH=-\log[H^+]'](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%27)
![6=-\log[H^+]'](https://tex.z-dn.net/?f=6%3D-%5Clog%5BH%5E%2B%5D%27)
![[H^+]'=1\times 10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D1%5Ctimes%2010%5E%7B-6%7D%20M)
..(2)
At pH = 6, the concentration of ions is 
.
On dividing (1) by (2).
![\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} M}{1\times 10^{-6}}=1\times 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BH%5E%2B%5D%27%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-11%7D%20M%7D%7B1%5Ctimes%2010%5E%7B-6%7D%7D%3D1%5Ctimes%2010%5E%7B-5%7D%20)
The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is 
.
Difference between the ions at both pH:
This means that Hydrogen ions in a solution at pH = 7 has 
ions fewer than in a solution at a pH = 6
The most important identity of an element is its atomic number or number of protons. Electrons and neutrons may vary, but protons stay constant. To start with, when an element is neutral, the number of protons and electrons are equal. When the isotope is a positive ion, that means that the element gives away 2 electrons. So, if the electrons is 48, then that means it was originally 48 - 2 = 46 electrons. From the periodic table, element 46 is Palladium (Pd).
