Electrochemistry

The conductivity of a solution depends on the amount of ions present per volume of the solution. When diluted, the concentration of the ions decreases which implies that the number of ions per volume decreases thus, in turn, conductivity decreases.

Given: 
[Ag⁺] = 0.002 M
[Ni²⁺] = 0.160 M
n = 2
(n = moles of e- from balanced redox reaction)
E⁰cell= 1.05 V
Now, using the Nernst equation, we get,

Given: 
For hydrogen electrode, pH = 10
n = 1
(n = moles of e- from balanced redox reaction)
On using the formula [H+] = 10– pH

⇒ [H+] = 10 − 10 M
We know,

For a substance to oxidise Fe²⁺ to Fe³⁺ ion, it must have high reduction potential than Fe³⁺. The reduction potential of Fe³⁺ to Fe²⁺ reaction is 0.77V, the substances which have reduction potentials higher than this value will oxidise Fe²⁺ ions. Comparing the values, from the table:

NO, because Zn is very reactive with Cu. It reacts with copper sulphate to form zinc sulphate i.e., Zn displaces Cu and metallic Cu is also formed.
The reaction is given as:
Zn + CuSO₄ ⇒ ZnSO₄ + Cu

To determine the standard electrode potential of the system Mg²⁺|Mg, connect it to the standard hydrogen electrode (SHE). Keep the Mg²⁺|Mg system as cathode and SHE as cathode. This is represented as shown below.
Pt (s) | H₂ (g, 1 bar)| H⁺ (aq, 1 M) |Mg²⁺ (aq, 1M)| Mg
The electrode potential of a cell is given by
E^? = E^? R – E^? L
Where,  
E^? R- Potential of the half-cell in the right side of the above representation
E^? L- Potential of the half-cell in the left side of the above representation
It is to be noted that the potential of the standard hydrogen electrode is zero.
Therefore, E^? L = 0
E^?  = E^? R – 0 
⇒ E^? = E^? R