Ncert Solutions Chemistry Class 12th

Metals with greater reactivity can be extracted electrolytically. Sodium, potassium, calcium, lithium, magnesium, aluminium which are present in the top of the reactivity series are extracted electrolytically. 

Current I = 0.5A
Time t = 2hrs = 2*60*60 = 7200 seconds
Charge Q = I * t
Q = 0.5*7200 = 3600 C
Charge carried by 1 mole of electrons (6.023*10²³electrons) is equal to 96487C.
No of electrons = 6.023*10²³ * 3600/96487
No of electrons = 2.25*10²² electrons

C = 0.025 mol L^-1
A_m = 46.1 Scm² mol L^-1
λ⁰ (H⁺) = 349.6 Scm² mol L^-1
λ⁰ (HCOO^-) = 54.6 Scm² mol L^-1
Λ⁰m (HCOOH) = Λ⁰ (H⁺) + Λ⁰ (HCOO^-) 
= 349.6 + 54.6
= 404.2 S cm² mol L^-1

^{Now, the degree of dissociation:}

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