Chapter Chosen

Electrochemistry

Book Chosen

Chemistry I

Subject Chosen

Chemistry

Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
Explain with examples the terms weak and strong electrolytes. How can these be distinguished?

Answer:
(i) Weak electrolytes : An electrolyte that ionizes partially in solution is called a weak electrolyte. The solution formed contains ions which are in equilibrium with un-ionised molecules, e.g., acetic acid dissolves in water to form H
3O+ and CH3COO+ ion. The solution contains H3O(hydronium ion), CH3COO (acetate ion) and unionised CH3COOH molecules.



The degree of ionisation of a weak electrolyte is much less than 1. These have low values of molar conductivities at high concentration. Degree of ionisation and molar conductivity both increases with dilution.

(ii) Strong electrolyte : An electrolyte which is almost completely ionised in solution is called a strong electrolyte. The degree of ionisation of a strong electrolyte is 1 or 100% (or nearly so). The solution formed contains ions which are in equilibrium with solid form of strong electrolyte.



Strong electrolyte

Weak electrolyte

1. These have higher molar conductivities at all concentrations.

2. λ°m values increase very slightly with dilution.

3. Degree of ionisation is very high at all concentration i.e., almost fully ionized.

4. Most of the salts like NaCl, KCl, NaNO3, BaCl2 and mineral acids like HCl, H2SO4, HNO3 and NaOH, KOH etc are common examples of strong electrolytes

1. These have much lower conductivities at high concentration.

2. λ°m values increase sharply with dilution.

3. Degree of ionisation is very low at high concentration and increases with dilution.

4. Salts like ammonium acetate, acetic acid, aq NH4OH, aqueous CO2 and organic acids and bases are common examples of weak electrolytes.

600 Views

Can you store copper sulphate solutions in a Zinc pot?

Answer:

No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.

Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
1384 Views

How would you determine the standard electrode potential of the system Mg2+/Mg?

Answer:

Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right  – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views

Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.

Answer:


oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-EFe3+Fe2+0 = 0.77V
Any substance which standard electrode potential is more than that of Fe+3 /F+2 can oxidise ferrous ions. 
(refer to the table given in book)

The EMF of the substance whose reduction potentials greater than 0.77V will oxidised ferrous ion.
for example Br2, Cl2,and F2 .

879 Views

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M)  Ni2+ (0.160 M) + 2Ag(s)
Given that EoCell = 1.05 V

Answer:

Ni(s)+2Ag+(aq)Ni2+(aq)+2As(s)
Ecell = Eocell + RT2FlnAg+(aq)2Ni2+ (aq)
or      Ecell = Eocell+0.0592 logAg+(aq)2Ni2+ (aq)
 
The equation is also written as

or    Ecell = Eocell-0.0592logNi2+(aq)Ag+ (aq)2         = 1.05 V - 0.0295 log 0.1600.002  

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.

1585 Views

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer:

For hydrogen electrode, ,
given that      pH = 10
use formula [H+] = 10–pH
so that  [H+] = 10−10 M
Electrode reaction will
H+  + e –  →1/2 H2
Use the formula


EH+/H2  = EoH+/H2 - RTnFlnH2[H+]2              = 0-8.314×2982×96500ln IH+2              = 0.05915 log [H+]              =-0.05915 pH             =-0.05915 × 10 =- 0.59 V
1279 Views