Define the following terms: (i) Cathodic protection, (ii) Electrochemical series, (iii) Cell constant, (iv) Equivalent conductivity, (v) Strong and weak electrolytes. from Chemistry Electrochemistry Class 12 Nagaland Board
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Class 10 Class 12
Define the following terms: (i) Cathodic protection, (ii) Electrochemical series, (iii) Cell constant, (iv) Equivalent conductivity, (v) Strong and weak electrolytes.

(i) Cathodic protection (CP) is a technique used to control the corrosion of a metal surface by making it the cathode of an electrochemical cell. A simple method of protection connects the metal to be protected to a more easily corroded "sacrificial metal" to act as the anode.
for example zinc is used to pervent iron
Zinc is more electro-positive than iron. Therefore, as long as zinc is there on the iron pipe, zinc acts as anode and the iron as cathode. As a result, rusting of iron is prevented.


(ii)Electrochemical series is a series of chemical elements arranged in order of their standard electrode potentials. The hydrogen electrode. H+(aq) + e- →← 1/2H2(g) is taken as having zero electrode potential. An electrode potential is, by definition, a reduction potential


(iii)The quanitty 1/A is called cell constant denoted by the symbol. G*. It depends on the distance between the electrodes and their area of cross -section and has the dimension of length-1 and can be calculated if l and A 
G* =l/A =Rk


(iv) A strong electrolyte is a solute that completely, or almost completely, ionizes or dissociates in a solution. While the specificconductance of a solution increases with concentration, the equivalent conductance decreases as the concentration increases. unit of equivalent conductance Ω-1cm-2equi-1


(v)electrolytes :A substance that when dissolved in water produced a solution that can conduct electric current.
there are two electrolytes 
1. strong 
2.weak

strong Electrolytes conduct current very efficiently.Completely ionized or dissociate when dissolved in water
a. Soluble Ionic compounds
b. Strong acids (HNO3(aq), H2SO4(aq), HCl(aq))

         HNO3--> H+ + NO3-     (100% ionization)

c. Strong bases (KOH and  NaOH)

      KOH -->K+    +OH -     (100% dissociation)


Weak electrolytes conduct only a small current
Slightly ionized in solution 
a. Weak acids (organic acids-->acetic, citric, butyric,malic, etc.)

                  HC2H3O <==> H+  + C2H3O2-

b. Weak bases (ammonia)

                        NH3 + H2O <==> NH4+ + OH-

 

 

132 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

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

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

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