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Class 10 Class 12

Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used are of platinum.
(i) An aqueous solution of AgNO3.
(ii) An aqueous solution of H2SO4.

(i)
Reaction in solution
AgNO3               ↔     Ag +    +        NO3
H2O             ↔     H+       +        OH
Reaction at cathode
Ag+   +          e       →      Ag
Reaction at anode
Due to platinum electrode self of ionization of water will take place
H2O   →     2H+    +        1/2O2(g)   +  2e
Hence Ag will deposit at cathode and O2 gas will generate at anode

(ii)
Reaction in solution
H2SO4                          ↔     2H +  +        SO42–
H2O             ↔     H+       +        OH
Reaction at cathode
H+      +          e       →      ½ H2
Reaction at anode
Due to platinum electrode self of ionization of water will take place
H2O   →     2H+    +        1/2O2(g)   +  2e
Hence H2 gas will generate at cathode and O2 gas will generate at anode

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

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

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

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

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

oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-
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 .

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M) $\to$ Ni2+ (0.160 M) + 2Ag(s)
Given that

$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{As}\left(\mathrm{s}\right)$

or

The equation is also written as

or

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.

Can you store copper sulphate solutions in a Zinc pot?