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Class 10 Class 12
Explain why electrolysis of aqueous solution of NaCl gives H2 at cathode and Cl2 at anode. Write overall reaction.

Sodium chloride and water ionize as follows:

At cathode: Both Na+ and H+ ions are present near the cathode. But the discharge potential of H+ is lower than that of Na+ ion. So H+ ions are discharged in preference to Na+ ions.

Thus H2 gas is liberated at the cathode and Na+ ions remain in the solution.
At the anode: Both Cl and OH ions are present near the anode. As the discharge potential of Cl ions is lower than that of OH- ions, so Cl ions are discharged in presence to OH ions.

Thus Cl
2 is liberated at anode and OH ions remain in the solution.
The overall reaction is:
NaCl(aq) + H2 O(l) → Na+ (aq) + OH– (aq)
$+\frac{1}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Cl}}_{2}\left(\mathrm{g}\right)$

3952 Views

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 .

879 Views

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
2203 Views

Can you store copper sulphate solutions in a Zinc pot?

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

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.

1585 Views

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