﻿ State and explain Faraday's law of electrolysis? from Chemistry Electrochemistry Class 12 Nagaland Board

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State and explain Faraday's law of electrolysis?

Faraday's first law. The amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

$\mathrm{W}\propto \mathrm{Q}$

or                               $\mathrm{W}\propto$It

or                                 W = ZIt

where W is the mass of substance produced at an electrode.
I is current in amperes
t is time in seconds for which current is passed.
Z is electro-chemical equivalent of substance.
Faraday's second law: It states that the masses of different substances liberated or dissolved by the same amount of electricity passed is directly proportional to their chemical equivalents. Or in other words “the same quantity of electricity will produce or dissolve chemically equivalent quantities of all substances.”
Mathematically,

Thus, we can say that the same quantity of electricity is required to produce one equivalent of any substance. It is called Faraday, F. It is equal to 96500 coulombs, and is equal to the charge on one mole of electrons.
321 Views

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

1279 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

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

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

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