Faradays first electrolysis Law states that the quantity of reaction taking place in terms of mass of ions formed or discharged from an electrolyte is proportional to the amount of electric current passed. Since electric current (ampere) is the number of coulombs (Q) flowing in one second,

Mass of the ions formed or reacted (m)∝ electric current α Q, or

m ∝ Q or

**m = ZQ**

Where, Z is a proportionality constant, called the chemical equivalent of the element.

For a flow of, 1 Coulomb of charges for one second, m = Z.

The proportionality constant is equal to the mass of the substance involved in the reaction. Z is the electrochemical equivalent mass of one coulomb charge.

**One coulomb of charge corresponds to a mass of one equivalent.**

**i) Electric current and Charge (Q)**

Electric current is measured in ampere and it is the charges flowing per unit time (seconds).

**I = **\(\frac{Q}{t}\)

Quantity of Charges flowing (Q) = It = ampere × seconds

**m = ZIt**

**ii) Number of electrons – Charge Q of electrons – Faraday -Equivalent mass of substances**

But the charges are associated with electrons. Every electron carries a charge of 1.6 × 10^{-19} coulombs.

Charges carried by one equivalent/ mole number of electrons(Q)= 6.02 × 10^{23} × (1.6 × 10^{-19}) coulombs

= 96485 coulombs ≈ 96500C = 1 Faraday = 1F

One equivalent (or mole or Avogadro’s) of electrons ∝ 96485 coulombs ≈ 96500C = 1 Faraday = 1F

**iii) Mass of substance undergoing electrolysis**

Faraday Law says, **m = Z ×****Q or m = Z I t.**

When one coulomb corresponds to one electrochemical equivalent mass (Z) of the substance,

one equivalent(or mole) of electrons flowing per second, will correspond to 96485 Equivalents mass.

This, **96485 electrochemical equivalents = Z × 96485=Equivalent weight of substance in gram**.

So, **Electrochemical equivalent of a substance = Z =** \(\frac{Equivalent\, weight\, of\, substance\, in\, gram}{96485}=\frac{E}{96485}\)

Equivalent weight of a substance is, related to its molecular weight.

Equivalent mass of a substance = \(\frac{Molecular\, weight\, of\, the\, substance}{Valency\, or\, charge}\)

For every, one mole of electrons or charges (1Faraday) or 96485ampere sec, passing through an

electrolyte, one equivalent mass of the electrolyte is reacted, discharged/deposited etc.

**One Avogadro’s number of electrons = Charge of 1 Faraday = 96485coulombs = 96485ampere sec**

** = 1equivalent mass (reacted/deposited/neutralized)**

Mass of the substance (gm) = Equivalent weight(gm) × \(\frac{Coulombs}{96485}\) = Equivalent weight(gm) × \(\frac{ampere,sec }{96485}\)

Or, **m = **\(\frac{EQ}{96485}=\frac{EIt}{96485}\)

**Example**

For example, on passing electric current through a copper sulphate electrolyte solution, copper ions gets discharged and deposit on the anode.

Cu^{2+} + 2e**–** →Cu

Each copper ion needs two electrons for its reduction reaction to copper atom. More the electrons passing through the copper sulphate, solution more will be the copper ions being, deposited on the anode. Hence there is a direct relationship between the mass of material being, reduced and the numbers of electrons flowing into the electrolyte.

Here, one copper ion needs 2 electrons;so, One mole of copper ions needs 2 moles of electrons.

m = 2 × equivalent weight of copper

**Understanding Faraday’s first law of electrolysis**

**1. When 0.1M MnO _{4}^{2-} is oxidized to MnO_{4}^{–}, the quantity of electricity required isa) 96500C b) 2 × 96500C c) 9650C d) 96.50C**

MnO_{4}^{2-} → MnO_{4}^{–}+ e^{–}

1mole of MnO_{4}^{2-} lose 1 mole of electrons or 96500C

⸪ 0.1 mole of MnO_{4}^{2-} lose 0.1mole of electrons or 9650C

Answer is (c)

**2. How much electricity in terms of Faraday is required to produce**

(i) 20.0 g of Ca from molten CaCl_{2}?

(ii) 40.0 g of Al from molten Al_{2}O_{3}?

i) Ca^{2+} + 2 e^{–} → Ca

One mole of Calcium ions gains two moles of electrons or 2 Faraday charge to produce one mole of

calcium.

Molecular weight of Calcium is 40.

So, 20 gm of calcium shall need one Faraday of electricity.

ii) Al^{3++} + 3 e^{–} → Al

One mole of Aluminum ions gains two moles of electrons or 3 Faraday charge to produce one mole of aluminim.

Molecular weight of Aluminum is 27.

So, 40 gm of Aluminum ion shall need \(\frac{40\times 3}{27}\) Faraday of electricity= 4.44F

**3. A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20**

**minutes. Calculate the weight of nickel reduced on the cathode.**

Mass of nickel deposited = \(\frac{E~I~t}{96500}\)

Equivalent eight of nickel = Atomic weight/2 =58.7/2

Mass = \(\frac{58.7\times 5\times 20\times 60}{2\times 96500}\) = 1.83gm

**4. The anodic half- cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of**

**PbSO4 electrolyzed in g during the process is ; (molar mass of PbSO4 = 303g/mole) **

a) 22.8 b) 15.2 c) 7.6 d) 11.4

In lead sulphate, lead is in 2+ state.

One mole (303) of lead sulphate, shall need 2 Faraday electricity.

So, 0.05 Faraday will electrolyze \(\frac{303\times 0.05}{2}\)g of lead sulphate = 7.6g

Answer (c)

**5. A cell, Ag | Ag+ || Cu2+ | Cu, initially contains 1 M Ag+ and 1 M Cu2+ ions. What will be the increase in**

**potential of the cell after passing an electric current of 9.65amperes for one hour?**

On passing electric current into the cell, the following reaction takes place.

**–** Cu → Cu^{2+} + 2^{e–}** ; **Ag^{+}+ e^{–} → Ag

i) Change in the concentration of copper and silver ions

Mass deposited= \(\frac{E\, I\,t}{96500}\)

Mass of copper ions additionally formed = \(\frac{63.54\times 9.65\times 60\times 60}{2\times 96500}\) = 11.43g = 0.18M

Concentration of copper half-cell = 1-0.18 = 0.82M

Mass of silver ions lost = \(\frac{108\times 9.65\times 60\times 60}{96500}\) = 38.9g = 0.36

Concentration of silver half-cell = 1-0.36 = 0.64M

ii) Cell potential, Ei = E° – \(\frac{2.303\times RT}{nF}{{log }_{10}}\frac{Ag+}{Cu2+}\)

Cell potential before passage of current = Ei = E° – \(\frac{2.303\times RT}{nF}{{log }_{10}}\frac{1}{1}=E{}^circ\)

Cell potential after passage of current = Ei = E° – \(\frac{2.303\times RT}{nF}{{log }_{10}}\frac{{{0.64}^{2}}}{0.82}\)

Change in the cell potential = \(\frac{0.0591}{2}{{log }_{10}}\frac{{{0.64}^{2}}}{0.82}=0.009V\)