Quantitative Aspects Of Electrolysis

Faraday's Laws of Electrolysis

While investigating the phenomena of electrolysis, M. Faraday discovered certain relationships between the quantity of electricity passing through an electrolyte and the amount of any material liberated at the electrode. The quantity of electricity is equal to the product of the current strength and the time for which it is passed. The results obtained by Faraday may be considered in the form of the two laws of electrolysis, as follows.

I. The amount of any substance deposited or dissolved at a particular electrode is proportional to the quantity of electricity passed i.e. w ∝ Q

or w ∝ I × t

orw = ZIt

where w is the mass of substance deposited or liberated in gram, Q is the amount of charge passed in coulombs, I is current strength in amperes, t is the time for which current flows through the cell in seconds and Z is the electrochemical equivalent.

Electrochemical equivalent is the mass of substance deposited in grams either by passing one coulomb of charge or by passing one ampere of current for one second.

One faraday is the quantity of charge carried by one mole of electrons.

∴ 1F = 1.6 ×1019× 6.023 ×1023≈96500 Coulombs

or one Faraday is the quantity of charge which must be passed in order to deposit or dissolve one gram equivalent of any substance.

Let E be the equivalent mass of a substance deposited at any electrode. Since 96500 coulombs of charge is required to deposit E g of substance, therefore one coulomb of charge would deposit g, which is nothing but electrochemical equivalent.

An important application of above formula is in electroplating of metals.

As mass of substance deposited, w = V × d = A ×l× d

where A is the area of the article to be electroplated, l is the thickness of layer deposited and d is the density of metal deposited.

An important use of this equation lies in its application for the measurement of quantity of electricity or of current strength or the time taken for discharge of an ion or cathode area, or thickness of deposited layer or density of metal deposited.

II. The amounts of different substances deposited or dissolved by the same quantity of electricity are proportional to their chemical equivalent weights.

From the first law,

It can be rearranged as = Number of equivalents

 Constant [as charge passed is same]

This means that if the quantity of charge passed through cells connected in series is same, the amount of substance deposited at various electrodes in terms of equivalents would also be same.

Another way of stating this law is that the same quantity of electricity will produce chemically equivalent quantities of all substances resulting from the process. Moreover, since 96500 coulombs will yield one equivalent of silver, a direct consequence of Faraday's second law is that during electrolysis 96500 coulombs of electricity will yield one equivalent of any substance.

For example, consider the given reactions at two different cathodes in two different electrolytic cells connected in series.

Let us assume x moles of electrons is passed through the two cells. The mass of Ag and Cu deposited are 108x and 31.75x respectively. We can see that 108 and 31.75 g/equivalent are the chemical equivalent weights of the Ag and Cu respectively.

∴ w = Number of moles of electrons × Equivalent mass

 Number of moles of electrons = Constant

Since the moles of electrons passed are constant, thus the equivalents of Ag and Cu deposited at the cathode of their respective cell will also be same.

[Note: It must be clear that anode is the electrode at which oxidation takes place and cathode is the electrode at which reduction takes place. Do not relate the sign (positive or negative) of the electrode with the nature of electrode.]

There is little doubt that Faraday's laws are exact. They have been found to hold at high and low temperatures, as well as under normal conditions, for non−aqueous solutions and fused salts, as well as for aqueous solutions, provided the correct equivalent weights are employed, there are no exceptions to these laws. The laws also applies to chemical changes at the anode and cathode which do not involve deposition on the electrode. For such reactions, law may still be employed if the appropriate equivalent weight is used. For example, in the reduction of ferric to ferrous ions, or the reverse oxidation process, the equivalent weight is equal to the atomic weight of iron, i.e. 56.

Apparent exceptions to Faraday's laws sometimes arise when two or more processes occur simultaneously at an electrode. For example, the deposition of a metal, such as zinc or nickel, may be accompanied by the evolution of hydrogen. The quantity of metal deposited at the cathode is then not in accordance with the results from this law. However, if allowance is made for the fact that part of the current is utilized in an alternative process, the laws are found to be obeyed.

Related Topic

Qualitative Aspects of Electrolysis

Electrolysis

Example 1

Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 16 minutes. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.

Solution

Total number of faradays passed = = 1.9896 × 10⁻⁵

Moles of Cu²⁺ ions deposited =

Since absorbance was reduced to 50% of its original value, the initial moles of Cu²⁺ would be two times the moles of Cu²⁺ reduced.

∴ Initial moles of Cu²⁺ =  x 2 = 1.9896 × 10⁻⁵

∴ The concentration of CuSO₄ in the solution

 = 1.9896 × 10−5× 4 = 7.958 × 10⁻⁵ mol/litre

Example 2

The passage of a constant current through a dilute solution of sulfuric acid, with platinum electrodes, for 1 hr resulted in the liberation of 336 ml of mixed hydrogen and oxygen, reduced to S.T.P. Calculate the strength of the current.

Solution

Let the current of I amperes is passed through the solution for 1 hour.

∴ Charge passed = I × 60 × 60 = 3600 I

Moles of electrons passed =

At anode: 2OH− ½O₂ + H₂O + 2e−

At cathode: 2H⁺ + 2e− H₂

Moles of O₂ released at anode = ×1/4

Volume of O₂ released at STP = × 1/4× 22400 = 208.9 I

Moles of H₂ released at cathode = ×1/2

Volume of H₂ released at STP = × 1/2× 22400 = 417.8 I

Total volume of gas released at STP = 208.9 I + 417.8 I = 336 ml

∴ I = 0.536 Amp.

Example 3

In an electrolysis experiment electric current was passed for 5 hrs through two cells connected in series. The first cell contains gold salt and second contains CuSO4 solution. 9.83 g of gold was deposited in the first cell. If the oxidation number of Au is +3, find the amount of copper deposited in the second cell. Also calculate the magnitude of the current in amperes.

Solution

At. weight of gold is 197

Eq. wt. of gold = =65.66

As we know that

∴wCu = 4.7533 g

By Faraday’s first law of electrolysis

w = Z × i × t =

4.7533 = × i × 5 × 60 × 60

∴ i = amp

= 0.8026 amp

Example 4

An electric current is passed through two electrolytic cells connected in series, one containing a solution of silver nitrate and the other solution of sulphuric acid. What volume of oxygen measured at 25°C and 750 mm of Hg would be liberated from H₂SO₄ if (i) 1 mole and (ii) 8 × 1022 ions of Ag+ are deposited from the silver nitrate solution?

Solution

(i) 1 mol of Ag+ = 1 equivalent of Ag+

∴ Equivalent of O2, i.e. 8 g of O2 will be liberated = = 5.6 litres at NTP

∴ V1 (Volume of oxygen at 25°C and 750 mm)

= = 6.2 litres

(ii) 8 × 10²² ions of Ag+ = mol

Equivalent of O₂ = = 0.1329g

For one equivalent volume of O₂ evolved = 6.2 litre

∴ 0.1329 g of O2 = 6.2 × 1329 litre = 0.823 litre