Electrolytic Cells And Electrolysis

IONIC THEORY

Chemical substance which dissolve in water and furnishes ions are called electrolytes. The phenomenon of the production of ions in solution is called dissociation or ionisation.

Electrolytes:

(i) Strong:

Concentration of dissociated ions largely dominate on the concentration of undissociated molecule.

\begin{aligned}  NaCl \,\,\,\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{{a}^{+}}\,\,+\,\,C{{l}^{-}} \\  \text{Negligible\hspace{2cm}Very large} \\ \text{concentration\hspace{1.2cm}concentration} \\ \end{aligned}

(ii) Weak:

Lesser concentration of dissociated ions

\begin{align}  C{{H}_{3}}COOH\,\,\,\,\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,C{{H}_{3}}CO{{O}^{-}}\,\,+\,\,{{H}^{+}} \\  L\arg e\,\,concentration\,\,\,\,\,\,\,small\,\,concentration \\ \end{align}

Degree of dissociation $(\alpha )$: $\frac{Molesofdissociatedmolecules/ions}{Totalmolesofmolecules/ions}$

$Rangeof\alpha \Rightarrow 0<\alpha <1$

Factors affecting of degree of dissociation:

(i) Nature of electrolytes

(ii) Nature of solvent

(iii) Presence of other solute (common ion effect)

(iv) Dilution

(v) Temperature

ELECTROLYSIS

Process in which electrolyte is decomposed into its constituents by passing electricity through its aqueous solution or fused (molten) state.

$Electrolyte=C^{+}+A^{-}$

$NaCl\rightleftharpoons N{a}^{+}+C{l}^{-}$

Reaction occurring at cathode $\to$ Reduction reaction (Gain of $e^{-}$)

$N{a}^{+}+e^{-}(fromcathode)\to Na$

Reaction occurring at anode $\to$ Oxidation reaction (Loss of $e^{-}$)

$C{l}^{-}\to Cl+e^{-}$(Go to anode) (Primary reaction)

$\begin{array}{r}Cl+Cl\to C{l}_2(g)(Secondaryreaction)\\ molecule\end{array}$

PREFERENTIAL DISCHARGING THEORY

If more than one type of ions are available during electrolysis, then that ion is discharged first at respective electrodes which requires least energy i.e. discharging potential.

(i) Electrolysis of sodium chloride solution:

Ions produced during electrolysis:

$NaCl\rightleftharpoons N{a}^{+}+C{l}^{-}$

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

At cathode, $H^{+}$ions are discharged in preference to $N{a}^{+}$ ions as the discharge potential of $H^{+}$ ions is lower than $N{a}^{+}$ions. Similarly at anode, $C{l}^{-}$ions are discharged in preference to $O{H}^{-}$ ions.

At cathode:

$2H^{+}+2e^{-}\to H_2$

At anode:

$2C{l}^{-}\to C{l}_2+2e^{-}$

Thus $N{a}^{+}andO{H}^{-}$ remain in solution.

(ii) Electrolysis of copper sulphate solution using platinum electrodes:

$CuS{O}_4\rightleftharpoons C{u}^{++}+S{O}_4^{--}$

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

At cathode:

$C{u}^{++}+2e^{-}\to Cu$

At anode:

$2O{H}^{-}\to H_{2}O+\frac{1}{2}{O}_2+2e^{-}$

(iii) Electrolysis of sodium sulphate solution using inert electrodes:

$N{a}_{2}S{O}_4\rightleftharpoons 2N{a}^{+}+S{O}_4^{--}$

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

At cathode:

$2H^{+}+2e^{-}\to H_2$

At anode:

$2O{H}^{-}\to H_{2}O+\frac{1}{2}{O}_2+2e^{-}$

(iv) Electrolysis of copper sulphate solution using copper electrodes :

$CuS{O}_4\rightleftharpoons C{u}^{++}+S{O}_4^{--}$

At cathode, copper is deposited.

$C{u}^{++}+2e^{-}\to Cu$

At anode, Cu – electrode oxidised to $C{u}^{++}$ ions which dissolve equivalent amount of copper at the anode.

$Cu\to C{u}^{++}+2e^{-}$

$Cu+S{O}_4^{--}\to CuS{O}_4+2e^{-}$

During electrolysis, copper is transferred from anode to cathode.

Illustration 1. The pH of a solution of NaCl is 7. This solution is electrolysed. What would be the pH of the solution after electrolysis?

Solution: Greater than 7, due to the formation of OH^- ions.

FARADAY'S LAW OF ELECTROLYSIS

(i) Faraday's first law:

The amount of substance deposited on respective electrodes is directly proportional to the quantity of electricity passed.

i.e. $w\propto Q$

or, $w=Z\times i\times t$

where, w = weight of deposited substance

i = amp

t = time in seconds

If i = 1 amp and t = 1 sec

w = z,

Where z is electrochemical equivalent.

Illustration 2.

A vanadium electrode is oxidized electrically. If the mass of the electrode decreases by 114 mg during the passage of 650. coulombs, what is the oxidation state of the vanadium product?

(A) +1

(B) +2

(C) +3

(D) +4

Solution: $m=\frac{E}{nF}Q$ Hence, n = 3

ELECTROCHEMICAL EQUIVALENT (Z)

It is the mass of substance deposited by one coulomb of charge.

(ii) Faraday's second law:

When same amount of electricity passed through different electrolytes, then deposited mass of respective electrodes will be in the ratio of their equivalent masses.

$\frac{Massofhydrogen}{Massofcopper}=\frac{E_H}{E_{Cu}}$

$\frac{MassofCopper}{MassofSilver}=\frac{E_{Cu}}{E_{Ag}}$

$\frac{Massofsilver}{Massofzinc}=\frac{E_{Ag}}{E_{Zn}}$

$\frac{Massofzinc}{Massofhydrogen}=\frac{E_{Zn}}{E_H}$

Illustration 3. Find the charge in coulombs of 1 g ion of N^3-.

Solution. One nitride ion carries three -ve charge

Charge of one ion of N^3- = 3 x1.6 x10^-19 coulomb

One g ion consist = 6.02 x 10²³ ions

$\therefore$Charge on one g – ion of N^-3 $=3\times 1.6\times {10}^{-19}\times 6.02\times {10}^{23}$ = 2.89 $\times$ 10⁵ CIllustration 4: How many moles of iron metal will be produced by passage of 4A of current through 1L of 0.1 M Fe³⁺ solution for 1 hour? Assume only iron Fe⁺³ + 3e^– $\to$ Fe is reduced.Solution: Total charge passed = (4 × 1 × 60 × 60) C = 14400 C = 0.149 FFirst of all ferric ion changes to ferrous and than ferrous changes to Fe.Fe⁺³ + e^– $\to$ Fe⁺², Fe⁺² + 2e^– $\to$ FeSo, charge used up in formation of Fe (II) = 0.1 FTherefore, remaining charge left = (0.149 – 0.1) F= 0.049 F Which is used in conversion of Fe⁺² to Fe So, number of moles of Fe produced = $\frac{1\times 0.049}{2}$ = 0.0245

GRAM EQUIVALENT WEIGHT OF SUBSTANCE
It is the mass in grams of deposited substance deposited by one Faraday (96500 coulomb) of electricity.1 coulomb = 1 ampere – second
1 Faraday = Charge carried by 1 mole of $e^{-}$$=\left(6.023\times {10}^{23}\right)\left(1.6\times {10}^{-19}\right)$ = 96500 coulomb

If w = z (deposited weight by one coulomb)W = E (deposited weight by one Faraday)

$\therefore z=\frac{E}{F}$

Faraday's law for gaseous electrolytic product.We know, w = zQ$=z\times l\times t$

$w=\frac{I\times t\times E}{96500}$ $\left(where;z=\frac{E}{96500}\right)$

For gas, Volume of evolved gas at S.T.P. $=\frac{I\times t\times Equivalentvolume}{96500}$

Illustration 5. The volume of gas measured at NTP, liberated at anode from the electrolysis of Na_­2SO₄ solution by a current of 5.0A passed for 3 min 13 s

(A)56 mL

(B)112 mL

(C)224 mL

(D)None of these

Solution: Quantity of charge passed;;= $\frac{5\times 193}{96500}$ = 0.01F;Mass of oxygen evolved = 0.01 equiv = 0.08 gm = $\frac{0.08}{32}$ = 0.0025 moleVolume of O₂ at NTP = 22400 x 0.0025 = 56 ml

llustration 6. A current of 0.20A is passed for 482s through 50.0 mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis?

(A)0.0159 M

(B)0.0199M

(C)0.10M

(D)0.030M

Solution: The quantity of charge passed = $\frac{0.20\times 482}{96500}F$ = $9.99\times {10}^{-4}F$The reaction at cathode is,

$2H_{2}O+2e\to H_2+2O{H}^{-}$1F = 1 mol of $O{H}^{-}$

$\therefore 9.99\times {10}^{-4}F\equiv 9.99\times {10}^{-4}\text{ mol of O}{\text{H}}^{\text{-}}$

$\therefore$Concentration of $O{H}^{-}$ in solution

After electrolysis = $\frac{9.99\times {10}^{-4}\text{ mol}}{\left(\frac{50}{1000}\right)L}=0.0199M$

Applications of electrolysis:

(1)Determination of equivalent masses of element

(2)Electrometallurgy

(3)Manufacturing of non – metals (By electrolysis process)

(4)Electro – refining of metals

(5)Manufacturing of compounds

(6)Electroplating

Calculation of thickness of coated layer during electrolysis:

If volume of coated layer $\left(a\times b\times c\right)c{m}^3$Mass of the deposited substance $=\left(a\times b\times c\right)\times density$or $\left(a\times b\times c\right)\times density=\frac{I\times t\times E}{96500}$

EFFICIENCY OF CURRENT DURING ELECTROLYSIS

% current efficiency $=\frac{Effectivecurrent}{Ammetercurrent}\times 100$

Illustration 7: A solution containing one mole per litre of each Cu(NO₃)₂; AgNO₃; Hg₂(NO₃)₂ ; Mg(NO₃)₂ is being electrolysed using inert electrodes. The values of standard electrode potentials (reduction potentials in volts are Ag/Ag⁺ =0.80 V, 2Hg/Hg₂⁺⁺ = 0.79V, Cu/Cu⁺⁺ = + 0.24 V, Mg/Mg⁺⁺ = –2.37 V. With increasing voltage, the sequence of deposition of metals on the cathode will be (A) Ag, Hg, Cu(B) Cu, Hg, Ag(C) Ag, Hg, Cu, Mg(D) Mg, Cu, Hg, Ag

Solution: Greater the value of standard reduction potential, greater will be it's tendency to undergo reduction. So the sequence of deposition of metals on cathode will be Ag, Hg, Cu. Here, magnesium will not be deposited because it's standard reduction potential is negative. So it has strong tendency to undergo oxidation. Therefore, on electrolysis of Mg(NO₃)₂ solution, H₂ gas will be evolved at cathode. Hence, (A) is correct.

Illustration 8: One coulomb of charge passes through solution of AgNO₃ and CuSO₄ connected in series and the conc. of two solutions being in the ratio 1:2. The ratio of weight of Ag and Cu deposited on Pt electrode is(A) 107.9 : 63.54(B) 54 : 31.77(C) 107.9 : 31.77(D) 54 : 63.54

Solution: Faraday's IInd Law $\frac{W}{E}$ = constantSo,

$\frac{W_{Ag}}{E_{Ag}}=\frac{W_{Cu}}{E_{Cu}}$

(Ag⁺ + e^– $\to$Ag, E_Ag =$\frac{M}{1}$)

$\therefore$$\frac{W_{Ag}}{W_{Cu}}=\frac{E_{Ag}}{E_{Cu}}=\frac{107.9}{63.54/2}$ (Cu⁺² + 2e^– $\to$ Cu, E_Cu =$\frac{M}{2}$)

= $\frac{107.9}{31.77}$

Hence, (C) is correct.