1. Derive / Derive an expression for / Explain Ostwald's dilution law
Ostwald’s dilution law relates the dissociation constant of the weak electrolyte with the degree of dissociation and the concentration of the weak electrolyte.
Consider the dissociation equilibrium of CH3COOH which is a weak electrolyte in water.
CH3COOH ⇌ CH3COO– + H+
Ka = [H+][CH3COO– ] / [CH3COOH]
α is the degree of dissociation which represents the fraction of total concentration of CH3COOH that exists in the completely ionised state. Hence (1 – α) is the fraction of the total concentration of CH3COOH, that exists in the unionised state. If ‘C’ is the total concentration of CH3COOH initially, then at equilibrium C α, C α and C (1 – α) represent the concentration of H+, CH3COO– and CH3COOH respectively.
CH3COOH
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CH3COO–
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H+
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Initial number of moles
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1
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0
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0
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Number of moles ionised
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α
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α
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α
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Number of moles remaining at equilibrium
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1 – α
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α
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α
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Equilibrium concentration
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(1– α).C
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α.C
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α.C
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= Cα2 / (1 - α)
If α is too small, then Ka = α2C and
α = √ Ka / C
Also [H+] = [CH3COO– ] = C.α
[H+] = C (Ka / C)1/2 = (Ka.C)1/2 = √Ka .C
Ka = α2 C / 1 – α is known as the Ostwald’s dilution law.
For weak bases, Kb = α2 C / 1 – α
and α = √ Kb / C at α = small values. Kb = dissociation constant for weak base.
Disadvantage:
Ostwald’s dilution law fails for strong electrolytes. For strong electrolytes, α tends to 1.0 and therefore Ka increases tremendously.
2. Derive Henderson Equation.
The pH of an acid buffer can be calculated from the dissociation constant, Ka of the weak acid and the concentrations of the acid and the salt used.
The dissociation expression of the weak acid, HA, may be represented as
HA ⇌ H+ + A–
and Ka = [H+][A–] / [HA]
or [H+] = Ka .[HA] / [A–] ... (1)
The weak acid is only slightly dissociated and its dissociation is further depressed by the addition of the salt (Na+ A–) which provides A– ions (Common ion effect). As a result the equilibrium concentration of the unionised acid is nearly equal to the initial concentration of the acid. The
equilibrium concentration [A–] is presumed to be equal to the initial concentration of the salt added since it is completely dissociated. Thus we can write the equation (1) as
[H+] = Ka x [acid] / [salt] ... (2)
Where [acid] is the initial concentration of the added acid and [salt] that of the salt used.
Taking negative logs of both sides of the equation (2), we have
– log [H+] = – log Ka – log [acid] / [salt] ... (3)
But – log [H+] = pH and – log Ka = pKa
Thus from (3) we have
pH = pKa – log [acid] / [salt] = pKa + log [salt] / [acid]
Hence, pH = pKa + log [salt] / [acid]
This relationship is called the Henderson-Hasselbalch equation or simply Henderson equation.
The Henderson-Hasselbalch equation for a basic buffer can be stated as:
pOH = pKb + log [salt] / [base]
3. Differentiate between electronic conduction and electrolytic conduction.
S.No
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Electronic conduction
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Electrolytic conduction
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1
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Pure metals and their solid solutions such as alloys are called as metallic conductors.
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Electrovalent (or) ionic compounds conduct electricity through their ions present in fused state or in dissolved state.
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2
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Free and mobile electrons of the metallic atoms or alloys are responsible for electrical conductance.
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Ions with positive and negative charges conduct electricity and move towards cathode and anode respectively.
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3
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Positive holes in the metals move in the opposite direction to electrons.
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Electrolysis occurs when electrical current is passed through electrolytic solutions.
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4
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There is no chemical change in the material when electricity is passed.
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Chemical change occurs.
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5
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There is only flow of electrical energy but there is no transfer of matter.
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There is actual transfer of matter since ions move towards respective electrodes.
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6
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Conductivity of metal decreases with increase in temperature due to the enhanced thermal vibration of metal atoms disrupting the movement of electrons passing through them.
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The conductivity of electrolytes increases with increase in temperature. This is due to increase with ionic mobility.
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1. When dissolved in water, neutral electrolyte molecules are split up into two types of charged particles. These particles were called ions and the process was termed ionisation.
The positively charged particles were called cations and those having negative charge were called anions.
A+ B– → A+ + B–
2. The ions present in solution constantly reunite to form neutral molecules. Thus there is a state of equilibrium between the undissociated molecules and the ions.
AB ⇌ A+ + B–
Applying the Law of Mass Action to the ionic equilibrium we have,
[A+][B–] / [AB] = K
Where, K is called the Dissociation constant.
3. The charged ions are free to move through the solution to the oppositely charged electrode. This is called as migration of ions. This movement of the ions constitutes the electric current through electrolytes. This explains the conductivity of electrolytes as well as the phenomenon of electrolysis.
4. The electrical conductivity of an electrolyte solution depends on the number of ions present in solution. Different ions move with different rates.
5. The properties of solution of electrolytes are the properties of ions. The solution of electrolyte as a whole is electrically neutral unless an electric field is applied to the electrodes dipped into it.
Examples: Presence of hydrogen ions (H+) renders the solution acidic while presence of hydroxide ions (OH–) renders the solution basic.
6. There are two types of electrolytes.
a) Strong electrolytes are those when dissolved in water are completely ionised into ions of positive and negative charges.
Examples: Al2(SO4)3 → 2Al3+ + 3SO42–
NaCl, KCl, AgNO3 etc.,
b) Weak electrolytes are those when dissolved in water are partially dissociated into ions and an equilibrium exists between the dissociated ions and the undissociated electrolyte.
Example: CH3COOH ⇌ CH3COO– + H+.
5. Explain Ostwald's theory of indicators.
1. Ostwald’s theory of indicators is based on Arrhenius theory.
2. The acid-base indicator is either a weak organic acid or a weak organic base.
3. They are partially ionised in solution.
4. The ionised and unionised forms have different colours.
5. The indicator exists predominantly in one of the two forms depending on the nature of the medium and hence there is colour change when the nature of the medium changes.
6. Phenolphthalein is a weak organic acid and it is partially ionised in solutions.
HPh ⇌ H+ + Ph–
Unionised form Ionised form
(Colourless) (Pink)
In acidic medium, excess H+ ions are present which suppress the dissociation of Hph due to common ion effect. Hence the indicator exists predominantly in unionised form and it is colourless.
In alkaline medium, the OH– ion neutralises H+ ion to form water. Consequently the dissociation of Hph is favoured and the indicator is predominantly in the ionised form and it is pink in colour.
7. Methyl orange is a weak base and its is partially ionised in solutions as
MeOH ⇌ Me+ + OH–
Unionised form Ionised form
(Yellow) (Pink)
In basic medium, excess OH– ions are present which suppress the dissociation of MeOH due to common ion effect. Hence the indicator is mostly in unionised form which is yellow.
In acidic medium, the H+ ions combine with OH– ions to form unionized water. Hence the indicator is mostly in ionised form and has pink colour.
6. Explain Quinonoid theory of indicators.
1. The colour change of an acid-base indicator arises as a result of structural change.
2. It is supposed that an indicator exists as an equilibrium mixture of two tautomeric forms namely, benzenoid and quinonoid forms.
3. One form exists in acidic solution and the other form in basic solution.
4. At least one of the tautomers is a weak acid or a weak base.
5. The two forms possess two different colours and as the pH of the solution containing the
indicator is changed, the solution shows a change of colour.
6. The colour change is due to the fact that one tautomer changes over to the other.
7. For example, phenolphthalein is tautomeric mixture of the two forms.
7. Explain the buffer action of acidic buffer with an example.
Acidic buffer consists of a weak acid together with a salt of the same acid.
Eg., Acetic acid and Sodium acetate (CH3COOH / CH3COONa).
The pH of the buffer is governed by the equilibrium
CH3COOH ⇌ CH3COO– + H+ ... (1)
The buffer solution has a large excess of CH3COO– ions produced by complete ionisation of sodium acetate,
CH3COONa → CH3COO– + Na+ ... (2)
1. Addition of HCl: When HCl is added to the buffer solution, the increase of H+ ions is counteracted by association with the excess of CH3COO– ions to form unionised CH3COOH. Thus the added H+ ions are neutralised and the pH of the buffer solution remains unchanged. However owing to the increased concentration of CH3COOH, the equilibrium (1) shifts slightly to the right to increase H+ ions. This explains the marginal increase of pH of the buffer solution on addition of HCl.
2. Addition of NaOH: When NaOH is added to the buffer solution, the additional OH– ions combine with H+ to give neutral H2O. Thus pH of the buffer solution is maintained almost constant.
8. What are the evidences in favour of Arrhenius theory of electrolytic dissociation?
1. The enthalpy of neutralisation of strong acid by strong base is a constant value and is equal to – 57.32 kJ. gm. equiv-1 because
a) Strong acids and strong bases are completely ionised in water and produce H+ and OH– ions respectively along with the counter ions.
b) The net reaction in the acid-base neutralisation is the formation of water from H+ and OH– ions.
H+ + OH– → H2O; ΔHro = – 57.32 kJ.mol-1.
2. The colour of certain salts or their solution is due to the ions present.
For example, copper sulphate is blue due to Cu2+ ions. Nickel salts are green due to Ni2+ ions. Metallic chromates are yellow due to CrO42– ions.
3. Ostwald’s dilution law, common ion effect and solubility product and other such concepts are based on Arrhenius theory.
4. Chemical reactions between electrolytes are almost ionic reactions. This is because these are essentially the reaction between oppositely charged ions.
For example, Ag+ + Cl– → AgCl ↓
5. Electrolytic solutions conduct current due to the presence of ions which migrate in the presence of electric field.
6. Colligative properties depend on the number of particles present in the solution. Electrolytic solution has abnormal colligative properties.
For example, 0.1 molal solution of NaCl has elevation of boiling point about twice that of 0.1 molal solution of non-electrolyte.
9. Explain the Buffer action of basic
buffer with an example
Basic buffer consists of a
weak base and its salt with a strong acid.
Eg., Ammonium hydroxide and Ammonium chloride
(NH4OH + NH4Cl)
The pH of the buffer is governed by the equilibrium
NH4OH ⇌ NH4+ + OH- ... (1)
The buffer solution has a large excess of NH4+ ions produced by complete
ionisation of Ammonium chloride,
NH4Cl → NH4+ + Cl- ... (2)
1. Addition of HCl: When HCl is
added to the buffer solution, the increase of H+ ions combine with NH4OH to form NH4+
and H2O. pH is retained.
2. Addition of NaOH:
When NaOH is
added to the buffer solution, the OH– ions combine
with NH4+ ions present in the buffer solution to give NH4OH and hence pH is maintained10. Explain the function of any one acid-base indicator on the basis of Ostwald’s theory.
PROBLEMS
1. What is the pH of a solution containing 0.5 M propionic acid and 0.5 M sodium propionate? The Ka of propionic acid is 1.34 x 10-5.
Ka of propionic acid = 1.34 x 10–5
∴ pKa = – log Ka = – log (1.34 x 10–5)
= 5 – log 1.34
= 5 – 0.1271
pKa = 4.8729
pH = pKa + log [salt] / [acid] ... Henderson - Hasselbalch equation
= 4.8729 + log 0.5 / 0.5
pH = 4.8729
Or
The dissociation equilibrium of propionic acid will be
C2H5COOH ⇌ C2H5COO– + H+
Ka = [C2H5COO–][ H+ ] / [C2H5COOH ]
= 0.5 x [ H+] / 0.5
Ka = [H+]
∴ pH = – log [H+] = – log Ka
= – log (1.34 x 10–5)
= 5 – log 1.34
= 5 – 0.1271
∴ pH = 4.8729
2. Find the pH of a buffer solution containing 0.30 mole per litre of CH3COONa and 0.15 mole per litre of CH3COOH. Ka for acetic acid is 1.8 x 10-5.
Ka = 1.8 x 10–5
pKa = – log Ka
pKa = – log (1.8 x 10–5)
= 5 – log 1.8
= 5 – log 1.8
= 5- 0.2553
pKa = 4.7447
pH = pKa + log[salt] / [acid] ... Henderson - Hasselbalch equation
= 4.7447 + log 0.30 / 0.15
pH = 4.7447 + log 2
= 4.7447 + 0.3010
pH = 5.0457
pH = 5.0457
3. Calculate the pH of the buffer solution containing 0.04 M NH4Cl and 0.02 M NH4OH. For NH4OH Kb is 1.8 x 10-5.
Kb = 1.8 x 10–5
pKb = – log Kb
pKb = – log (1.8 x 10–5)
= 5 – log 1.8
= 5- 0.2553
pKa = 4.7447
pOH = pKb + log[salt] / [base] ... Henderson - Hasselbalch equation
= 4.7447 + log 0.04 / 0.02
pOH = 4.7447 + log2
= 4.7447 + 0.3010 = 5.0457
pH + pOH = 14.00
pH = 14.00 – pOH
pH = 14.00 – 5.0457
pH = 8.9543
4. Equivalent conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 M acetic acid the equivalent conductance is 5.2 mho.cm2.gm.equiv–1. Calculate degree of dissociation, H+ ion concentration and dissociation constant of the acid.
4. Equivalent conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 M acetic acid the equivalent conductance is 5.2 mho.cm2.gm.equiv–1. Calculate degree of dissociation, H+ ion concentration and dissociation constant of the acid.
Equivalent conductivity of acetic acid at infinite dilution, λ∞ = 390.7
Equivalent conductance of acetic acid, λc = 5.2 mho.cm2.gm.equiv–1
Concentration of acetic acid, C = 0.1 M
Degree of dissociation, a = λc / λ∞
= 5.2 / 390.7
= 0.01333
= 1.33 x 10-2 = 1.33%
CH3COOH ⇌ H+ + CH3COO–
C (1 – a) Ca Ca
∴ H+ ion concentration, [H+] = Ca
= 0.1 x 0.0133
= 0.00133
= 1.33 x 10-2 M
Dissociation constant of acetic acid, K = a2C / 1 - a
= (0.0133)2 x 0.1 / (1 - 0.0133)
= 0.000017689 / 0.9867
= 1.79 x 10-5 M
5. Calculate / Find the pH of a buffer solution containing 0.2 / 0.20 mole per litre CH3COONa and 0.15 mole per litre CH3COOH. Ka for acetic acid is 1.8x10-5.
Ka = 1.8 x 10–5
pKa = – log Ka
pKa = – log (1.8 x 10–5)
= 5 – log 1.8
= 5- 0.2553
pKa = 4.7447
pH = pKa + log [salt] / [acid] ... Henderson - Hasselbalch equation
= 4.7447 + log 0.20 / 0.15
pH = 4.7447 + log 4 / 3
pH = 4.7447 + log 4 - log 3
pH = 4.7447 + 0.6021 – 0.4771
pH = 4.8697
6. An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being deposited?
6. An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being deposited?
Wt. of Copper / Wt. of Iodine = Eqvt. wt. Copper (31.7) / Eqvt. wt. of Iodine (127)
1.25 / x = 31.7 / 127
Wt. of Iodine, x = 1.25 x 127 / 31.7
Hence, Wt. of Iodine, x = 5.0 g
Also, Wt. of Copper / Wt. of Silver = Eqvt. wt. of Copper (31.7) / Eqvt. wt. of Silver (108)
1.25 / y = 31.7 / 108
Wt. of Silver, y = 108 x 1.25 / 317
Wt. of Silver, y = 4.26 g
7. The equivalent conductance of HCl, CH3COONa and NaCl at infinite dilution are 426.16, 91.0 and 126.45 ohm-l cm2 (gram equivalent)-1 respectively. Calculate the equivalent conductance / λ∞ of acetic acid.
Or
31.7 g of Copper (1g eqvt) is liberated by = 96,495 coulomb
1.25 g of Copper is liberated by = 96,495 x 1.25 / 31.7 coulomb
Quantity of electricity = 3805 coulomb
96,495 coulomb deposits 127 g of Iodine
∴ 3805 coulomb deposits = 127 x 3805 / 96,495
= 5.0 g of Iodine
96,495 coulomb deposits 108 g of Silver
∴ 3805 coulomb deposits = 108 x 3805 / 96,495
= 4.26 g of Silver
7. The equivalent conductance of HCl, CH3COONa and NaCl at infinite dilution are 426.16, 91.0 and 126.45 ohm-l cm2 (gram equivalent)-1 respectively. Calculate the equivalent conductance / λ∞ of acetic acid.
λ∞ CH3COOH = λ∞ CH3COONa + λ∞ HCl – λ∞ NaCl
λ∞ CH3COOH = 91.0 + 426.16 – 126.45 = 517.16 - 126.45
λ∞ CH3COOH = 390.71
8. 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrochemical equivalent of copper?
8. 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrochemical equivalent of copper?
Current, I = 0.2 ampere
Time, t = 50 minutes = 50 ´ 60 seconds
Quantity of electricity used, Q = I x t = 0.2 x 50 x 60 = 600 coulombs
Amount of copper deposited by 600 coulombs, m = 0.1978 g
Amount of copper deposited by 1 coulomb, Z = 0.1978 / 600g = 0.0003296 g
Electrochemical equivalent of copper, Z = m / I . t
=0.0003296 gC–1
= 3.296 x 10–4 gC–1
= 3.296 x 10–4 gC–1
= 3.206 x 10–7 kgC–1
Or
Electrochemical equivalent of copper, Z = m / I . t
(or) Z = m / Q
= 0.1978 g / 0.2 amp x 50 x 60 sec
= 0.1978 g / 600 C [ ∵ amp . sec = C
=0.0003296 = 3.296 x 10–4 gC–1
= 3.296 x 10–7 kgC–1
9. Calculate the pH of 0.1 M acetic acid / CH3COOH solution. Dissociation constant of acetic acid is 1.8 x 10-5 M.
For weak acids, [H+] = √Ka x C
= √1.8 x 10-5 x 0.1 = √1.8 x 10-6
= 1.34 x 10–3 M
∴ pH = – log [H+]
= – log (1.34 x 10–3)
= 3 – log1.34
= 3 – 0.1271
∴ pH = 2.8729
10. 0.04 N solution of a weak acid has a specific conductance 4 x10–4 mho.cm–1. The degree of dissociation of acid at this dilution is 0.05. Calculate the equivalent conductance of weak acid at infinite solution.
10. 0.04 N solution of a weak acid has a specific conductance 4 x10–4 mho.cm–1. The degree of dissociation of acid at this dilution is 0.05. Calculate the equivalent conductance of weak acid at infinite solution.
Specific conductance, κ = 4 x 10-4 mho.cm-1
Equivalent conductance of weak acid, λc = κ .1000 / C
= 4 x 10-4 x 1000 / 0.04
λc = 10 mho.cm2.eq-1
Degree of dissociation, α = 0.05
α = λc / λ∞
Equivalent conductance of weak acid at infinite solution, λ∞ = λc / α
λ∞ = 10 / 0.05
∴ λ∞ = 200 mho.cm2.gm.equiv.-1
11.Ionic conductances at infinite dilution of Al3+ and SO42– are 189 ohm–1cm2gm.equiv-1 and 160 ohm-1cm2gm.equiv.-1. Calculate equivalent and molar conductance of the electrolyte at infinite dilution.
11.Ionic conductances at infinite dilution of Al3+ and SO42– are 189 ohm–1cm2gm.equiv-1 and 160 ohm-1cm2gm.equiv.-1. Calculate equivalent and molar conductance of the electrolyte at infinite dilution.
Equivalent conductance of the electrolyte at infinite dilution,
λ∞ = 1/n+. λA+ + 1/m- . λB–
Where,
λ∞+ and λ∞– are the cationic and anionic equivalent conductances at infinite dilution
n+ and m– correspond the valency of cations and anions
The electrolyte is Al2(SO4)3
λ∞ Al2(SO4)3 = 1/3 λ∞ Al3+ + 1/2 λ∞ SO42–
λ∞ Al2(SO4)3 = 189 / 3 + 160 / 2
= 63 + 80
= 63 + 80
= 143 mho cm2 gm.equiv–1
Molar conductance of the electrolyte at infinite dilution,
μ∞ = γ+ μ∞+ + γ- μ∞-
Where,
μ∞- and μ∞+are the ionic conductances at infinite dilution
γ+ = number of cations and γ- = number of anions
μ∞ Al2(SO4)3 = 2 μ∞+ + 3 μ∞-
= (2 x 189) + (3 x 160)
= 378 + 480
= 858 mho cm2 mol–1
ONE MARKS THREE MARKS= (2 x 189) + (3 x 160)
= 378 + 480
= 858 mho cm2 mol–1
12. What current strength in amperes will be required to liberate 10 g of iodine from potassium iodide solution in one hour?
127 g of iodine (1g eqvt) is liberated by = 96,495 coulomb
10 g of iodine is liberated by = (96,495 x 10 / 127) coulomb
Let the current strength be = I
Time (in seconds) = 1 x 60 x 60 = 3600 seconds
The quantity of electricity used, Q = I x Time (in seconds)
Curernt strength, I = Q / t
= 96,495 coulomb x 10 / 127 x 3600 seconds
= 2.11 coulomb . second–1
or = 2.11ampere [ ∵ coulomb . second–1 = ampere
13. Calculate the potential of the following cell at 298 K
Zn / Zn2+ (a = 0.1) ║ Cu2+ (a = 0.01) / Cu
EoZn2+ / Zn = – 0.762 V
EoCu2+ / Cu = + 0.337 V
The overall cell reaction is : Zn + Cu2+ (a = 0.01) → Zn2+ (a = 0.1) + Cu
The cell potential given by Nernst equation
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