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Chemistry - Electrochemistry [Time : 90 Mins]
SOLUTION OF ASSIGNMENT
[Max. Marks : 35]
Note : 1. Electrochemistry chapter Weightage total 5 marks. 2. All questions are compulsory. 3. Draw neat and labelled diagram wherever necessary. 4. Figure to the right indicates full marks. 5. Answer to every new question must begin on a new page.
1.
What is meant by cell constant?
Solution :
Cell constants the ratio of distance between electrodes and area of cross – section. If is denoted by l/a. Its unit is cm-1.
[1 Mark]
2.
What does the negative value of E0Cell indicate?
Solution :
E0Cell is –ve means ∆G will be +ve, and the cell will not work.
3.
What are secondary cells?
Solution :
Secondary cells are those cells which are rechargeable i.e., the products can be charged back to reactants.
[1 Mark]
[1 Mark]
4.
How does cathodic protection of iron operate?
Solution :
More reactive metal coated on iron acts as anode, ie., loses electrons whereas iron acts as cathode. If any Fe2+ ions are formed will gain electron to form iron.
5.
[1 Mark]
Electrolysis of KBr(aq) gives Br2 at anode but of KF(aq) does not give F2. Give reason for disparity in behaviour.
Solution :
E oF
2 /F
+
has highest reduction potential, therefore, it reacts with water to liberate O2 and not F2. [1 Mark]
6.
Can you store copper sulphate solutions in a zinc pot?
Solution :
E0Zn2+ = - 0.76 V, E0 Cu2+/Cu = 0.34V.
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We have to check whether the following reaction takes place or not. Zn(s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
[1 Mark]
E0 Cell = E0 Cu2+/Cu -EZn2+, Zn = 0.34 V – (-0.76 V) =1.1 V Since E0 Cell is positive. The reaction takes place and we cannot store.
7.
Solution :
[1 Mark]
With the help of ionic equations describe what happen : When (i)
pH of a solution of dichromate ions is raised.
(ii)
Potassium mangnate is electro-chemically oxidized.
(i) Cr2O72+ + 2OH-
2CrO42-
(orange)
(yellow)
higher pH
where pH is increased, i.e., solution is more basic, orange coloured dichromate ion change to yellow [1 Mark]
coloured chromate ion. (ii) At anode
electrolysis MnO42- → MnO4-+ e-
purple When potassium manganate is electro-chemically oxidised i.e.,undergoes electrolytic oxidation, it forms purple colour KMnO4.
[1 Mark]
8.
How does molar conductance of a strong electrolyte vary with its concentration in solution?
Solution :
The molar conductivity of a strong electrolyte like KCl decreases slightly as the concentration is increased. This is because greater inter ionic attractions retard the motion of the ions and therefore molar conductivity decreases.
[1 Mark]
At low concentrations, the molar conductance of strong electrolyte can be expressed by the relationship. Λm = Λm∞ - k√ Where Λm = molar conductance at concentration C. Λm∞ = molar conductance at infinite dilution. K = Constant, C = Concentration of solution on molar scale.
9.
[1 Mark]
Define molar conductivity of an electrolytic solution. Mention the effect of temperature on molar conductivity?
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Solution :
Molar conductivity is defined as the electrolytic conductivity power of all the ions furnished by 1 mole of the electrolyte at a given concentration in solution. It is gives as Λm =
k Cm
where Λm = molar conductivity , k = Specific conductivity, conductivity, Cm = Concentration in molar per litre. [1 Mark] Effect of temperature on molar conductivity conducti : The speed of movement of ions increases with an increase in temperature. Therefore, molar conductivity increases with temperature.
10.
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Solution :
For hydrogen electrode. H+ + e- → ½ H2
[1 Mark]
[1 Mark]
From Nernst equation, EH+ / H2 = E 0 H+ / H2 −
=0−
11.
0.0591 1 log + n [H ]
0.0591 1 log −10 = −0.0591× 10 = −0.591V 1 10
[1 Mark]
Explain Kohlrausch's law of independent migration of ions. Mention one application ap of Kohlrausch's law?
Solution :
Kohlrausch's law of independent migration of ions: The molar conductivity of an electrolyte at infinite solution is the sum of the individual contributions of the anion and cation of the electrolyte.
where λ0+ and λ0- are the limiting molar conductivities of the cation and anion respectively and v+ and v- are the number of cations and anions formed from a formula unit of the electrolyte. For example, one formula unit of A12(SO4)3 gives two Al + ions and three sulphate ions. Therefore, [1 Mark] Determination of molar conductivity of weak electrolytes at infinity solution: Consider acetic acid CH3COOH as the example of electrolyte. Λ (CH3COOH) = λ° (CH3COO-) + λ° (H+)
...(1)
Λ°(HCl) = λ° (H+)+ λ°(Cl °(Cl-)
...(2)
Λ 0(NaCl) = λ0(Na+) + λ°(Cl λ )
...(3)
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[2 Marks]
12.
The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5 55 x 103 ohm. Calculate its resistivity, conductivity and molar conductivity.
Solution :
a = πr2 =3.14 × (0.5 cm)2 = 0.785 cm2 or
= 87.135 Ω cm Conductivity,
[1 Mark]
Molar conductivity, = 0.01148 S cm-1
[1 Mark]
[1 Mark]
13.
Calculate the potential (emf) of the cell Cd | Cd2+ (0.10M) || H+(0.20M)|Pt, H2 (0.5 atm) (Given E° for Cd2+ /Cd = - 0.403 V, R = 8.314 JKT-1 mol-1, F = 96,00 C mol-1).
Ans.:
The cell reaction is Cd + 2H+ (0.20M) → Cd2+ (0.10M) + H2 (0.5 atm) Ecell = E0H+/1/2H2 - E0Cd2+/Cd = 0 - (-0.403) = 0.403 V
[1 Mark]
Applying Nernst equation to the cell reaction,
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= 0.403 −
0.0591 0.1 × 0.5 log = 0.403 − 0.02955 × 0.09691 = 0.403 − 0.002863 = 0.4001V 2 (0.2)2 [2 Mark]
14.
From the following molar conductivities at infinite dilution. Λ°m for Ba(OH)2 = 257.6 Ω mol-1 cm2 mol-1 Λ°m for BaCl2 = 240 6 Ω mol-1 cm2 mol-1 Λ°m for NH4C; = 129.8 Ω mol-1 cm2 mol-l Calculate Λ°m for NH4OH.
Solution :
[2 Marks] (ii)
= 238.3 Ω mol-1 cm2 mol-1
15.
[1 Mark]
Show that the electrical work obtainable from a galvanic cell is given by the expression. ∆G0 = -nFE0cell.
Ans.:
The cell potential is simply related to the free energy change for the reaction. In an electro-chemical electro cell the system does work by transferring the electric charge through an external circuit. The free energy change ∆G G is equal to electrical work done. ∆G = Welectrical
[1 Mark]
For a reaction occurring in a electro-chemical electro chemical cell whose electrodes differ in potential by Ecell, the work done when amount of charge nF is transferred is given by Welect = -nFEcell ... ∆G = -nFEcell Under standard state conditions
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∆G0 = -nFE0cell.
[2 Marks]
16.
What happens during corrosion of a metal? State the electro-chemical basis of corrosion of iron.
Solution :
Metal reacts with gases present in atmosphere to form surface compounds in corrosion. The electrochemical basis of corrosion of iron is as follows: H2O + CO2 → H2CO3; H2CO3 → 2H+ + CO32- ; H2O → 2H+ + OHIron is contact with the dissolved CO2 and O2 undergoes oxidation as follows: Fe → Fe2+ + 2e-
Eooxidation = + 0.44 V
The electrons lost by iron are taken up by the H+ ions present on the surface of the metal which were produced by the dissociation of H2CO3 and H2O.
[2 Mark]
H+ + e- → H 2H2 + O2→ 2H2O O2 + 4H+ + 4e- → 2H2O
[Eored = 1.23V]
O2 + 2H2O + 4e- → 4OH-
[1.5 Mark]
2Fe + O2 + 4H+ → 2Fe2+(aq) + 2H2O(l) E0cell = 1.67 V 2Fe2+ + O2 + 4H2O → 2Fe2O3 + 8H+(aq) Fe2O3 + XH2O → Fe2O3 .X H2O
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[1.5 Mark]
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