Practice Start Test

Electrochemistry

NEET Chemistry · 91 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
A increase in ionic mobility of ions
B 100% ionisation of electrolyte at normal dilution
C increase in both i.e., number of ions and ionic mobility of ions
D increase in number of ions.
Correct Answer
Option A
Solution

Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions increases which in turn increases equivalent conductance of the solution.

Q42
For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25 o C. The value of standard Gibb's energy, Δ\Delta G o will be (F = 96500 C mol −-1 )
A −- 89.0 kJ
B −- 89.0 J
C −- 44.5 kJ
D −- 98.0 kJ
Correct Answer
Option A
Solution

The cell reaction Cu + 2Ag + →\to Cu 2+ + 2Ag We know,

Δ\Delta

G° = – nFE° cell = – 2 × 96500 × 0.46 = – 88780 J = – 88.780 kJ = – 89 kJ

Q43
Al 2 O 3 is reduced by electrolysis at low potentials and high currents. If 4.0 × \times 10 4 amperes of current is passed through molten Al 2 O 3 for 6 hours, what mass of aluminium is product? (Assume 100% current efficiency, at mass of Al = 27 g mol −-1 ).
A 8.1 × \times 10 4 g
B 2.4 × \times 10 5 g
C 1.3 × \times 10 4 g
D 9.0 × \times 10 3 g
Correct Answer
Option A
Solution

E = Z × 96500 ⇒\Rightarrow

273{{27} \over 3}

= Z ×\times 96500 ⇒\Rightarrow Z =

996500{9 \over {96500}}

Now applying the formula, W = Z × I × t W =

996500{9 \over {96500}}

×\times 4 ×\times 10 4 ×\times 6 ×\times 60 ×\times 60 = 8.1 × 10 4 g

Q44
Given : (i) Cu 2+ + 2e −- → \to Cu, E o = 0.337 V (ii) Cu 2+ + e −- → \to Cu + , E o = 0.153 V Electrode potential, E o for the reaction, Cu + + e −- → \to Cu, will be
A 0.90 V
B 0.30 V
C 0.38 V
D 0.52 V
Correct Answer
Option D
Solution

For the reaction, Cu 2+ + 2e −- →\to Cu, E o = 0.337 V

Δ\Delta

G o = - nFE o = – 2 × F × 0.337 = – 0.674 F ......(i) For the reaction, Cu 2+ + e −- →\to Cu + , E o = 0.153 V

Δ\Delta

G o = - nFE o = – 1 × F × – 0.153 = 0.153 F On adding eqn (i) & (ii) Cu 2+ + e −- →\to Cu +

Δ\Delta

G o = –0.521 F = –nFE° ⇒\Rightarrow E° = 0.52 V

Q45
The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm 2 and at infinite dilution is 400 mho cm 2 . The dissociation constant of this acid is
A 1.25 × \times 10 −-6
B 6.25 × \times 10 −-4
C 1.25 × \times 10 −-4
D 1.25 × \times 10 −-5
Correct Answer
Option D
Solution

The degree of dissociation (α\alpha) α\alpha =

8400{8 \over {400}}

= 2 ×\times 10 -2 From Ostwald’s dilution law for weak monobasic acid, we have k c = Cα\alpha 2 =

132×(2×10−2)2{1 \over {32}} \times {\left( {2 \times {{10}^{ - 2}}} \right)^2}

= 1.25 ×\times 10 -5

Q46
Kohlrausch's law states that at
A Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
B Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
D Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.
Correct Answer
Option A
Solution

Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions. λ\lambda ∞\infty = λ\lambda a + λ\lambda c where, λ\lambda a = equivalent conductance of the anion λ\lambda c = equivalent conductance of the cation.

Q47
Standard free energies of formation (in kJ/mol) at 298 K are −-237.2, −- 394.4 and −-8.2 for H 2 O (l), CO 2(g) and pentane (g) respectively. The value of E o cell for the pentane-oxygen fuel cell is
A 1.0968 V
B 0.0968 V
C 1.968 V
D 2.0968 V
Correct Answer
Option A
Solution

At Anode: C 5 H 12 + 10H 2 O →\to 5CO 2 + 32H + + 32e - At Cathode: 8O 2 + 32H + + 32e - →\to 16H 2 O ------------------------------------------------- C 5 H 12 (g) + 8O 2 (g) →\to 5CO 2 (g) + 6H 2 O(l)

Δ\Delta

G = 5×

Δ\Delta

G CO 2 + 6

Δ\Delta

G (H 2 O) – [

Δ\Delta

G (C 5 H 12 ) +8 ×

Δ\Delta

G O 2 ] = 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0) = – 3387 kJ mol –1 = – 3387 × 10 3 J mol –1

Δ\Delta

G = - nFE o cell From the overall equation we find n = 32 – 3387 × 10 3 = -32 ×\times 96500 ×\times E o cell E o cell =

−3387×103−32×96500{{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}

= 1.0968 V

Q48
On the basis of the following E o values, the strongest oxidizing agent is [Fe(CN) 6 ] 4−- → \to [Fe(CN) 6 ] 3−- + e −- ; E o = −-0.35 V Fe 2+ → \to Fe 3+ + e −- ; E o = −-0.77 V
A Fe 3+
B [Fe(CN) 6 ] 3−-
C [Fe(CN) 6 ] 4−-
D Fe 2+
Correct Answer
Option A
Solution

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN) 6 ] 4−- →\to [Fe(CN) 6 ] 3−- + e −- ; E o = −-0.35 V Fe 2+ →\to Fe 3+ + e −- ; E o = −-0.77 V Higher the +ve reduction potential, stronger will be the oxidising agent.

Oxidising agent oxidises other compounds and gets itself reduced easily.

Q49
The equilibrium constant of the reaction: Cu (s) + 2Ag + (aq) → \to Cu 2+ (aq) + 2Ag (s) ; E o = 0.46 V at 298 K is
A 2.0 × \times 10 10
B 4.0 × \times 10 10
C 4.0 × \times 10 15
D 2.4 × \times 10 10
Correct Answer
Option C
Solution

RT ln K = nFE° ln K =

nFE∘RT{{nFE^\circ } \over {RT}}

=

2×0.460.0591{{2 \times 0.46} \over {0.0591}}

⇒\Rightarrow K = 4 ×\times 10 15

Q50
The efficiency of a fuel cell is given by
A Δ\Delta G/Δ\Delta S
B Δ\Delta G/Δ\Delta H
C Δ\Delta S/Δ\Delta G
D Δ\Delta H/Δ\Delta G
Correct Answer
Option B
Solution

Efficiency of a fuel cell (ϕ\phi) =

ΔGΔH×100{{\Delta G} \over {\Delta H}} \times 100

Generally, fuel cells are expected to have an efficiency of 100 percent.

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