Electrochemistry — Page 3 | NEET Chemistry

Q21

The molar conductivity of a 0.5 mol/dm 3 solution of AgNO 3 with electrolytic conductivity of 5.76

\times

10

-

3 S cm

-

1 at 298 K is

A 2.88 S cm 2 /mol

B 11.52 S cm 2 /mol

C 0.086 S cm 2 /mol

D 28.8 S cm 2 /mol

Correct Answer

Option B

Solution

{\Lambda _m} = {{\kappa \times 1000} \over {Molality}}

=

{{5.76 \times {{10}^{ - 3}} \times 1000} \over {0.5}}

= 11.52 S cm 2 /mol

Q22

The pressure of H 2 required to make the potential of H 2 -electrode zero in pure water at 298 K is

A 10

-

10 atm

B 10

-

4 atm

C 10

-

14 atm

D 10

-

12 atm

Correct Answer

Option C

Solution

2H + (aq) + 2e – $\to$ H 2 (g) E cell = E o cell -

{{0.0591} \over 2}\log {{{p_{{H_2}}}} \over {{{\left[ {{H^ + }} \right]}^2}}}

$\Rightarrow$ 0 = 0 -

{{0.0591} \over 2}\log {{{p_{{H_2}}}} \over {{{\left[ {{{10}^{ - 7}}} \right]}^2}}}

$\Rightarrow$

{{{p_{{H_2}}}} \over {{{\left[ {{{10}^{ - 7}}} \right]}^2}}} = 1

$\Rightarrow$ P H 2 = 10 –14 atm

Q23

Aqueous solution of which of the following compounds is the best conductor of electric current ?

A Hydrochloric acid, HCl

B Ammonia, NH 3

C Fructose, C 6 H 12 O 6

D Acetic acid, C 2 H 4 O 2

Correct Answer

Option A

Solution

HCl completely dissociates to give H + and Cl - ions, hence act as very good electrolyte.

While others are non- electrolytes.

Q24

A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as

A dynamo

B Ni-Cd cell

C fuel cell

D electrolytic cell

Correct Answer

Option C

Solution

A device that converts energy of combustion of fuels, directly into electrical energy is known as fuel cell.

Q25

When 0.1 mol MnO

_4^{2 - }

is oxidised the quantity of electricity required to completely oxidise MnO

_4^{2 - }

to MnO

_4^ -

is

A 96500 C

B 2

\times

96500 C

C 9650 C

D 96.50 C

Correct Answer

Option C

Solution

\mathop {MnO_4^{2 - }}\limits_{0.1\,mole}^{ - 6}

$\to$

\mathop {MnO_4^ - }\limits^{ - 7}

+ e - Quantity of electricity required = 0.1F = 0.1 × 96500 = 9650 C

Q26

The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O 2 at STP will be

A 5.4 g

B 10.8 g

C 54.0 g

D 108.0 g

Correct Answer

Option D

Solution

We know, from Faraday’s second law

{{{W_{Ag}}} \over {{E_{Ag}}}} = {{{W_{{O_2}}}} \over {{E_{{O_2}}}}}

$\Rightarrow$

{{{W_{Ag}}} \over {108}} = {{{{5600} \over {22400}} \times 32} \over 8}

$\Rightarrow$

{{{W_{Ag}}} \over {108}} = {8 \over 8}

$\Rightarrow$

{{W_{Ag}} = 108}

g

Q27

How many gram of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1 Faraday = 96,500 C; Atomic mass of Co = 59 u)

A 4.0

B 20.0

C 40.0

D 0.66

Correct Answer

Option B

Solution

W =

{{IEt} \over {96500}}

=

{{10 \times 109 \times 60 \times 59} \over {96500}}

= 20

Q28

Consider the half-cell reduction reaction Mn 2+ + 2e

-

\to

Mn, E o =

-

1.18 V Mn 2+

\to

Mn 3+ + e

-

, E o =

-

1.51 V The

E

o for the reaction 3 Mn 2+

\to

Mn o + 2Mn 3+ , and possibility of the forward reaction are respectively

A

-

4.18 V and yes

B + 0.33 V and yes

C + 2.69 V and no

D

-

2.69 V and no

Correct Answer

Option D

Solution

Mn 2+ + 2e $-$ $\to$ Mn, E o = $-$ 1.18 V Mn 2+ $\to$ Mn 3+ + e $-$ , E o = $-$ 1.51 V 3Mn 2+ $\to$ Mn o + 2Mn 3+ ,

\Delta

E o = - 1.81 - 1.51 = $-$ 2.69 V Since

\Delta

E° is negative, $\therefore$

\Delta

G = –nFE°,

\Delta

G will have positive value so, forward reaction is not possible.

Q29

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

A 0.118 V

B 1.18 V

C 0.059 V

D 0.59 V

Correct Answer

Option D

Solution

Oxidation half reaction is H 2 $\to$ 2H + + 2e – If pH = 10, [H + ] = 10 –10 We know, Nernst Equation E cell = E° cell -

{{0.0591} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}} \over {{P_{{H_2}}}}}

For hydrogen electrode E° cell = 0 E cell = -

{{0.0591} \over 2}\log {{{{\left( {{{10}^{ - 10}}} \right)}^2}} \over 1}

= 0.0591log10 10 E cell = 0.591 V

Q30

At 25 o C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm

-

1 cm 2 mol

-

1 and at infinite dilution its molar conductance is 238 ohm

-

1 cm 2 mol

-

1 . The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

A 4.008%

B 40.800%

C 2.080%

D 20.800%

Correct Answer

Option A

Solution

Degree of dissociation( $\alpha$ ) =

{{9.54} \over {238}}

= 0.04008 = 4.008%