Electrochemistry — Page 2 | NEET Chemistry

Q11

The molar conductance of NaCl, HCl and CH 3 COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm 2 mol

-

1 respectively. The molar conductance of CH 3 COOH at infinite dilution is. Choose the right option for your answer.

A 540.48 S cm 2 mol

-

1

B 201.28 S cm 2 mol

-

1

C 390.71 S cm 2 mol

-

1

D 698.28 S cm 2 mol

-

1

Correct Answer

Option C

Solution

According to Kohlrausch law of independent migration of ions.

\Lambda _m^o(C{H_3}COOH)

= \Lambda _m^o(C{H_3}COONa) + \Lambda _m^o(HCl) - \Lambda _m^o(NaCl)

= 91.0 S cm 2 mol $-$ 1 + 426.16 S cm 2 mol $-$ 1 $-$ 126.45 S cm 2 mol $-$ 1 = 390.71 S cm 2 mol $-$ 1

Q12

The molar conductivity of 0.007 M acetic acid is 20 S cm 2 mol

-

1 . What is the dissociation constant of acetic acid? Choose the correct option. [

\Lambda _{{H^ + }}^o

= 350 S cm 2 mol

-

1

\Lambda _{C{H_3}CO{O^ - }}^o

= 50 S cm 2 mol

-

1 ]

A 2.50

\times

10

-

5 mol L

-

1

B 1.75

\times

10

-

4 mol L

-

1

C 2.50

\times

10

-

4 mol L

-

1

D 1.75

\times

10

-

5 mol L

-

1

Correct Answer

Option D

Solution

\Lambda _m^{}

= 20 S cm 2 mol $-$ 1 According to Kohlrausch’s law,

\Lambda _{m\,\,C{H_3}COOH}^o = \Lambda _{C{H_3}CO{O^ - }}^o + \Lambda _{m\,\,{H^ + }}^o

= 50 + 350 = 400 S cm 2 mol $-$ 1 Degree of dissociation,

\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{20} \over {400}} = {1 \over {20}}

{K_a} = {{C{\alpha ^2}} \over {1 - \alpha }} \simeq C{\alpha ^2}

As $\alpha$ << 1 so 1 - $\alpha$ $\approx$ 1

= 7 \times {10^{ - 3}} \times {\left( {{1 \over {20}}} \right)^2}

= 7 \times {10^{ - 3}} \times {1 \over 4} \times {10^{ - 2}}

= 1.75 \times {10^{ - 5}}

mol L -1

Q13

On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be :

A Oxygen gas

B H 2 S gas

C SO 2 gas

D Hydrogen gas

Correct Answer

Option A

Solution

During the electrolysis of dil. sulphuric acid using Pt electrodes following reaction will take place. At cathode :

4H_{(aq)}^ + + 4{e^ - } \to 2{H_{2(g)}}

At anode :

2{H_2}{O_{(I)}} \to {O_{2(g)}} + 4H_{(aq)}^ + + 4{e^ - }

At the anode oxygen gas will be released.

Q14

The number of Faradays(F) required to produce 20 g of calcium from molten CaCl 2 (Atomic mass of Ca = 40 g mol -1 ) is :

A 2

B 3

C 4

D 1

Correct Answer

Option D

Solution

1 equivalent of any substance is deposited by 1F of charge. We have 20g calcium. No. of equivalent =

{{given\;mass} \over {equivalent\;mass}}

Ca 2+ + 2e $-$ $\to$ Ca(s) v.f. = 2 According to Faraday's 1st law Charge passed in Faradey = g. equivalent of product =

{{20} \over {40}} \times 2

= 1 F So, 1 F of charge is required.

Q15

For the cell reaction 2Fe 3+ (aq) + 2I – (aq)

\to

2Fe 2+ (aq) + I 2 (aq)

{E_{cell}^\Theta }

= 0.24 V at 298 K. The standard Gibbs energy (

\Delta

r G o ) of the cell reaction is : [Given that Faraday constant F = 96500 C mol –1 ]

A 46.32 kJ mol –1

B 23.16 kJ mol –1

C –46.32 kJ mol –1

D –23.16 kJ mol –1

Correct Answer

Option C

Solution

Here, n = 2

\Delta

G o = -nFE o = – 2 × 96500 × 0.24 = – 46320 J mol –1 = – 46.32 KJ mol –1

Q16

For a cell involving one electron

E_{cell}^\Theta

= 0.59 V at 298 K, the equilibrium constant for the cell reaction is : [Given that

{{2.303RT} \over F}

= 0.059 V at T = 298 K ]

A 1.0

\times

10 30

B 1.0

\times

10 10

C 1.0

\times

10 2

D 1.0

\times

10 5

Correct Answer

Option B

Solution

We know, E cell =

E_{cell}^\Theta

-

{{2.303RT} \over {nF}}\log Q

$\Rightarrow$ E cell =

E_{cell}^\Theta

-

{{0.059} \over n}\log Q

(At equilibrium, E cell = 0 and Q = K eq ) $\Rightarrow$ 0 =

E_{cell}^\Theta

-

{{0.059} \over 1}\log {K_{eq}}

$\Rightarrow$

\log {K_{eq}} = {{E_{cell}^\Theta } \over {0.059}}

=

{{0.59} \over {0.059}}

= 10 $\Rightarrow$ K eq = 10 10

Q17

In the electrochemical cell :

Zn\left| {ZnS{O_4}\left( {0.01\,M} \right)} \right|

\left| {CuS{O_4}\left( {1.0M} \right)} \right|Cu,

the emf of this Daniell cell is E 1 . When the concentration of ZnSO 4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E 2 . From the followings, which one is the relationship between E 1 and E 2 ? (Given, RT/F = 0.059)

A E 1 < E 2

B E 1 > E 2

C E 2 = 0 1 E 1

D E 1 = E 2

Correct Answer

Option B

Solution

E cell = E o -

{{0.059} \over n}\log {{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}

E 1 = E o -

{{0.059} \over 2}\log {{0.01} \over 1}

= E o -

{{0.059} \over 2}\left( { - 2} \right)\log 10

E 1 = E o + 0.059 E 2 = E o -

{{0.059} \over 2}\log {1 \over {0.01}}

$\Rightarrow$ E 2 = E o -

{{0.059} \over 2}\log 100

$\Rightarrow$ E 2 = E o - 0.059 $\therefore$ E 1 > E 2

Q18

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is

A 55 minutes

B 110 minutes

C 220 minutes

D 330 minutes

Correct Answer

Option B

Solution

At cathode : 2Na + + 2e – $\to$ 2Na At anode : 2Cl – $\to$ Cl 2 + 2e – ---------------------------------------------- Net reaction: 2Na + + 2Cl – $\to$ 2Na + Cl 2 From Faraday’s first law of electrolysis, w = Z $\times$ I $\times$ t =

{E \over {96500}}

$\times$ I $\times$ t No. of moles of Cl 2 gas × Mol. wt. of Cl 2 gas =

{{Eq.\,wt.\,of\,C{l_2}\,gas \times I \times t} \over {96500}}

$\Rightarrow$ 0.10 $\times$ 71 =

{{35.5 \times 3 \times t} \over {96500}}

$\Rightarrow$ t =

{{0.10 \times 71 \times 96500} \over {35.5 \times 3}}

= 6433.33 sec = 107.22 min

\simeq

110 min

Q19

The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60

\times

10

-

19 C)

A 6

\times

10 23

B 6

\times

10 20

C 3.75

\times

10 20

D 7.48

\times

10 23

Correct Answer

Option C

Solution

Q = I × t Q = 1 × 60 = 60 C Now, 1.60 × 10 –19 C $\equiv$ 1 electron $\therefore$ 60 C $\equiv$

{{60} \over {1.6 \times {{10}^{ - 19}}}}

= 3.75 $\times$ 10 20 electrons

Q20

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because

A zinc is lighter than iron

B zinc has lower melting point than iron

C zinc has lower negative electrode potential than iron

D zinc has higher negative electrode potential than iron

Correct Answer

Option D

Solution

The reduction potential values are E° Zn 2+ /Zn = – 0.76 V E° Fe 2+ /Fe = – 0.44 V Thus, due to higher negative electro potential value of zinc than iron, iron cannot be coated on zinc.