Electrochemistry — Page 10 | JEE Chemistry

Q91

For a spontaneous reaction the ∆G , equilibrium constant (K) and

E_{cell}^o

will be respectively

A -ve, >1, +ve

B +ve, >1, -ve

C -ve, <1, -ve

D -ve, >1, -ve

Correct Answer

Option A

Solution

NOTE : For spontaneous reaction

\Delta G

should be negative. Equilibrium constant should be more than one

\left( {\Delta G = - 2.303\,RT\,\log {K_c},\,\,} \right.

If

{K_c} = 1\,\,

then

\,\,\,\Delta G = 0;\,\,

If

{K_c} < 1

then

\left. {\Delta G = + ve} \right).\,\,

Again

\,\,\,\Delta G = - nFE_{cell}^ \circ

E_{cell}^ \circ \,\,\,

must be

+ve

to have

\Delta G - ve.

Q92

The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25oC. The equilibrium constant of the reaction is (F = 96,500 C mol-1: R = 8.314 JK-1 mol-1)

A 1.0

\times

101

B 1.0

\times

1030

C 1.0

\times

1010

D 1.0

\times

105

Correct Answer

Option C

Solution

E_{cell}^o = E_{cell}^o - {{0.059} \over n}\log \,{K_c}

or

\,\,\,\,\,\,

0 = 0.591 - {{0.0591} \over 1}\log {K_c}

or

\,\,\,\,\,\,

\log \,{K_c} = {{0.591} \over {0.0591}} = 10

or

\,\,\,\,\,\,

{K_c} = 1 \times {10^{10}}

Q93

The highest electrical conductivity of the following aqueous solutions is of :

A 0.1 M acetic acid

B 0.1 M chloroacetic acid

C 0.1 M fluoroacetic acid

D 0.1 M difluoroacetic acid

Correct Answer

Option D

Solution

Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity.

The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid

>

fluoro acetic acid

>

chloro acitic acid

>

acetic acid.

Q94

When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is :

A 9.81 g

B 10.9 g

C 98.1 g

D 109.0 g

Correct Answer

Option A

Solution

Moler mass of p $-$ aminophenol = 6 $\times$ 12 + 7 + 14 + 16 = 109 gmol $-$ 1 Eq. weight (E) =

{W \over Q}

$\times$ 96500 $\Rightarrow$

\,\,\,

W =

{{E\,Q} \over {96500}}

$\Rightarrow$

\,\,\,

W =

{{E\,I\,t} \over {96500}}

$\Rightarrow$

\,\,\,

W =

{{109} \over 4} \times {{9.65 \times 3600} \over {96500}}

$\Rightarrow$

\,\,\,

W = 9.81 gm

Q95

\wedge _m^ \circ

for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is- 5

\times

10–5 S cm–1, degree of dissociation of HA is -

A 0.50

B 0.125

C 0.25

D 0.75

Correct Answer

Option B

Solution

\wedge _m^ \circ

(HA) =

\wedge _m^ \circ

(HCl) +

\wedge _m^ \circ

(NaA) $-$

\wedge _m^ \circ

(NaCl) = 425.9 + 100.5 $-$ 126.4 = 400 S cm2 . mol $-$ 1

\wedge _m^ c

=

{{K \times 1000} \over M}

=

{{5 \times {{10}^{ - 5}} \times 1000} \over {{{10}^{ - 3}}}}

= 50 S cm2 mol $-$ 1 $\alpha$ =

{{\Lambda _m^c} \over {\Lambda _m^o}}

=

{{50} \over {400}}

= 0.125

Q96

For lead storage battery pick the correct statements A. During charging of battery,

\mathrm{PbSO}_{4}

on anode is converted into

\mathrm{PbO}_{2}

B. During charging of battery,

\mathrm{PbSO}_{4}

on cathode is converted into

\mathrm{PbO}_{2}

C. Lead storage battery consists of grid of lead packed with

\mathrm{PbO}_{2}

as anode D. Lead storage battery has

\sim 38 \%

solution of sulphuric acid as an electrolyte Choose the correct answer from the options given below:

A A, B, D only

B B, C only

C B, C, D only

D B, D only

Correct Answer

Option D

Solution

Let's review each statement: A.

During charging of battery, PbSO₄ on anode is converted into PbO₂.

This statement is incorrect.

During charging, PbSO₄ on the anode (lead plate) is converted to Pb, not PbO₂.

B.

During charging of battery, PbSO₄ on cathode is converted into PbO₂.

This statement is correct.

During charging, PbSO₄ on the cathode (lead dioxide plate) is converted back to PbO₂.

C.

Lead storage battery consists of grid of lead packed with PbO₂ as anode.

This statement is incorrect.

The anode in a lead-acid battery is made of lead (Pb), not lead dioxide (PbO₂).

The cathode is made of lead dioxide (PbO₂).

D.

Lead storage battery has ∼38% solution of sulphuric acid as an electrolyte.

This statement is correct.

The electrolyte in a lead-acid battery is a solution of about 35-38% sulfuric acid (H₂SO₄) in water.

Therefore, the correct answer is: B, D only.

Q97

Given below are two statements : Statement (I) : Fusion of

\mathrm{MnO}_2

with

\mathrm{KOH}

and an oxidising agent gives dark green

\mathrm{K}_2 \mathrm{MnO}_4

. Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is false but Statement II is true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

Statement I is indeed true. When

\mathrm{MnO}_2

(manganese dioxide) is fused with

\mathrm{KOH}

(potassium hydroxide) and an oxidising agent such as

\mathrm{KNO}_3

(potassium nitrate), it forms dark green potassium manganate (

\mathrm{K}_2\mathrm{MnO}_4

). The reaction can be represented as follows:

\mathrm{2MnO}_2 + 4KOH + \mathrm{O}_2 \rightarrow \mathrm{2K}_2\mathrm{MnO}_4 + 2H_2\mathrm{O}

This process involves the oxidation of manganese dioxide to manganate ion (

\mathrm{MnO}_4^{2-}

) in an alkaline medium. Statement II is also true. The manganate ion (

\mathrm{MnO}_4^{2-}

), when subjected to electrolytic oxidation in an alkaline medium, can indeed be converted into permanganate ion (

\mathrm{MnO}_4^-

), which is characterized by a deep purple color.

This conversion is an example of an electron-loss (oxidation) process at the anode of an electrolytic cell.

The reaction can be represented as follows:

\mathrm{MnO}_4^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_4^- + 2\mathrm{OH}^- + 2e^-

Thus, both statements I and II are accurate, making the correct answer: Option D: Both Statement I and Statement II are true.

Q98

The molar conductivities

\wedge _{NaOAc}^o

and

\wedge _{HCl}^o

and at infinite dilution in water at 25oC are 91.0 and 426.2 Scm2/mol respectively. To calculate

\wedge _{HOAc}^o

, the additional value required is

A

\wedge _{{H_2}O}^o

B

\wedge _{KCl}^o

C

\wedge _{NaOH}^o

D

\wedge _{NaCl}^o

Correct Answer

Option D

Solution

\Lambda _{C{H_3}COOH}^o

is given by the following equation

\Lambda _{C{H_3}COOH}^o = \left( {\Lambda _{C{H_3}COONa}^o + \Lambda _{HCl}^o} \right) - \left( {\Lambda _{NaCl}^o} \right)

Hence

\Lambda _{NaCl}^ \circ

is required.

Q99

Given the equilibrium constant: KC of the reaction : Cu(s) + 2Ag+ (aq)

\to

Cu2+ (aq) + 2Ag(s) is 10

\times

1015, calculate the E

_{cell}^0

of this reaciton at 298 K [2.303

{{RT} \over F}

at 298 K = 0.059V]

A 0.4736 mV

B 0.04736 V

C 0.4736 V

D 0.04736 mV

Correct Answer

Option C

Solution

We know,

\Delta

Go = -RTln(KC) ....(1) Also

\Delta

Go = -nF

E_{cell}^o

....(2) $\therefore$ -nF

E_{cell}^o

= -RTln(KC) $\Rightarrow$

E_{cell}^o

=

{{RT} \over {nF}}\ln \left( {{K_C}} \right)

=

2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)

=

{{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)

( n = no of electron transferred = 2 ) = 0.059 $\times$

{{16} \over 2}

= 0.059 $\times$ 8 = 0.472 V

Q100

For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below : .tg .tg Mx+ (aq)/M(s) Au3+(aq)/Au(s) Ag+(aq)/Ag(s) Fe3+(aq)/Fe2+ (aq) Fe2+(aq)/Fe(s) E0Mx+/M/(V) 1.40 0.80 0.77

-

0.44 If

E_{z{n^{2 + }}/zn}^0

=

-

0.76 V, which cathode will give maximum value of Eocell per electron transferred?

A Ag+/Ag

B Fe3+/Fe2+

C Au3+/Au

D Fe2+/Fe

Correct Answer

Option A

Solution

Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s) --------------------------------------- Anode Cathode Eocell = Eocathode – Eoanode (i) For Ag+/Ag : Eocell = 0.80 – (– 0.76) = 1.56 V No of electrons transferred = LCM of valency factor of two electrode Valency factor of Zn(s) |Zn2+ = 2 Valency factor of Ag+/Ag = 1 LCM of 1 and 2 = 2 $\therefore$ No of electrons transferred = 2 $\therefore$ Eocell per electron =

{{1.56} \over 2}

= 0.78 (ii) For Fe3+/Fe2+ : Eocell = 0.77 – (– 0.76) = 1.53 V No of electrons transferred = LCM of valency factor of two electrode Valency factor of Zn(s) |Zn2+ = 2 Valency factor of Fe3+/Fe2+ = 1 LCM of 2 and 1 = 2 $\therefore$ No of electrons transferred = 2 $\therefore$ Eocell per electron =

{{1.53} \over 2}

= 0.76 (iii) For Au3+/Au : Eocell = 1.40 – (– 0.76) = 2.16 V No of electrons transferred = LCM of valency factor of two electrode Valency factor of Zn(s) |Zn2+ = 2 Valency factor of Au3+/Au = 3 LCM of 2 and 3 = 6 $\therefore$ No of electrons transferred = 6 $\therefore$ Eocell per electron =

{{2.16} \over 6}

= 0.36 (iv) For Fe2+/Fe : Eocell = –0.44 – (– 0.76) = 0.32 V No of electrons transferred = LCM of valency factor of two electrode Valency factor of Zn(s) |Zn2+ = 2 Valency factor of Fe2+/Fe = 2 LCM of 2 and 2 = 2 $\therefore$ No of electrons transferred = 2 $\therefore$ Eocell per electron =

{{0.32} \over 2}

= 0.16 Eocell is maximum for EoAg+(aq)/Ag(s) .