Electrochemistry — Page 8 | NEET Chemistry

Q71

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.

Q72

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

Q73

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) .

Q74

Match with . (s) \to CdO(s) + 2Ni{(OH)_2}(s) + {H_2}O(l)$$

List - I		List - II
(B)	$Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)$	(I)	Primary battery
(C)	$2PbS{O_4}(s) + 2{H_2}O(l) \to Pb(s) + Pb{O_2}(s) + 2{H_2}S{O_4}(aq)$	(II)	Discharging of secondary battery
(D)	$2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l)$	(III)	Fuel cell
()		(IV)	Charging of secondary battery

A

(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})

B

(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{III})

C

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})

D

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})

Correct Answer

Option C

Solution

(a) $\mathrm{Cd}(\mathrm{s})+2 \mathrm{Ni}(\mathrm{OH})_3(\mathrm{~s}) \rightarrow \mathrm{CdO}(\mathrm{s})+2 \mathrm{Ni}(\mathrm{OH})_2(\mathrm{~s})$ $+\mathrm{H}_2 \mathrm{O}(l)$ Discharge of secondary Battery (b) $\mathrm{Zn}(\mathrm{Hg})+\mathrm{HgO}$ (s) $\rightarrow \mathrm{ZnO}(\mathrm{s})+\mathrm{Hg}(l)$ (Primary Battery Mercury cell) (c) $2 \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_2(\mathrm{~s})+$ $2 \mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})$ (Charging of secondary Battery) (d) $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)$ (Fuel cell)

Q75

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : An aqueous solution of

\mathrm{KOH}

when used for volumetric analysis, its concentration should be checked before the use. Reason (R) : On aging,

\mathrm{KOH}

solution absorbs atmospheric

\mathrm{CO}_{2}

. In the light of the above statements, choose the correct answer from the options given below :

A (A) is correct but (R) is not correct

B (A) is not correct but (R) is correct

C Both (A) and (R) are correct but (R) is not the correct explanation of (A)

D Both (A) and (R) are correct and (R) is the correct explanation of (A)

Correct Answer

Option D

Solution

In volumetric analysis, the concentration of a solution is a crucial factor in determining the accuracy of the results.

In the case of an aqueous solution of KOH, the concentration can change over time due to the absorption of atmospheric CO2.

This occurs because KOH is a basic (alkaline) solution and reacts with carbon dioxide (CO2) from the air to form potassium carbonate (K2CO3).

The reaction between KOH and CO2 is given by : KOH + CO2 $\to$ K2CO3 This reaction will change the concentration of KOH in the solution, which in turn will impact the accuracy of the results obtained in volumetric analysis.

For example, if the concentration of KOH is lower than it was when the solution was prepared, then more solution will be needed to reach the endpoint in a reaction, and the results will be inaccurate.

That's why it is important to check the concentration of the KOH solution before using it for volumetric analysis, as stated in the assertion (A).

And the reason (R) provides the explanation for why the concentration should be checked.

Hence, both the assertion and the reason are correct and the correct answer is (D).

Both (A) and (R) are correct and (R) is the correct explanation of (A).

Q76

Match with (Parmeter) (Unit)

List - I		List - II
(a)	Cell constant	(i)	$S\,c{m^2}mo{l^{ - 1}}$
(b)	Molar conductivity	(ii)	Dimensionless
(c)	Conductivity	(iii)	${m^{ - 1}}$
(d)	Degree of dissociation of electrolyte	(iv)	${\Omega ^{ - 1}}{m^{ - 1}}$ Choose the most appropriate answer from the options given below :

A (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

B (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)

C (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)

D (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

Correct Answer

Option A

Solution

Cell constant =

\left( {{l \over A}} \right)

$\Rightarrow$ Units = m $-$ 1 Molar conductivity (

\Lambda

m) $\Rightarrow$ Units = Sm2 mole $-$ 1 Conductivity (K) $\Rightarrow$ Units = S m $-$ 1 Degree of dissociation ( $\alpha$ $\to$ Dimensionless $\therefore$ (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)

Q77

The equation that is incorrect is :

A

{\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}

B

{\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}

C

{\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}

D

{\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}

Correct Answer

Option B

Solution

Left hand side :

{\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}}

=

{\left( {\Lambda _m^0} \right)_{Br^-}} - {\left( {\Lambda _m^0} \right)_{I^-}}

....(1) Right hand side :

{\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}

=

{\left( {\Lambda _m^0} \right)_{K^+}} - {\left( {\Lambda _m^0} \right)_{Na^+}}

....(2) According to Kohlrausch's law option (B) is incorrect as (1) and (2) are not equal.

Q78

The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25oC are given below:

\wedge _{C{H_3}COONa}^o

= 91.0 S cm2/equiv

\wedge _{HCl}^o

= 426.2 S cm2/equiv What additional information/quantity one needs to calculate

\wedge ^o

of an aqueous solution of acetic acid?

A

\wedge ^o

of chloroacetic acid (C/CH2COOH)

B

\wedge ^o

of NaCl

C

\wedge ^o

of CH3COOK

D The limiting equivalent conductance of

{H^ + }( \wedge _{{H^ + }}^o)

Correct Answer

Option B

Solution

NOTE : According to Kohlrausch's law, molar conductivity of weak electrolyte acetic acid

\left( {C{H_3}COOH} \right)

can be calculated as follows:

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

$\therefore$ Value of

{\Lambda ^o}_{NaCl}

should also be known for calculating value of

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

Q79

.tg .tg Electrolyte: KCl KNO3 HCl NaOAc NaCl

{ \wedge ^\infty }(Sc{m^2}mo{l^{ - 1}}):

149.9 145 426.2 91 126.5 Calculate

\wedge _{HOAc}^\infty

Using appropriate molar conductances of the electrolytes listed above at infinite dilution in H2O at 25oC

A 517.2

B 552.7

C 390.7

D 217.5

Correct Answer

Option C

Solution

A_{HCl}^\infty = 426.2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

A_{AcONa}^\infty = 91.0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

A_{NaCl}^\infty = 126.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

A_{AcOH}^\infty = \left( i \right) + \left( {ii} \right) - \left( {iii} \right)

= \left[ {426.2 + 91.0 - 126.5} \right]

= 390.7

Q80

Calculate the standard cell potential in (V) of the cell in which following reaction takes place : Fe2+(aq) + Ag+(aq)

\to

Fe3+(aq) + Ag (s) Given that

E_{A{g^ + }/Ag}^o = xV

E_{Fe^{2+ }/Fe}^o = yV

E_{Fe^{3+ }/Fe}^o = zV

A x + 2y - 3z

B x - z

C x - y

D x + y - z

Correct Answer

Option A

Solution

Standard emf,

{E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0

..............(1) Given

E_{Fe^{2+ }/Fe}^o = yV

$\therefore$ Fe2+ + 2e- $\to$ Fe ..........(2)

{E^0}

= y and

\Delta {G^0}

= -2Fy Also given

E_{Fe^{3+ }/Fe}^o = zV

$\therefore$ Fe3+ + 3e- $\to$ Fe .........(3)

{E^0}

= z and

\Delta {G^0}

= -3Fz Performing (2) - (1), we get Fe3+ + e- $\to$ Fe2+ $\therefore$

\Delta {G^0}

= -3Fz + 2Fy $\Rightarrow$ -(1)F

{E^0}

= -3Fz + 2Fy $\therefore$

E_{F{e^{3 + }}|F{e^{2 + }}}^0

= 3z - 2y From Equation (1),

{E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0

= x - (3z - 2y) = x + 2y - 3z