Electrochemistry — Page 11 | JEE Chemistry

Q101

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)

Q102

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).

Q103

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)

Q104

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.

Q105

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}

Q106

.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

Q107

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

Q108

The

{\left( {{{\partial E} \over {\partial T}}} \right)_P}

of different types of half cells are as follows: .tg .tg A B C D

1 \times {10^{ - 4}}

2 \times {10^{ - 4}}

0.1 \times {10^{ - 4}}

0.2 \times {10^{ - 4}}

(Where E is the electromotive force) Which of the above half cells would be preferred to be used as reference electrode?

A A

B B

C C

D D

Correct Answer

Option C

Solution

A cell with less variation in EMF with temperature is preferred as a reference electrode because it can be used for a wider range of temperatures without much derivation from standard value so a cell with less

{\left( {{{\partial E} \over {\partial T}}} \right)_P}

is preferred.

Q109

In a hydrogen – oxygen fuel cell, combustion of hydrogen occurs to :

A generate heat

B remove adsorbed oxygen from electrode surfaces

C produce high purity water

D create potential difference between the two electrodes

Correct Answer

Option D

Solution

In

{H_2} - {O_2}

fuel cell, the combustion of

{H_2}

occurs to create potential difference between the two electrodes

Q110

Based on the data given below :

\begin{array}{ll} \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^0=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \end{array}

the strongest reducing agent is :

A

\mathrm{Cl}^{-}

B

\mathrm{MnO}_4^{-}

C

\mathrm{Cr}

D

\mathrm{Mn}^{2+}

Correct Answer

Option C

Solution

To determine the strongest reducing agent among the options, we need to consider which species most readily donates electrons (i.e., is most easily oxidized).

One standard method is to compare the standard reduction potentials of the corresponding half-reactions.

A lower (more negative) standard reduction potential indicates that the reduced species is less stable and more prone to oxidation.

Here’s the data provided:

\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7H_2O,\quad E^\circ = +1.33\,V

\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V

\mathrm{MnO}_4^- + 8H^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4H_2O,\quad E^\circ = +1.51\,V

\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V

Now, let’s consider the species from the Options: Option A: Cl⁻ The relevant half-reaction is:

\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V.

The reduced form here is Cl⁻. Its corresponding oxidation (the reverse reaction) would have an electrode potential of

E^\circ_{\text{oxidation}} = -1.36\,V.

Option B: MnO₄⁻ In the given half-reaction, MnO₄⁻ acts as the oxidizing agent (it is reduced to Mn²⁺).

Therefore, MnO₄⁻ is not a good reducing agent.

Option C: Cr (metallic chromium) The relevant half-reaction is:

\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V.

Here, Cr is the reduced species. Reversing this reaction gives:

\mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^-,

with an effective oxidation potential of

E^\circ_{\text{oxidation}} = +0.74\,V.

Option D: Mn²⁺ In the MnO₄⁻/Mn²⁺ couple, Mn²⁺ is the reduced form. Its reverse (oxidation) potential would be:

E^\circ_{\text{oxidation}} = -1.51\,V.

When comparing the oxidation potentials (remember, a higher oxidation potential means the species is more willing to lose electrons), we have: Cl⁻:

E^\circ_{\text{ox}} = -1.36\,V

Cr:

E^\circ_{\text{ox}} = +0.74\,V

Mn²⁺:

E^\circ_{\text{ox}} = -1.51\,V

Among these, metallic Cr stands out because its corresponding reduction half-reaction has the most negative potential ( $-0.74\,V$ ).

This negative value indicates that Cr (in its elemental form) is unstable relative to its oxidized form ( $\mathrm{Cr}^{3+}$ ) and, therefore, is the most eager to lose electrons.

In other words, Cr is the strongest reducing agent.

Thus, the answer is: Option C:

\mathrm{Cr}