Electrochemistry — Page 6 | JEE Chemistry

Q51

Let CNaCl and CBaSO4 be the conductances (in S) measured for saturated aqueous solutions of NaCl and BaSO4, respectively, at a temperature T. Which of the following is false?

A Ionic mobilities of ions from both salts increase with T.

B CNaCl(T2) > CNaCl(T1) for T2 > T1

C CBaSO4(T2) > CBaSO4(T1) for T2 > T1

D CNaCl >> CBaSO4 at a given T

Correct Answer

Option D

Solution

BaSO4 is sparingly soluble salt but NaCl is completely soluble salt so it will produce more number of ions.

That is why Conductance (NaCl) > Conductance (BaSO4)

Q52

Given below are two statements : Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte). Statement II : Molar conductivity decreases with decrease in concentration of electrolyte. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is true but Statement II is false.

B Statement I is false but Statement II is true.

C Both Statement I and Statement II are true.

D Both Statement I and Statement II are false.

Correct Answer

Option D

Solution

Ion

{H^ + }

{K^ + }

C{l^ - }

C{H_3}CO{O^ - }

\Lambda _{m\,Sc{m^2}/mole}^\infty

349.8 73.5 76.3 40.9 So,

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

= 349.8 + 40.9 = 390.7 Scm2/mole

\Lambda _{m\,\,KCl}^\infty = \Lambda _{m\,\,({K^ + })}^\infty + \Lambda _{m\,\,(C{l^ - })}^\infty

= 73.5 + 76.3 = 149.3 Scm2/mole So, Statement I is wrong or false.

As the concentration decreases, the dilution increases which increases the degree of dissociation, thus increasing the no. of ions, which increases the molar conductance.

So Statement II is false.

Image

Q53

The correct order of reduction potentials of the following pairs is A. Cl2/Cl

-

B. I2/I

-

C. Ag+/Ag D. Na+/Na E. Li+/Li Choose the correct answer from the options given below.

A A > C > B > D > E

B A > B > C > D > E

C A > C > B > E > D

D A > B > C > E > D

Correct Answer

Option A

Solution

\mathrm{E}_{\mathrm{C}_{2} / \mathrm{CI}}^{\circ}=+1.36 \mathrm{~V}

\mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{0}=+0.54 \mathrm{~V}

\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V}

\mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\circ}=-2.71 \mathrm{~V}

\mathrm{E}_{\mathrm{L}^{+} / \mathrm{Li}}=-3.05 \mathrm{~V}

Q54

In which of the following half cells, electrochemical reaction is pH dependent?

A

Pt\,|\,F{e^{3 + }},\,F{e^{2 + }}

B

MnO_4^ - \,|M{n^{2 + }}

C

Ag\,|\,AgCl\,|C{l^{ - 1}}

D

{1 \over 2}{F_2}\,|{F^ - }

Correct Answer

Option B

Solution

Reduction of $\mathrm{MnO}_{4}^{-}$ is $\mathrm{pH}$ dependent. In acidic medium

\mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}

In neutral medium

\mathrm{MnO}_{4}^{-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{4+}

In basic medium

\mathrm{MnO}_{4}^{-}+\mathrm{e} \longrightarrow \mathrm{Mn}^{6+}

So, according to $\mathrm{pH}$ , the reaction and potential of cell changes.

Q55

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is

{\Lambda _{m1}}

and that of 20 moles another identical cell heaving 80 mL NaCl solution is

{\Lambda _{m2}}

. The conductivities exhibited by these two cells are same. The relationship between

{\Lambda _{m2}}

and

{\Lambda _{m1}}

is

A

{\Lambda _{m2}}

= 2

{\Lambda _{m1}}

B

{\Lambda _{m2}}

=

{\Lambda _{m1}}

/ 2

C

{\Lambda _{m2}}

=

{\Lambda _{m1}}

D

{\Lambda _{m2}}

= 4

{\Lambda _{m1}}

Correct Answer

Option A

Solution

\Lambda_{\mathrm{m}_{1}}=\frac{\mathrm{k}_{1} \times 1000}{\mathrm{M}_{1}}=\frac{\mathrm{k} \times 1000}{\frac{10}{0.02}}

\Lambda_{\mathrm{m}_{2}}=\frac{\mathrm{k}_{2} \times 1000}{\frac{20}{0.08}}

It is given that

\mathrm{k}_{1}=\mathrm{k}_{2}

\mathrm{k}_{1}=\frac{\Lambda_{\mathrm{m}_{1}}}{2} \quad \quad \mathrm{k}_{2}=\frac{\Lambda_{\mathrm{m}_{2}}}{4}

Applying the given condition on conductivity.

\begin{gathered} \frac{\Lambda_{\mathrm{m}_{1}}}{2}=\frac{\Lambda_{\mathrm{m}_{2}}}{4} \\ \Lambda_{\mathrm{m}_{2}}=2 \Lambda_{\mathrm{m}_{1}} \end{gathered}

Q56

Given below are two statements : Statement I : For KI, molar conductivity increases steeply with dilution Statement II : For carbonic acid, molar conductivity increases slowly with dilution In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is true but Statement II is false

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

Statement I is false : KI is a strong electrolyte which gets completely dissociate on dissolution.

On dilution, the molar conductivity almost remain constant.

Statement II is false : carbonic acid is a weak electrolyte which do not gets completely dissociate on dissolution.

On dilution the molar conductivity of weak electrolyte increases sharply.

$\therefore$ Both Statements are false.

Q57

Which one of the following statements is correct for electrolysis of brine solution?

A

\mathrm{O}_{2}

is formed at cathode

B

\mathrm{H}_{2}

is formed at anode

C

\mathrm{Cl}_{2}

is formed at cathode

D

\mathrm{OH}^{-}

is formed at cathode

Correct Answer

Option D

Solution

Cathode : $\mathrm{H}_2 \mathrm{O}_{(l)}+e^{-} \longrightarrow \dfrac{1}{2} \mathrm{H}_{2(g)}+\mathrm{OH}_{(a q)}^{-}$ Anode $: \mathrm{Cl}_{(a q)}^{-} \longrightarrow \dfrac{1}{2} \mathrm{Cl}_{2(g)}+e^{-}$

Q58

The standard electrode potential

\mathrm{(M^{3+}/M^{2+})}

for V, Cr, Mn & Co are

-

0.26 V,

-

0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate

\mathrm{H_2}

from a dilute acid are :

A

\mathrm{Mn^{2+}}

and

\mathrm{Co^{2+}}

B

\mathrm{V^{2+}}

and

\mathrm{Mn^{2+}}

C

\mathrm{V^{2+}}

and

\mathrm{Cr^{2+}}

D

\mathrm{Cr^{2+}}

and

\mathrm{Co^{2+}}

Correct Answer

Option C

Solution

Metal cation with $(-)$ value of reduction potential $\left(\mathrm{M}^{+3} / \mathrm{M}^{+2}\right)$ or with $(+)$ value of oxidation potential $\left(\mathrm{M}^{+2} / \mathrm{M}^{+3}\right)$ will liberate $\mathrm{H}_{2}$ Therefore they will reduce $\mathrm{H}^{+}$ i. $\mathrm{e~V}^{+2}$ and $\mathrm{Cr}^{+2}$

Q59

The reaction

\dfrac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})

occurs in which of the given galvanic cell.

A

\mathrm{Ag}|\mathrm{AgCl}(\mathrm{s})| \mathrm{KCl}\left(\mathrm{sol}^{\mathrm{n}}\right)\left|\mathrm{AgNO}_{3}\right| \mathrm{Ag}

B

\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{~g})\right| \mathrm{HCl}\left(\mathrm{sol}^{\mathrm{n}}\right)|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}

C

\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{~g})\right| \mathrm{HCl}\left(\mathrm{sol}^{\mathrm{n}}\right)\left|\mathrm{AgNO}_{3}\left(\mathrm{sol}^{\mathrm{n}}\right)\right| \mathrm{Ag}

D

\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{~g})\right| \mathrm{KCl}\left(\mathrm{sol}^{\mathrm{n}}\right)|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}

Correct Answer

Option B

Solution

The provided reaction is:

\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})

This reaction involves the following half-reactions: Oxidation of hydrogen gas to H+ ions:

\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + e^-

Reduction of AgCl to Ag:

\mathrm{AgCl}(\mathrm{s}) + e^- \rightleftharpoons \mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}^{-}(\mathrm{aq})

Looking at the options provided: Option A: Doesn't involve H2 gas, so it can't be correct.

Option B: This includes the necessary elements - H2, AgCl, and Ag.

Option C: Doesn't involve AgCl, so it can't be correct.

Option D: Also includes the necessary elements - H2, AgCl, and Ag.

However, looking closely, we can see that Option B represents the galvanic cell for this reaction.

The reaction requires the oxidation of H2 to H+, which occurs at the anode.

The reaction also requires the reduction of AgCl to Ag and Cl-, which occurs at the cathode.

In Option B, the anode (on the left) is where H2 is being oxidized to H+.

The cathode (on the right) is where AgCl is reduced to Ag and Cl-.

The salt bridge or ion exchange component is HCl, which allows for the flow of ions to balance charge in the cell.

Therefore, the reaction occurs in the galvanic cell represented by Option B.

Q60

The standard electrode potential of

\mathrm{M}^{+} / \mathrm{M}

in aqueous solution does not depend on

A Ionisation of a gaseous metal atom

B Sublimation of a solid metal

C Ionisation of a solid metal atom

D Hydration of a gaseous metal ion

Correct Answer

Option C

Solution

The standard electrode potential (E°) of a metal ion M+/M in aqueous solution is calculated under standard conditions and relates to the tendency of the M+ ion to gain an electron to form the neutral metal atom, M.

This is indeed a redox (reduction-oxidation) process, and the standard electrode potential is defined for the reduction half-reaction.

E° values are based on the energies involved in the processes of ionization and hydration.

However, the standard electrode potential does not depend on the state (solid, liquid, gas) of the atom being ionized.

Ionization of a solid metal atom (Option C) would not directly impact the electrode potential as it is not inherently part of the process of reduction at the electrode that the E° values are measuring.

The correct answer should be Option C : Ionisation of a solid metal atom.