Electrochemistry — Page 8 | JEE Chemistry

Q71

The reaction;

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

occurs in which of the following galvanic cell :

A

\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text{soln.) }}\left|\mathrm{AgNO}_{3(\mathrm{aq})}\right| \mathrm{Ag}

B

\mathrm{Ag}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{KCl}_{\text{(soln.) }}\left|\mathrm{AgNO}_{3 \text{ (aq.) }}\right| \mathrm{Ag}

C

\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{KCl}_{(\text{soln.) }}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}

D

\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text{soln. })}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}

Correct Answer

Option D

Solution

To determine which galvanic cell corresponds to the given reaction:

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

we need to examine the setups of the given cells and check if they produce the same reactants and products as in the above reaction.

Analyzing the reaction, we observe: Hydrogen gas (

\mathrm{H}_{2(\mathrm{~g})}

) is involved at one electrode and splits into

\mathrm{H}_{(\mathrm{aq})}^{+}

ions. Solid silver chloride (

\mathrm{AgCl}_{(\mathrm{s})}

) decomposes at another electrode into

\mathrm{Ag}_{(\mathrm{s})}

and

\mathrm{Cl}_{(\mathrm{aq})}^{-}

ions. Let's examine each option: Option A:

\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text{soln.) }}\left|\mathrm{AgNO}_{3(\mathrm{aq})}\right| \mathrm{Ag}

This cell setup does not directly involve

\mathrm{AgCl}_{(\mathrm{s})}

solid separating into

\mathrm{Ag}_{(\mathrm{s})}

and

\mathrm{Cl}_{(\mathrm{aq})}^{-}

. Hence, it does not match the reaction given. Option B:

\mathrm{Ag}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{KCl}_{(\text{(soln.) }}\left|\mathrm{AgNO}_{3 \text{ (aq.) }}\right| \mathrm{Ag}

This involves silver electrodes and does not have a setup for reaction with hydrogen gas or the splitting of

\mathrm{H}_{2(\mathrm{~g})}

into

\mathrm{H}_{(\mathrm{aq})}^{+}

ions. So, this does not match either. Option C:

\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{KCl}_{(\text{soln.) }}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}

While this configuration includes

\mathrm{AgCl}_{(\mathrm{s})}

and

\mathrm{H}_{2(\mathrm{~g})}

, it uses potassium chloride solution instead of hydrochloric acid solution, which alters the chemistry and does not perfectly match the required reaction mechanism.

Option D:

\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text{soln. })}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}

This configuration fits the reaction given. Here hydrogen gas at the platinum electrode splits into

\mathrm{H}_{(\mathrm{aq})}^{+}

ions in the hydrochloric acid solution. On the other side, the solid silver chloride separates into

\mathrm{Ag}_{(\mathrm{s})}

and

\mathrm{Cl}_{(\mathrm{aq})}^{-}

ions. Hence, this accurately matches the reaction. Therefore, the correct answer is: Option D

Q72

Molar ionic conductivities of divalent cation and anion are

57 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}

and

73 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}

respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:

A

187 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

B

260 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

C

65 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

D

130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Correct Answer

Option D

Solution

The compound with divalent cation

\left(\mathrm{A}^{2+}\right)

and anion

\left(B^{2-}\right)

will be

A B

. Molar conductivity of its solution will be

57+73=130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Q73

The quantity of silver deposited when one coulomb charge is passed through

\mathrm{AgNO}_3

solution :

A

0.1 \mathrm{~g}

atom of silver

B 1 chemical equivalent of silver

C

1 \mathrm{~g}

of silver

D 1 electrochemical equivalent of silver

Correct Answer

Option D

Solution

To determine the quantity of silver deposited when one coulomb of charge is passed through

\mathrm{AgNO}_3

solution, we need to refer to Faraday's laws of electrolysis, especially to the concept of the electrochemical equivalent.

The electrochemical equivalent (ECE) of a substance is the amount of the substance that is deposited or dissolved at an electrode during electrolysis by one coulomb of charge.

The electrochemical equivalent of a substance can be calculated using the formula:

E = \frac{M}{nF}

where: $E$ is the electrochemical equivalent of the substance in grams per coulomb ( $\mathrm{g/C}$ ). $M$ is the molar mass of the substance in grams per mole ( $\mathrm{g/mol}$ ). $n$ is the number of electrons involved in the oxidation or reduction reaction (the valency). $F$ is the Faraday constant, approximately $96485 \, \mathrm{C/mol}$ , which is the charge of one mole of electrons.

In the case of silver ( $\mathrm{Ag}$ ) being deposited from $\mathrm{AgNO}_3$ , when silver ion ( $\mathrm{Ag}^+$ ) is reduced to metallic silver ( $\mathrm{Ag}$ ), the reaction involves one electron ( $n = 1$ ).

The molar mass of silver (Ag) is approximately $107.868 \, \mathrm{g/mol}$ .

Using this information:

E = \frac{107.868 \, \mathrm{g/mol}}{1 \cdot 96485 \, \mathrm{C/mol}} \approx 0.001118 \, \mathrm{g/C}

This means that for every coulomb of charge passed through the solution, $0.001118 \, \mathrm{g}$ of silver is deposited.

Given the options: Option A: $0.1$ g atom of silver - This is incorrect because the calculation shows a much smaller amount is deposited per coulomb.

Option B: 1 chemical equivalent of silver - This does not directly relate to the amount deposited per coulomb.

Option C: $1 \mathrm{~g}$ of silver - This is incorrect as it overestimates the amount deposited by a large margin.

Option D: 1 electrochemical equivalent of silver - This is correct, as it directly relates to the amount of a substance deposited by one coulomb of charge, based on the formula and calculation shown.

Therefore, the correct answer is Option D: 1 electrochemical equivalent of silver.

Q74

How can an electrochemical cell be converted into an electrolytic cell ?

A Applying an external opposite potential greater than

\mathrm{E}_{\text{cell }}^0

B Exchanging the electrodes at anode and cathode.

C Applying an external opposite potential lower than

\mathrm{E}^0{ }_{\text{cell }}

D Reversing the flow of ions in salt bridge.

Correct Answer

Option A

Solution

An electrochemical cell can be converted into an electrolytic cell by applying an external potential greater than

E_{\text{cell }}^{\circ}

.

The entire cell reaction will be reversed.

The oxidised species will be reduced and the reduced species will be oxidised.

Q75

A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is __________ . [Given : molar mass of aluminium and chlorine are

27 \mathrm{~g} \mathrm{~mol}^{-1}

and

35.5 \mathrm{~g} \mathrm{~mol}^{-1}

respectively. Faraday constant

\left.=96500 \mathrm{C} \mathrm{~mol}^{-1}\right]

A 1.660 g

B 1.007 g

C 0.336 g

D 0.441 g

Correct Answer

Option C

Solution

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

The mass of aluminium deposited can be calculated using Faraday’s laws of electrolysis.

The steps are as follows: Compute the total charge passed:

Q = I \cdot t = 2\,\text{A} \cdot (30 \times 60\,\text{s}) = 3600\,\text{C}

Determine the moles of aluminium deposited.

Since three electrons are required to deposit one mole of aluminium, the number of moles is given by:

n(\text{Al}) = \frac{Q}{3F} = \frac{3600}{3 \times 96500} \approx 0.01242\,\text{mol}

Finally, calculate the mass of aluminium using its molar mass ( $M = 27\,\text{g/mol}$ ):

m(\text{Al}) = n(\text{Al}) \times M = 0.01242 \times 27 \approx 0.335\,\text{g}

Thus, the amount of aluminium deposited at the cathode is approximately

0.336\, \text{g}

Q76

Match with : (Applications) (Batteries/Cell)

List - I		List - II
(A)	Transistors	(I)	Anode - Zn/Hg; Cathode - HgO + C
(B)	Hearing aids	(II)	Hydrogen fuel cell
(C)	Inverters	(III)	Anode - Zn; Cathode - Carbon
(D)	Apollo space ship	(IV)	Anode - Pb; Cathode - Pb \| PbO2

A (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

B (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

C (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

D (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Correct Answer

Option A

Solution

For each application listed in List-I, match it with the appropriate battery or cell type from List-II based on their characteristics: (A) Transistors: Generally use a dry cell or Leclanché cell, where the anode is zinc (Zn) and the cathode is a carbon rod surrounded by manganese dioxide (MnO₂).

Therefore, (A) Transistors match with (III) Anode - Zn; Cathode - Carbon.

(B) Hearing aids: Operate using a mercury cell.

The anode in this cell is a zinc-mercury amalgam, and the cathode is composed of mercury oxide (HgO) with carbon.

Thus, (B) Hearing aids match with (I) Anode - Zn/Hg; Cathode - HgO + C.

(C) Inverters: Use a lead-acid storage cell.

In this type of cell, the anode consists of lead (Pb), and the cathode is a lead dioxide (PbO₂) grid.

As a result, (C) Inverters match with (IV) Anode - Pb; Cathode - Pb | PbO₂.

(D) Apollo space ship: Utilizes a hydrogen-oxygen fuel cell.

The hydrogen is oxidized at the anode, while oxygen is reduced at the cathode.

Therefore, (D) Apollo space ship matches with (II) Hydrogen fuel cell.

Summary of matches: A - III B - I C - IV D - II This corresponds with Answer Option A.

Q77

For a Mg | Mg2+ (aq) || Ag+ (aq) | Ag the correct Nernst Equation is :

A

\mathrm{E}_{\text{cell }}=\mathrm{E}_{\text{cell }}^{\mathrm{o}}-\dfrac{\mathrm{RT}}{2 \mathrm{~F}} \ln \dfrac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}

B

\mathrm{E}_{\text{cell }}=\mathrm{E}_{\text{cell }}^{\mathrm{o}}-\dfrac{\mathrm{RT}}{2 \mathrm{~F}} \ln \dfrac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]}

C

\mathrm{E}_{\text{cell }}=\mathrm{E}_{\text{cell }}^{\mathrm{o}}-\dfrac{\mathrm{RT}}{2 \mathrm{~F}} \ln \dfrac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Mg}^{2+}\right]}

D

\mathrm{E}_{\text{cell }}=\mathrm{E}_{\text{cell }}^{\mathrm{o}}+\dfrac{\mathrm{RT}}{2 \mathrm{~F}} \ln \dfrac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}

Correct Answer

Option D

Solution

\mathop {Mg\left| {M{g^{2 + }}(aq)} \right|}\limits_{(Anode)} \mathop {\left| {A{g^ + }(aq)} \right|Ag}\limits_{(Cathode)}

Nernst equation (general)

E_{Cell}=E_{Cell}^0-\frac{R T}{n F} \ln Q

For the cell, the chemical equation can be written as

\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\text{aq })}^{+} \longrightarrow \mathrm{Mg}_{(\text{aq })}^{2+}+2 \mathrm{Ag}(\mathrm{~s})

$n$ (number of electrons transferred $)=2$

{E_{cell}} = E_{cell}^o - {{RT} \over {2F}}\ln {{\left[ {M{g^{2 + }}} \right]{{\left[ {Ag} \right]}^2}} \over {\left[ {Mg} \right]{{\left[ {A{g^ + }} \right]}^2}}}

{E_{cell}} = E_{cell}^o - {{RT} \over {2F}}\ln {{\left[ {M{g^{2 + }}} \right]} \over {{{\left[ {A{g^ + }} \right]}^2}}}

{E_{cell}} = E_{cell}^o + {{RT} \over {2F}}\ln {{{{\left[ {A{g^ + }} \right]}^2}} \over {\left[ {M{g^{2 + }}} \right]}}

Concentration of solid species

[\mathrm{Mg}]=[\mathrm{Ag}]=1

Correct answer : Option D)

{E_{cell}} = E_{cell}^o + {{RT} \over {2F}}\ln {{{{\left[ {A{g^ + }} \right]}^2}} \over {\left[ {M{g^{2 + }}} \right]}}

Q78

The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?

A Molar conductivity increases sharply with increase in concentration.

B Molar conductivity decreases sharply with increase in concentration.

C A small increase in molar conductivity is observed at infinite dilution.

D A small decrease in molar conductivity is observed at infinite dilution.

Correct Answer

Option B

Solution

The molar conductivity of a weak electrolyte is plotted against the square root of its concentration: $c$ -concentration of weak electrolyte Weak electrolytes give curved decreasing graph.

1) Molar conductivity increases sharply with increase in concentration.

Incorrect.

With increase in concentration, molar conductivity decreases and it is lean from the graph of molar conductivity vs $\sqrt{c}$ .

2) Molar conductivity decreases sharply with increase in concentration.

Correct From the graph it is clean that, molar conductivity decreases sharply with increase in concentration.

3) A small increase in molar conductivity is observed at infinite dilution.

Incorrect There is no small increase in molas conductivity with increase in concentration for a weak electrolyte.

And there is no small increase in mola conductivity.

Significant increase occurs.

4) A small decrease in molar conductivity is observed at infinite dilution.

Incorrect For a weak electrolyte, the molar conductivity significantly increases with dilution, especially at infinite dilution.

Correct answer: Option 2) Molen conductivity decreases sharply with increase in concentration.

Q79

Given below are two statements : Statement (I) : Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode. Statement (II) : The rate of corrosion is more in alkaline medium than in acidic medium. 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 C

Solution

Let’s examine both statements one by one and see whether they hold true for typical corrosion processes (e.g., iron rusting): Statement (I) "Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode."

Corrosion is electrochemical True.

Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and reduction at another region (the cathode) on the metal’s surface.

Pure metal as anode, impurity as cathode In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) — that is, it serves as the anode.

The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode region where reduction (e.g., oxygen reduction) occurs.

This difference in electrode potentials between the pure region and the impurity region drives the corrosion cell.

Hence, Statement (I) is generally considered true in the usual context of corrosion (like rusting of iron).

Statement (II) "The rate of corrosion is more in alkaline medium than in acidic medium."

For most common metals (like iron), acidic media typically enhance corrosion because abundant H⁺ ions (and possibly other acidic species) can accelerate the oxidation/dissolution of the metal.

In mildly alkaline or neutral environments, metals often form protective oxide or hydroxide layers that can slow down further corrosion.

While certain strong bases can attack specific metals (e.g., Al in strong NaOH), in general for iron and many other metals, corrosion is more severe in acidic environments than in alkaline ones.

Thus, Statement (II) is false under normal corrosion scenarios (e.g., rusting of iron).

Conclusion Statement (I): True Statement (II): False Therefore, the correct choice (matching these truth values) is: (C) Statement I is true but Statement II is false.

Q80

\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0

In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of

\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}

is :

A 1.2 V

B 2.1 V

C 1.4 V

D 1.7 V

Correct Answer

Option D

Solution

\begin{aligned} & \Delta \mathrm{G}_4^{\mathrm{o}}=\Delta \mathrm{G}_1^{\mathrm{o}}+\Delta \mathrm{G}_2^{\mathrm{o}} \\ \Rightarrow & -\mathrm{n}_4 \mathrm{FE}_4^{\mathrm{o}}=-\mathrm{n}_1 \mathrm{FE}_1^0-\mathrm{n}_2 \mathrm{FE}_2^{\mathrm{o}} \\ \Rightarrow & +4 \mathrm{E}_4^{\mathrm{o}}=3 \times 2+(1 \times 0.8) \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=\frac{6.8}{4} \mathrm{~V} \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=1.7 \mathrm{~V} \end{aligned}