Electrochemistry — Page 9 | NEET Chemistry

Q81

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.

Q82

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

Q83

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}

Q84

When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are :

A cathode = pure zinc, anode = pure copper

B cathode = impure sample, anode = pure copper

C cathode = impure zinc, anode = impure sample

D cathode = pure copper, anode = impure sample

Correct Answer

Option D

Solution

Pure metal always deposits at cathode.

Q85

The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.

A

E^o_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +1.15 \text{ V}

B

E^o_{\text{Al}^{3+}/\text{Al}} = -1.66 \text{ V}

C

E^o_{\text{Pb}^{4+}/\text{Pb}^{2+}} = +1.67 \text{ V}

D

E^o_{\text{Tl}^{3+}/\text{Tl}} = +1.26 \text{ V}

Correct Answer

Option C

Solution

Oxidizing capacity is an element's tendency to donate electrons, or the atmosphere's ability to oxidize compounds.

A higher standard reduction potential indicates a stronger oxidizing capacity.

A substance with a higher standard reduction potential is move likely to gain electrons and act as an oxidizing agent in a redon reaction.

The more positive the reduction potential, the stronger the oxidizing power of a species.

From the given options, the higher standard reduction potential is for lead $(\mathrm{Pb}) .(+1.67)$ . so, the p-block ion with strongest oxidizing capacity is Pb (lead).

Answer: Option 3) $E_{p_b{ }^{4+} / \mathrm{Pb}^{2+}}^0=+1.67 \mathrm{~V}$

Q86

\mathrm{O}_2

gas will be evolved as a product of electrolysis of : (A) an aqueous solution of

\mathrm{AgNO}_3

using silver electrodes. (B) an aqueous solution of

\mathrm{AgNO}_3

using platinum electrodes. (C) a dilute solution of

\mathrm{H}_2 \mathrm{SO}_4

using platinum electrodes. (D) a high concentration solution of

\mathrm{H}_2 \mathrm{SO}_4

using platinum electrodes. Choose the correct answer from the options given below :

A (B) and (C) only

B (B) and (D) only

C (A) and (D) only

D

(A)

and

(C)

only

Correct Answer

Option A

Solution

(A) An aqueous solution of AgNO $_3$ using silver electrodes. Cathode - Reduction -

Ag_{(aq)}^ + + {e^ - } \to Ag(s)

Anode - Oxidation -

Ag(s) \to Ag_{(aq)}^ + + {e^ - }

Solid silver will be deposited at the cathode.

Solid anode (silver) will dissolve, releasing silver ions into the solution.

So, there is no formation of O $_2$ gas in this electrolysis.

(B) An aqueous solution of AgNO $_3$ using platinum electrodes.

Cathode - Reduction -

Ag_{(aq)}^ + + {e^ - } \to Ag(s)

Anode - Oxidation -

2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }

When platinum electrodes are used, Ag $^+$ from solution is reduced and deposited at cathode whereas O $_2$ is produced at the anode.

(C) A dilute soution of H $_2$ SO $_4$ using platinum electrodes.

Cathode - Reduction -

2{H^ + } + 2{e^ - } \to {H_2}

Anode - Oxidation -

2{H_2}O \to {O_2} + 4{H^ + } + 4{e^ - }

H $_2$ gas is produded at the cathode and O $_2$ gas is produced at the anode.

(D) a high concentration solution of H $_2$ SO $_4$ using platinum electrodes.

O $_2$ gas is not formed in this case.

Cathode - Reduction - The substance formed is H $_2$ gas.

Anode - Oxidation - The substance formed is not O $_2$ gas.

So, statements (B) and (C) are correct.

Q87

EMF of a cell in terms of reduction potential of its left and right electrodes is :

A E = Eleft - Eright

B E = Eleft + Eright

C E = Eright - Eleft

D E = -(Eright + Eleft)

Correct Answer

Option C

Solution

{E_{cell}} = \,\,

Reduction potential of cathode (right)

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

$-$ Reduction potential of anode (left)

= {E_{right}} - {E_{left}}.

Q88

The standard electrode potential

{E^o }

and its temperature coefficient

\left( {{{d{E^o }} \over {dT}}} \right)

for a cell are 2V and

-

5

\times

10

-

4 VK

-

1 at 300 K respectively. The cell reaction is Zn(s) + Cu2+ (aq)

\overset{\,}\longrightarrow

Zn2+ (aq) + Cu(s) The standard reaction enthalpy (

\Delta

rH

{^o }

) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]

A

-

412.8

B

-

384.0

C 192.0

D 206.4

Correct Answer

Option A

Solution

\Delta

G = -nFEcell = -2 $\times$ 96500 $\times$ 2 = -386 kJ

\Delta

S = nF

\left( {{{d{E^o }} \over {dT}}} \right)

= 2 $\times$ 96500 $\times$ ( $-$ 5 $\times$ 10 $-$ 4) = -96.5 kJ At 298 K T

\Delta

S = 298 $\times$ (–96.5 J) = – 28.8 kJ at constant T (=248 K) and pressure

\Delta

G =

\Delta

H – T

\Delta

S $\Rightarrow$

\Delta

H =

\Delta

G + T

\Delta

S = -386 - 28.8 = -412.8 kJ

Q89

The reaction at cathode in the cells commonly used in clocks involves.

A oxidation of

\mathrm{Mn}

from +2 to +7

B reduction of

\mathrm{Mn}

from +4 to +3

C oxidation of

\mathrm{Mn}

from +3 to +4

D reduction of

\mathrm{Mn}

from +7 to +2

Correct Answer

Option B

Solution

In the cells commonly used in clocks, specifically alkaline batteries, the reaction at the cathode involves the reduction of manganese dioxide (MnO2).

The manganese in MnO2 is initially in the +4 oxidation state, and it gets reduced to the +3 oxidation state.

Thus, the correct reaction occurring at the cathode can be represented as follows: The correct answer is: Option B reduction of

\mathrm{Mn}

from +4 to +3

Q90

For the electro chemical cell

\mathrm{M}\left|\mathrm{M}^{2+}\right||\mathrm{X}| \mathrm{X}^{2-}

If

\mathrm{E}_{\left(\mathrm{M}^{2+} / \mathrm{M}\right)}^0=0.46 \mathrm{~V}

and

\mathrm{E}_{\left(\mathrm{x} / \mathrm{x}^{2-}\right)}^0=0.34 \mathrm{~V}

. Which of the following is correct?

A

\mathrm{E}_{\mathrm{cell}}=0.80 \mathrm{~V}

B

\mathrm{M}+\mathrm{X} \rightarrow \mathrm{M}^{2+}+\mathrm{X}^{2-}

is a spontaneous reaction

C

\mathrm{E}_{\text{cell }}=-0.80 \mathrm{~V}

D

\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}

is a spontaneous reaction

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}_{\mathrm{X} \mid \mathrm{X}^{2-}}^{\circ}-\mathrm{E}_{\mathrm{M}^{2+} \mid \mathrm{M}}^{\circ} \\ & =0.34-0.46 \\ & =-0.12 \mathrm{~V} \text{ (Non-spontaneous) } \end{aligned}

So, reverse reaction will be spontaneous.

\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}