Electrochemistry — Page 9 | JEE Chemistry

Q81

Standard electrode potentials for a few half cells are mentioned below :

\begin{aligned} & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{~V}, \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V} \end{aligned}

Which one of the following cells gives the most negative value of

\Delta \mathrm{G}^{\circ}

?

A

\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(1 \mathrm{M})\right| \mathrm{Ag}

B

\mathrm{Ag}\left|\mathrm{Ag}^{+}(1 \mathrm{M})\right|\left|\mathrm{Mg}^{2+}(1 \mathrm{M})\right| \mathrm{Mg}

C

\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M})\right|\left|\mathrm{Mg}^{2+}(1 \mathrm{M})\right| \mathrm{Mg}

D

\mathrm{Cu}\left|\mathrm{Cu}^{2+}(1 \mathrm{M})\right|\left|\mathrm{Ag}^{+}(1 \mathrm{M})\right| \mathrm{Ag}

Correct Answer

Option A

Solution

\begin{gathered} \because \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ \text{ Option }(1) \mathrm{E}^{\circ}=0.8+0.76 \\ =1.56 \mathrm{~V} \\ \therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 1.56 \\ =-3.12 \mathrm{~V} \\ \text{ Option (2) } \mathrm{E}^{\circ}=-2.37+0.76 \\ =-1.61 \mathrm{~V} \\ \therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times(-1.61) \\ =+3.22 \mathrm{~V} \\ \text{ Option (3) } \mathrm{E}^{\circ}=-2.37-0.8 \\ =-3.17 \mathrm{~V} \\ \begin{array}{c} \therefore \Delta \mathrm{G}^{\circ}=-2 \end{array} \\ =\mathrm{F} \times(-3.17) \\ =+6.34 \\ \text{ Option }(4) \mathrm{E}^{\circ}=0.8-0.34 \\ =0.46 \mathrm{~V} \\ \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 0.46 \\ =-0.92 \mathrm{~V} \end{gathered}

Q82

For the given cell

\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{s})}

The standard cell potential of the above reaction is Given:

\begin{array}{lr} \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^\theta=\mathrm{xV} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{yV} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{zV} \end{array}

A

x+2 y-3 z

B

x+2 y

C

y-2 x

D

x+y-z

Correct Answer

Option A

Solution

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

\begin{aligned} & \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\ & -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2^0 \\ & \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy} \end{aligned}

Also $\Delta \mathrm{G}_2^0=-\mathrm{nFE}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0$

\begin{aligned} & 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0\right) \\ & \mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0=2 \mathrm{y}-3 \mathrm{z} \end{aligned}

$\mathrm{E}_{\text{Cell }}^0$ for reaction will be

\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}^0}^0+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0 \\ & =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z} \end{aligned}

Option (2)

Q83

In the cell Pt

\left| {\left( s \right)} \right|

H2(g, 1 bar)

\left| {HCl\left( {aq} \right)} \right|

AgCl

\left| {\left( s \right)} \right|

Ag(s)|Pt(s) the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :

\left\{ {} \right.

Given,

{{2.303RT} \over F} = 0.06V

at

\left. {298} \right\}

A 0.94 V

B 0.40 V

C 0.76 V

D 0.20 V

Correct Answer

Option D

Solution

Anode : H2(g) $\to$ 2H+(aq) + 2e- Cathode : AgCl(s) + e- $\to$ Ag(s) + Cl-(aq) --------------------------------------------------------------- H2(g) + 2AgCl(s) $\to$ 2Ag(s) + 2H+(aq) + 2Cl-(aq) From Nernst equation we know, Ecell = E0cell -

{{0.06} \over n}\log Q

Here, Ecell = E0cell -

{{0.06} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}{{\left[ {C{l^ - }} \right]}^2}} \over {{P_{{H_2}}}}}

$\Rightarrow$ 0.92 = E0AgCl/AgCl - -

0.03\log {{{{\left[ {{{10}^{ - 6}}} \right]}^2}{{\left[ {{{10}^{ - 6}}} \right]}^2}} \over 1}

$\Rightarrow$ E0AgCl/AgCl = 0.92 - 0.72 = 0.2 V

Q84

Correct order of limiting molar conductivity for cations in water at 298 K is :

A

\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}

B

\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}

C

\mathrm{Mg}^{2+}>\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}

D

\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}

Correct Answer

Option B

Solution

The limiting molar conductivities of ions are important in determining how well they conduct electricity in solution.

Here are the values for some common cations at 298 K: H⁺: 349.8 S cm² mol⁻¹ Na⁺: 50.11 S cm² mol⁻¹ K⁺: 73.52 S cm² mol⁻¹ Ca²⁺: 119 S cm² mol⁻¹ Mg²⁺: 106.12 S cm² mol⁻¹ Based on these values, the correct order of limiting molar conductivity for these cations is: $\stackrel{\oplus}{\text{H}}^+ > \text{Ca}^{2+} > \text{Mg}^{2+} > \text{K}^+ > \text{Na}^+$ This order reflects the decreasing ability of each ion to conduct electricity in water under the given conditions.

Q85

Given below are two statements : 1 M aqueous solutions of each of Cu(NO3)2, AgNO3, Hg2(NO3)2, Mg(NO3)2 are electrolysed using inert electrodes. Given: E0Ag+/Ag = 0.80 V, E0Hg22+/Hg = 0.79 V, E0Cu2+/Cu = 0.24 V and E0Mg2+/Mg = -2.37 V.Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.Statement (II) : Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode.In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are incorrect

B Statement I is incorrect but Statement II is correct

C Statement I is correct but Statement II is incorrect

D Both Statement I and Statement II are correct

Correct Answer

Option C

Solution

Let’s analyze the situation step by step. We have 1 M aqueous solutions of:

\text{AgNO}_3

with

\text{Ag}^+/ \text{Ag} \quad E^0 = +0.80\,V

\text{Hg}_2(\text{NO}_3)_2

with

\text{Hg}_2^{2+}/ \text{Hg} \quad E^0 = +0.79\,V

\text{Cu(NO}_3)_2

with

\text{Cu}^{2+}/ \text{Cu} \quad E^0 = +0.24\,V

\text{Mg(NO}_3)_2

with

\text{Mg}^{2+}/ \text{Mg} \quad E^0 = -2.37\,V

In an electrolytic cell with inert electrodes, reduction occurs at the cathode.

The ion with the highest (most positive) reduction potential will be reduced (or deposited) first.

Therefore, as we increase the applied voltage (making the cathode sufficiently negative), the metals will begin to deposit in the order of their reduction potentials.

The order of reduction (deposition) will be: First, silver (

+0.80\,V

) Next, mercury (

+0.79\,V

) Then, copper (

+0.24\,V

) So, Statement (I) which says “With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu” is correct.

For magnesium, the reduction potential is very negative (

-2.37\,V

).

In aqueous solution, before reaching such a negative potential, water will be reduced.

The cathodic reaction for water typically is:

2\,\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\,\text{OH}^-

which leads to the evolution of hydrogen gas, not oxygen.

(Oxygen is produced at the anode during the oxidation of water.)

Hence, Statement (II) erroneously states that “oxygen gas will be evolved at the cathode” when in fact, if magnesium were to be reduced (which it won’t be), water reduction would yield hydrogen gas.

Thus, Statement (II) is incorrect.

Summary: Statement (I) is correct.

Statement (II) is incorrect.

Therefore, the most appropriate answer is: Option C: Statement I is correct but Statement II is incorrect.

Q86

On charging the lead storage battery, the oxidation state of lead changes from

x_1

to

y_1

at the anode and from

x_2

to

y_2

at the cathode. The values of

x_1, y_1, x_2, y_2

are respectively :

A

0,+2,+4,+2

B

+2,0,0,+4

C

+4,+2,0,+2

D

+2,0,+2,+4

Correct Answer

Option D

Solution

For charging of lead storage battery cell reaction is

2 \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Pb}(\mathrm{~s})+\mathrm{PbO}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})

At anode $\mathrm{PbSO}_4$ reduced back to Pb and at cathode $\mathrm{PbSO}_4$ oxidised back to $\mathrm{PbO}_2$ .

\begin{aligned} \because ~& \mathrm{x}_1=+2, \mathrm{y}_1=0 \\ \mathrm{x}_2 & =+2, \mathrm{y}_2=4 \end{aligned}

Q87

The standard cell potential

\left(\mathrm{E}_{\text{cell }}^{\ominus}\right)

of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V . The standard half cell reduction potential for

\mathrm{O}_2\left(\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}\right)

is 1.229 V . Choose the correct statement :

A Reactants are fed at one go to each electrode.

B Oxygen is formed at the anode.

C The standard half cell reduction potential for the reduction of

\mathrm{CO}_2\left(\mathrm{E}_{\mathrm{CO}_2 / \mathrm{CH}_3 \mathrm{OH}}^{\circ}\right)

is 19 mV

D Reduction of methanol takes place at the cathode.

Correct Answer

Option C

Solution

To determine the reaction potentials at the anode and cathode in the methanol fuel cell, we start with the given standard cell potential: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$ Given: $E_{\text{cell}}^{\circ} = 1.21 \, \text{V}$ $E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} = 1.229 \, \text{V}$ Substituting the values, $1.21 = 1.229 - E_{\text{anode}}^{\circ}$ Solving for $E_{\text{anode}}^{\circ}$ : $E_{\text{anode}}^{\circ} = 1.229 - 1.21 = 0.019 \, \text{V}$ Since this is a fuel cell based on the oxidation of methanol, methanol undergoes oxidation at the anode, while oxygen is reduced at the cathode.

Therefore: Anode Reaction: Oxidation of methanol occurs here, contributing to the fuel cell’s electrical output.

Cathode Reaction: Reduction of oxygen occurs at the cathode, where the $E_{\text{O}_2/\text{H}_2\text{O}}^{\circ}$ potential is utilized.

Thus, we verify that the half-cell reduction potential for oxidation of methanol by CO $_2$ is indeed 19 mV, aligning with the calculated $E_{\text{anode}}^{\circ}$ .

The processes correctly identify the roles of anode and cathode in this fuel cell.

Q88

Consider the following standard electrode potentials (Eo in volts) in aqueous solution : .tg .tg Element M3+ /M M+ /M A1 -1.66 + 0.55 T1 +1.26 - 0.34 Based on these data, which of the following statements is correct ?

A T1+ is more stable than A13+

B A1+ is more stable than A13+

C T1 + is more stable than A1+

D T13+ is more stable than A13+

Correct Answer

Option C

Solution

The standard electrode potential E0 for M+/M become negative for Tl which indicates that Tl+ is more stable than Al+.

This can also be be explained by inert pair effect.

The atoms of this group have an outer electronic configuration of s2p1.

The two s electrons of Tl do not participate in bonding due to poor shielding of d electrons.

Q89

Given that the standard potentials (Eo) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the Eo of Cu2+/Cu+ :

A - 0.182 V

B - 0.158 V

C 0.182 V

D +0.158 V

Correct Answer

Option D

Solution

Cu+2 + 2e– $\to$ Cu Eo = 0.34 V ....(

1) Cu+ + e– $\to$ Cu Eo = 0.522 V Cu $\to$ Cu+ + e– Eo = -0.522 V ...(

2) By adding (1) and (2) we get, Cu+2 + e– $\to$ Cu+ Eocell =

{{0.34 \times 2 + 1 \times \left( { - 0.522} \right)} \over 1}

= 0.158 V

Q90

Consider the following Eo values

E_{F{e^{3 + }}/F{e^{2 + }}}^o

= 0.77 V;

E_{S{n^{2 + }}/S{n}}^o

= -0.14 V Under standard conditions the potential for the reaction Sn(s) + 2Fe3+(aq)

\to

2Fe2+(aq) + Sn2+(aq) is :

A 1.68 V

B 0.63 V

C 0.91 V

D 1.40 V

Correct Answer

Option C

Solution

F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\Delta {G^ \circ }

\,\,\,\,\,\,\,\,\,\, = - 1 \times F \times 0.77

S{n^{2 + }} + 2{e^ - } \to Sn\left( s \right)\Delta {G^ \circ }

= - 2 \times F\left( { - 0.14} \right)

Sn\left( s \right) + 2F{e^{3 + }}\left( {aq} \right)

\,\,\,\,\,\,\,\,\,\, \to 2F{e^{2 + }}\left( {aq} \right) + S{n^{2 + }}\left( {aq} \right)

$\therefore$ Standard potential for the given reaction or

E_{cell}^o = E_{Sn/S{n^{2 + }}}^o + E_{F{e^{3 + }}/F{e^{2 + }}}^0

= 0.14 + 0.77 = 0.91\,V