Electrochemistry — Page 7 | NEET Chemistry

Q61

For the disproportionation of copper 2Cu +

\to

Cu 2+ + Cu, E o is (Given E o for Cu 2+ /Cu is 0.34 V and E o for Cu 2+ /Cu + is 0.15 V.)

A 0.49 V

B

-

0.19 V

C 0.38 V

D

-

0.38 V

Correct Answer

Option C

Solution

Cu 2+ + 2e – $\to$ Cu; E° 1 = 0.34 V .....(

1) Cu 2+ + e – $\to$ Cu + ; E° 2 = 0.15 V.....(

2) Cu + + e – $\to$ Cu; E° 3 = ? .....(

3) $\therefore$

\Delta

G o 1 = -2 $\times$ 0.34 $\times$ F and

\Delta

G o 2 = -1 $\times$ 0.15 $\times$ F and

\Delta

G o 3 = -1 $\times$ E° 3 $\times$ F Also,

\Delta

G o 1 =

\Delta

G o 2 +

\Delta

G o 3 $\Rightarrow$ -0.68F = -0.15F - E° 3 $\times$ F $\Rightarrow$ E° 3 = 0 68 - 0 15 = 0 53 V $\therefore$ E° cell = 0.53 - 0.15 = 0.38 V

Q62

Equivalent conductances of Ba 2+ and Cl

-

ions are 127 and 76 ohm

-

1 cm

-

1 eq

-

1 respectively. Equivalent conductance of BaCl 2 at infinite dilution is

A 139.5

B 101.5

C 203

D 279

Correct Answer

Option A

Solution

According to Kohlrausch’s law, the equivalent conductance of BaCl 2 at infinite dilution, $\lambda$ $\infty$ of BaCl 2 =

{1 \over 2}

$\lambda$ $\infty$ of Ba 2+ + $\lambda$ $\infty$ of Cl - $\Rightarrow$ $\lambda$ $\infty$ of BaCl 2 =

{1 \over 2} \times 127 + 76

= 139 5

Q63

For the redox reaction Zn(s) + Cu2+(0.1 M)

\to

Zn2+(1M) + Cu(s) taking place in a cell,

E_{cell}^o

is 1.10 volt. Ecell for the cell will be (

2.303{{RT} \over F}

= 0.0591)

A 1.80 volt

B 1.07 volt

C 0.82 volt

D 2.14 volt

Correct Answer

Option B

Solution

{E_{cell}} = {E^ \circ }_{cell} + {{0.059} \over n}\log {{\left[ {C{u^{ + 2}}} \right]} \over {\left[ {Z{n^{ + 2}}} \right]}}

= 1.10 + {{0.059} \over 2}\log \left[ {0.1} \right]

= 1.10 - 0.0295

= 1.07V

Q64

The Gibbs energy for the decomposition of Al2O3 at 500oC is as follows :

{2 \over 3}A{l_2}{O_3}

\to

{4 \over 3}Al + {O_2}

,

{\Delta _r}G

= + 966 kJ mol–1 The potential difference needed for electrolytic reduction of Al2O3 at 500oC is at least :

A 4.5 V

B 3.0 V

C 2.5 V

D 5.0 V

Correct Answer

Option C

Solution

\Delta G = - nFE

or

E = {{\Delta G} \over { - nF}} = {{966 \times {{10}^3}} \over {4 \times 36500}}

= - 2.5\,\,V

$\therefore$ The potential difference needed for the reduction

=2.5

V

Q65

The emf of cell

\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}

is

0.83 \mathrm{~V}

at

298 \mathrm{~K}

. It could be increased by :

A increasing concentration of

\mathrm{Tl}^{+}

ions

B increasing concentration of

\mathrm{Cu}^{2+}

ions

C increasing concentration of both

\mathrm{Tl}^{+}

and

\mathrm{Cu}^{2+}

ions

D decreasing concentration of both

\mathrm{Tl}^{+}

and

\mathrm{Cu}^{2+}

ions

Correct Answer

Option B

Solution

To determine how the emf of the cell,

\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}

, can be increased, we can use the Nernst equation. The Nernst equation for this electrochemical cell is given by:

E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log \left( \frac{[\mathrm{Tl}^{+}]}{[\mathrm{Cu}^{2+}]} \right)

where:

E_{\text{cell}}

is the emf of the cell.

E_{\text{cell}}^\circ

is the standard emf of the cell.

n

is the number of moles of electrons transferred in the cell reaction; here,

n = 2

.

[\mathrm{Tl}^{+}]

is the concentration of thallium ions.

[\mathrm{Cu}^{2+}]

is the concentration of copper ions.

Given that the emf of the cell can be expressed in terms of the concentration of the ions involved, we can see that increasing the concentration of

[\mathrm{Cu}^{2+}]

or decreasing the concentration of

[\mathrm{Tl}^{+}]

will affect the logarithmic term in the Nernst equation:

E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{2} \log \left( \frac{0.001}{0.01} \right)

To increase the emf (

E_{\text{cell}}

) of the cell, it is beneficial to have a less negative (or more positive) correction term.

This can be achieved by: Increasing the concentration of

[\mathrm{Cu}^{2+}]

ions: This makes the term inside the logarithm smaller (since we are dividing by a larger number), which in turn makes the logarithmic term less negative.

Therefore, the correct answer is: Option B: increasing concentration of

\mathrm{Cu}^{2+}

ions

Q66

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

Q67

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.

Q68

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

Q69

For lead storage battery pick the correct statements A. During charging of battery,

\mathrm{PbSO}_{4}

on anode is converted into

\mathrm{PbO}_{2}

B. During charging of battery,

\mathrm{PbSO}_{4}

on cathode is converted into

\mathrm{PbO}_{2}

C. Lead storage battery consists of grid of lead packed with

\mathrm{PbO}_{2}

as anode D. Lead storage battery has

\sim 38 \%

solution of sulphuric acid as an electrolyte Choose the correct answer from the options given below:

A A, B, D only

B B, C only

C B, C, D only

D B, D only

Correct Answer

Option D

Solution

Let's review each statement: A.

During charging of battery, PbSO₄ on anode is converted into PbO₂.

This statement is incorrect.

During charging, PbSO₄ on the anode (lead plate) is converted to Pb, not PbO₂.

B.

During charging of battery, PbSO₄ on cathode is converted into PbO₂.

This statement is correct.

During charging, PbSO₄ on the cathode (lead dioxide plate) is converted back to PbO₂.

C.

Lead storage battery consists of grid of lead packed with PbO₂ as anode.

This statement is incorrect.

The anode in a lead-acid battery is made of lead (Pb), not lead dioxide (PbO₂).

The cathode is made of lead dioxide (PbO₂).

D.

Lead storage battery has ∼38% solution of sulphuric acid as an electrolyte.

This statement is correct.

The electrolyte in a lead-acid battery is a solution of about 35-38% sulfuric acid (H₂SO₄) in water.

Therefore, the correct answer is: B, D only.

Q70

Given below are two statements : Statement (I) : Fusion of

\mathrm{MnO}_2

with

\mathrm{KOH}

and an oxidising agent gives dark green

\mathrm{K}_2 \mathrm{MnO}_4

. Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is false but Statement II is true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

Statement I is indeed true. When

\mathrm{MnO}_2

(manganese dioxide) is fused with

\mathrm{KOH}

(potassium hydroxide) and an oxidising agent such as

\mathrm{KNO}_3

(potassium nitrate), it forms dark green potassium manganate (

\mathrm{K}_2\mathrm{MnO}_4

). The reaction can be represented as follows:

\mathrm{2MnO}_2 + 4KOH + \mathrm{O}_2 \rightarrow \mathrm{2K}_2\mathrm{MnO}_4 + 2H_2\mathrm{O}

This process involves the oxidation of manganese dioxide to manganate ion (

\mathrm{MnO}_4^{2-}

) in an alkaline medium. Statement II is also true. The manganate ion (

\mathrm{MnO}_4^{2-}

), when subjected to electrolytic oxidation in an alkaline medium, can indeed be converted into permanganate ion (

\mathrm{MnO}_4^-

), which is characterized by a deep purple color.

This conversion is an example of an electron-loss (oxidation) process at the anode of an electrolytic cell.

The reaction can be represented as follows:

\mathrm{MnO}_4^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_4^- + 2\mathrm{OH}^- + 2e^-

Thus, both statements I and II are accurate, making the correct answer: Option D: Both Statement I and Statement II are true.