Electrochemistry — Page 6 | NEET Chemistry

Q51

E o Fe 2+ /Fe =

-

0.441 V and E o Fe 3+ /Fe 2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe 3+

\to

3Fe 2+ will be

A 0.111 V

B 0.330 V

C 1.653 V

D 1.212 V

Correct Answer

Option D

Solution

At anode : Fe $\to$ Fe 2+ + 2e - ; E o = -0.441 V At cathode : Fe 3+ + e – $\to$ Fe 2+ ; E o = 0.771 V Fe + 2Fe 3+ $\to$ 3Fe 2+ ; E o = ?

To get the above equation, (ii) × 2 – (i) 2Fe 3+ + 2e – $\to$ 2Fe 2+ ; E o = 0.771 V Fe $\to$ Fe 2+ + 2e - ; E o = - 0.441 V ------------------------------------------------ Fe + 2Fe 3+ $\to$ 3Fe 2+ E o = 0.771 + 0.441 = 1.212 V

Q52

A hypothetical electrochemical cell is shown below.

A\left| {{A^ + }\left( {xM} \right)} \right|\left| {{B^ + }\left( {yM} \right)} \right|B

The emf measured is + 0.20 V. The cell reaction is

A A + B +

\to

A + + B

B A + + B

\to

A + B +

C A + + e

-

\to

A; B + + e

-

\to

B

D the cell reaction cannot be predicted.

Correct Answer

Option A

Solution

From the given expression: At anode : A $\to$ A + + e – At cathode : B + + e – $\to$ B Overall reaction is : A + B + $\to$ A + + B

Q53

4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al 3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H ions in solution by the same quantity of electric charge will be

A 44.8 L

B 22.4 L

C 11.2 L

D 5.6 L

Correct Answer

Option D

Solution

Faraday second law of electrolysis

{{{M_{Al}}} \over {{M_H}}} = {{{E_{Al}}} \over {{E_H}}}

$\Rightarrow$

{{4.5} \over {{M_H}}} = {{{{27} \over 3}} \over 1}

$\Rightarrow$ M H = 0.5 g We know, Volume of 2 g H 2 at STP = 22.4 L $\therefore$ Volume of 0.5 g H 2 at STP =

{{22.4 \times 0.5} \over 2}

= 5.6 L

Q54

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is (Atomic mass : Al = 27)

A 270 kg

B 540 kg.

C 90 kg

D 180 kg

Correct Answer

Option C

Solution

2Al 2 O 3 + 3C $\to$ 4Al + 3CO 2 From the above equation, 3 mol × 12 g mol –1 = 36 g of carbon is consumed to give 4 mol × 27 g mol –1 = 108 g of Al So, 108 g of Al is produced by 36 g of carbon $\therefore$ 270000 g of Al is produced by =

{{36} \over {108}} \times 270000

of C = 90000 g of C = 90 kg of C

Q55

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25 o C. The equilibrium constant of the reaction would be

A 2.0

\times

10 11

B 4.0

\times

10 12

C 1.0

\times

10 2

D 1.0

\times

10 10

Correct Answer

Option D

Solution

We know, from Nernst Equation E cell = E o cell -

{{2.303RT} \over {nF}}{\log _{10}}K

At equilibrium E cell = 0 $\therefore$ 0 = E o cell -

{{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K

$\Rightarrow$ 0 = 0.295 -

{{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K

$\Rightarrow$ 0.295 =

{{0.0591} \over 2}{\log _{10}}K

$\Rightarrow$

{\log _{10}}K

= 10 $\Rightarrow$ K = 1 $\times$ 10 10

Q56

The e.m.f. of a Daniell cell at 298 K is E 1 . When the concentration of ZnSO 4 is 1.0 M and that of CuSO 4 is 0.01 M, the e.m.f. changed to E 2 . What is the relationship between E 1 and E 2 ?

A E 1 > E 2

B E 1 < E 2

C E 1 = E 2

D E 2 = 0

\ne

E 1

Correct Answer

Option A

Solution

Cell reaction is, Zn + Cu 2+ $\to$ Zn 2+ + Cu E cell = E o cell -

{{RT} \over {nF}}\ln {{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}

Greater the factor

{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}

, less is the EMF. Hence E 1 > E 2

Q57

On the basis of the information available from the reaction, 4/3Al + O 2

\to

2/3Al 2 O 3 ,

\Delta

G =

-

827 kJ mol

-

1 of O 2 , the minimum e.m.f. required to carry out an electrolysis of Al 2 O 3 is (F = 96500 C mol

-

1 )

A 2.14 V

B 4.28 V

C 6.42 V

D 8.56 V

Correct Answer

Option A

Solution

4/3Al + O 2 $\to$ 2/3Al 2 O 3 ,

\Delta

G = $-$ 827 kJ mol $-$ 1 For 1 mol of Al, n = 3 $\therefore$ For

{4 \over 3}

mol of Al, n =

3 \times {4 \over 3} = 4

As

\Delta

G = - nFE o $\Rightarrow$ – 827 × 10 3 J = – 4 × E° × 96500 $\Rightarrow$ E o =

{{827 \times {{10}^3}} \over {4 \times 96500}}

= 2.14 V

Q58

In electrolysis of NaCl when Pt electrode is taken then H 2 is liberated at cathode while with Hg cathode it forms sodium amalgam

A Hg is more inert than Pt

B More voltage is required to reduce H + at Hg than at Pt

C Na is dissolved in Hg while it does not dissolve in Pt

D Conc. of H + ions is larger when Pt electrode is taken.

Correct Answer

Option B

Solution

In electrolysis of NaCl when Pt electrode is taken then H 2 liberated at cathode while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H + at Hg than Pt.

Q59

Standard electrode potentials are Fe 2+ /Fe [E o =

-

0.44] and Fe 3+ /Fe 2+ [ E o = 0.77]; If Fe 2+ , Fe 3+ and Fe blocks are kept together, then

A Fe 3+ increases

B Fe 3+ decreases

C Fe 2+ /Fe 3+ remains unchanged

D Fe 2+ decreases.

Correct Answer

Option B

Solution

The metals having higher negative values of their electrode potential can displace metals having lower values from their salt solutions.

Q60

Cell reaction is spontaneous when

A

\Delta

G o is negative

B

\Delta

G o is positive

C

\Delta

E o red is positive

D

\Delta

E o red is negative

Correct Answer

Option A

Solution

For spontaneous reaction

\Delta

G o = – ve and E o cell = + ve as

\Delta

G o = – nFE o cell where, n = number of electrons taking part E o = emf of cell F = Faraday constant