Electrochemistry — Page 4 | JEE Chemistry

Q31

Given below are the half-cell reactions: Mn2+ + 2e-

\to

Mn; Eo = -1.18 V 2(Mn3+ + e-

\to

Mn2+); Eo = +1.51 V The Eo for 3Mn2+

\to

Mn + 2Mn3+ will be :

A – 0.33 V; the reaction will not occur

B – 0.33 V; the reaction will occur

C – 2.69 V; the reaction will not occur

D – 2.69 V; the reaction will occur

Correct Answer

Option C

Solution

(a)

\,\,\,\,\,

M{n^{2 + }} + 2{e^ - } \to Mn;\,\,{E^ \circ } = - 1.18V;\,...\left( i \right)

(b)

\,\,\,\,\,

M{n^{3 + }} + e \to M{n^{2 + }};\,\,{E^ \circ } = - 1.51V;\,...\left( {ii} \right)

Now multiplying equation

(ii)

by two and subtracting from equation

(i)

3M{n^{2 + }} \to M{n^ + } + 2M{n^{3 + }};

{E^ \circ } = {E_{{O_X}}} + {E_{{\mathop{\rm Re}\nolimits} d.}} = - 1.18 + \left( { - 1.51} \right) = - 2.69V

[

-ve

value of

EMF

(i.e.,

\Delta G = + ve

) shows that the reaction is non-spontaneous ]

Q32

Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu)

A 63.5 g

B 2 g

C 127 g

D 0 g

Correct Answer

Option A

Solution

C{u^{2 + }} + 2{e^ - }\overset{\,}\longrightarrow Cu

2F\,\,i.e.\,\,2 \times 96500\,C

deposit

Cu=1

mol

=63.5

g

Q33

What will occur if a block of copper metal is dropped into a beaker containing a solution of 1M ZnSO4?

A The copper metal will dissolve and zinc metal will be deposited.

B The copper metal will dissolve with evolution of hydrogen gas.

C The copper metal will dissolve with evolution of oxygen gas.

D No reaction will occur.

Correct Answer

Option D

Solution

In the reaction of copper with zinc ions, zinc is more active (easily oxidised) than copper; thus, no reaction will take place.

Q34

Identify the correct statement :

A Iron corrodes in oxygen-free water.

B Iron corrodes more rapidly in salt water because its electrochemical potential is higher.

C Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential.

D Corrosion of iron can be minimized by forming an impermeable barrier at its surface.

Correct Answer

Option D

Solution

The two conditions necessary for the corrosion of iron to take place are presence of moisture and oxygen.

Factors that catalyse the process of rusting are the presence of carbon dioxide, acids and impurities.

It can be minimised by introducing of a barrier film between surface of iron and atmosphere.

This can be done by (i) painting the surface. (ii) coating the surface with oil or grease. (iii) electroplating iron with non-corrosive metals such as nickel or chromium. (iv) covering the surface of iron with layer of more active metal with higher oxidation potential like zinc.

Q35

Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :

A 2.24 L

B 4.48 L

C 6.72 L

D 8.96 L

Correct Answer

Option D

Solution

The reaction is

2C{H_3}COOK + 2{H_2}O\overset{{Electrolysis}}\longrightarrow C{H_3} - C{H_3} + 2C{O_2} + {H_2} + 2KOH

At the anode (Oxidation): At the cathode (Reduction):

2{H_2}O\overset{{ + 2{e^ - }}}\longrightarrow 2O{H^ - } + 2\mathop H\limits^ \bullet

2\mathop H\limits^ \bullet \overset{{}}\longrightarrow {H_2}

Total number of moles of gases = moles of C2H6 + moles of CO2 + moles of H2

n = {{0.2} \over 2} + {{0.2} \over 1} + {{0.2} \over 2} = 0.4

V = {{nRT} \over p}

= {{(0.4 \times 0.0821 \times 273)} \over 1} = 8.96

L

Q36

The electrode potential of M2+/M of 3d-series elements shows positive value for :

A Zn

B Fe

C Cu

D Co

Correct Answer

Option C

Solution

In the electrode potential series, only copper have positive value for electrode potential because copper has lower tendency than hydrogen to form ions.

So, if standard hydrogen electrode (ECell = 0) is connected to copper half-cell, the copper with be relatively less negative or less number of electrons.

E_{(C{u^{2 + }}/Cu)}^o

= + 0.34 V;

E_{(F{e^{2 + }}/Fe)}^o

= $-$ 0.41 V

E_{(C{o^{2 + }}/Co)}^o

= $-$ 0.28 V;

E_{(Z{n^{2 + }}/Zn)}^o

= $-$ 0.76 V $\therefore$ Electrode potential of Cu

E_{(C{u^{2 + }}/Cu)}^o

show positive value.

Q37

What is the standard reduction potential (Eo) for Fe3+

\to

Fe ? Given that : Fe2+ + 2e

-

\to

Fe;

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

=

-

0.47 V Fe3+ + e

-

\to

Fe2+;

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

= +0.77 V

A

-

0.057 V

B + 0.057 V

C + 0.30 V

D

-

0.30 V

Correct Answer

Option A

Solution

For the given reaction : Fe3+ + e $-$ $\to$ Fe2+;

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

= +0.77 V

\Delta

Go = -nFEo $\Rightarrow$

\Delta G_1^o = - \left( 1 \right)F\left( {0.77} \right)

= -0.77F For the following reaction : Fe2+ + 2e $-$ $\to$ Fe;

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

= $-$ 0.47 V

\Delta G_2^o = - \left( 2 \right)F\left( {-0.47} \right)

= 0.47 $\times$ 2F Overall reaction : Fe3+ + 3e $-$ $\to$ Fe;

\Delta G_3^o = - 3FE_3^o

$\Rightarrow$

\Delta G_1^o

+

\Delta G_2^o

=

3FE_3^o

$\Rightarrow$ -0.77F + 0.47 $\times$ 2F =

3FE_3^o

$\Rightarrow$

E_3^o = - {{0.17} \over 3}

= -0.057 V

Q38

Given

E_{C{l_2}/C{l^ - }}^o

= 1.36 V,

E_{C{r^{3 + }}/Cr}^o

= - 0.74 V

E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o

= 1.33 V,

E_{Mn{O_4}^ - /Mn ^{2+}}^o

= 1.51 V Among the following, the strongest reducing agent is :

A Mn2+

B Cr3+

C Cl–

D Cr

Correct Answer

Option D

Solution

E_{C{l_2}/C{l^ - }}^o

= 1.36 V,

E_{C{r^{3 + }}/Cr}^o

= - 0.74 V

E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o

= 1.33 V,

E_{Mn{O_4}^ - /Mn ^{2+}}^o

= 1.51 V More negative the E° value of the species, more stronger is the reducing agent.

Since Cr3+ is having least reducing potential, so Cr would be strongest reducing agent.

Q39

When an electric currents passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is :

A 1.0

B 0.5

C 0.1

D 2.0

Correct Answer

Option A

Solution

Reaction at cathode : 2H+ + 2e $-$ $\to$ H2 We know, $\omega$ = zIt =

{{EIt} \over {96500}}

< no. of moles of H2 =

{{112} \over {22400}}

\therefore\,\,\,

mass (w) of H2 =

{{112} \over {22400}}

$\times$ 2

\therefore\,\,\,\,

{{112} \over {22400}}

$\times$ 2 =

{{1 \times I \times 965} \over {96500}}

$\Rightarrow$

\,\,\,\,

I

= 1 A

Q40

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)

A 1.6 hours

B 6.4 hours

C 0.8 hours

D 3.2 hours

Correct Answer

Option D

Solution

Required reaction : B2H6 + 3O2 $\to$ B2 O3 + 3 H2 O Here molar mass of B2H6 =10.8 $\times$ 2 + 6 = 27.6 gm Given weight of B2H6 = 27.66 g

\therefore\,\,\,\,

No of moles of B2H6 =

{{27.6} \over {27.66}} \simeq 1

mole. For combustion of 1 mole B2H6 3 moles O2 required. This 3 mole of O2 is obtained by electrolysis of H2O. 2H2O(

l

) $\to$ O2 (g) + 4 H+ (aq) + 4 e $-$ From Faradays law of electrolysis, moles $\times$ nf =

{{It} \over {96500}}

Here moles of O2 = 3.

Nf of O2 = 4 (in H2 change of O = $-$ 2 and in O2 change of 0 = O.

So change in charge = 2 . for two atoms of O2 change in charge = 2 $\times$ 2 = 4)

\therefore\,\,\,\,

3 $\times$ 4 =

{{100 \times t} \over {96500}}

\Rightarrow \,\,\,\,

t = 12 $\times$ 965 sec.

\Rightarrow \,\,\,\,

t =

{{12 \times 965} \over {60 \times 60}}

hr = 3.2 hr