Electrochemistry — Page 5 | JEE Chemistry

Q41

Given CO3+ + e–

\to

CO2+ ; Eo = + 1.81 V Pb4+ + 2e–

\to

Pb2+ ; Eo = + 1.67 V Ce4+ + e–

\to

Ce3+ ; Eo = + 1.61 V Bi3+ + 3e–

\to

Bi ; Eo = + 0.20 V Oxidizing power of the species will increase in the order :

A Co3+ < Ce4+ < Bi3+ < Pb4+

B Co3+ < Pb4+ < Ce4+ < Bi3+

C Ce4+ < Pb4+ < Bi3+ < Co3+

D Bi3+ < Ce4+ < Pb4+ < Co3+

Correct Answer

Option D

Solution

Higher the value of Standard reduction potential of metal ion electrode, the more strong oxidizing agent is that metal ion electrode and have greater oxidising power.

So, Bi3+ < Ce4+ < Pb4+ < Co3+

Q42

Consider the statements S1 and S2 S1 : Conductivity always increases with decrease in the concentration of electrolyte. S2 : Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is :

A Both S1 and S2 are wrong

B S1 is correct and S2 is wrong

C Both S1 and S2 are correct

D S1 is wrong and S2 is correct

Correct Answer

Option D

Solution

We know conductivity (k) =

{G \over V}

V = volume When concentration decreases volume increases and when volume increases then conductivity (k) decreases.

So, we can say S1 is incorrect.

We know that,

{\lambda _m} = {k \over c}

where $\lambda$ m = molar conductivity k = conductivity c = concentration So, when concentration decreases molar conductivity increases.

So, we can say S2 is correct.

Q43

A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?

A 0.10

B 0.15

C 0.20

D 0.05

Correct Answer

Option D

Solution

Cathode reaction : Ni+2 + 2e- $\to$ Ni(s) $\therefore$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.

So from 0.1 F or 0.1 mole of electrons

{1 \over 2} \times 0.1

= 0.05 mole of Ni is deposited at the cathode.

Q44

The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is : Zn(s) + Cu2+ (aq)

\to

Zn2+ (aq) + Cu (s), E° = 2 V at 298 K (Faraday's constant, F = 96000 C mol–1)

A 384

B –192

C –384

D 192

Correct Answer

Option C

Solution

Here Zn is losing two electrons and Cu is gaining two electrons.

So only two electrons are involved in the reaction. $\therefore$ n = 2

\Delta

Go = - nFEo = -2 $\times$ 96000 $\times$ 2 = -384 kJ/mol

Q45

Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is :

A PbO

B PbO2

C PbSO4

D Pb3O4

Correct Answer

Option B

Solution

PbO2 is strong oxidizing agent because Pb+4 is not stable and can be easily reduced to Pb+2.

PbO2 is used in lead storage batteries.

It is also amphoteric in nature.

Q46

Given that

{E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V

;

{E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V

{E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V

{E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V

The strongest oxidizing agent is :

A O2

B Au3+

C Br2

D

{S_2}O_8^{2 - }

Correct Answer

Option D

Solution

Which electrode have higher value of standard reduction potential (SRP), that electrode will be strongest oxidizing agent.

Tendency to gain electrone is called standard reduction potential.

When tendency to gain electron is more then that electrode will have more oxidizing power.

Here given SRP's are :

{E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V

;

{E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V

{E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V

{E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V

You can see

{E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05

, have highest SRP so

{S_2}O_8^{2 - }

is the strongest oxidizing agent.

Q47

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction Zn(s) + Cu2+ (aq)

\rightleftharpoons

Zn2+(aq) + Cu(s) at 300 K is approximately, (R = 8 JK

-

1mol

-

1, F = 96000 C mol

-

1)

A e

-

80

B e

-

160

C e320

D e160

Correct Answer

Option D

Solution

\Delta

Go = $-$ RT lnk = $-$ nFEocell lnk =

{{n \times F \times {E^o}} \over {R \times T}} = {{2 \times 96000 \times 2} \over {8 \times 300}}

lnk = 160 k = e160

Q48

Consider the following reduction processes : Zn2+ + 2e–

\to

Zn(s) ; Eo = – 0.76 V Ca2+ + 2e–

\to

Ca(s); Eo = –2.87 V Mg2+ + 2e–

\to

Mg(s) ; Eo = – 2.36 V Ni2 + 2e–

\to

Ni(s) ; Eo = – 0.25 The reducing power of the metals increases in the order :

A Ca < Mg < Zn < Ni

B Ni < Zn < Mg < Ca

C Zn < Mg < Ni < Ca

D Ca < Zn < Mg < Ni

Correct Answer

Option B

Solution

Higher the oxidation potential better will be reducing power.

Q49

For the given cell : Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s) change in Gibbs energy (

\Delta

G) is negative, if :

A C2 =

\sqrt 2

C1

B C2 =

{{{C_1}} \over {\sqrt 2 }}

C C1 = 2C2

D C1 = C2

Correct Answer

Option A

Solution

Given

\Delta

G < 0 $\therefore$ -nFEcell < 0 $\Rightarrow$ Ecell > 0 We know, Ecell =

E_{cell}^0

-

{{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)

= 0 -

{{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)

$\therefore$ -

{{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)

> 0 $\Rightarrow$

\ln \left( {{{{C_1}} \over {{C_2}}}} \right)

< 0 $\Rightarrow$ C1 < C2 By checking option, we can see C2 =

\sqrt 2

C1 satisfy the condition C1 < C2.

Q50

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be [

E_{A{g^ + }/Ag}^0

= 0.80 V,

E_{A{u^ + }/Au}^0

= 1.69 V ]

A Silver and gold in equal mass proportion

B Silver and gold in proportion to their atomic weights

C Only gold

D Only silver

Correct Answer

Option C

Solution

Millimoles of Au+ = 0.1 × 250 = 25 Mole of Au+ =

{{25} \over {1000}}

=

{1 \over {40}}

Charge passed = I × t = 1 × 15 × 60 = 900 C moles of e– passed =

{{900} \over {96500}}

=

{9 \over {965}}

Only gold will be deposited as quantity of charge passed is less than the amount of Au+ present.