1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is :
A
9.81 g
B
10.9 g
C
98.1 g
D
109.0 g

Explanation


Moler mass of p$$-$$aminophenol

=   6 $$ \times $$ 12 + 7 + 14 + 16

= 109 gmol$$-$$1

Eq. weight (E) = $${W \over Q}$$ $$ \times $$ 96500

$$ \Rightarrow $$$$\,\,\,$$ W = $${{E\,Q} \over {96500}}$$

$$ \Rightarrow $$$$\,\,\,$$ W = $${{E\,I\,t} \over {96500}}$$

$$ \Rightarrow $$$$\,\,\,$$ W = $${{109} \over 4} \times {{9.65 \times 3600} \over {96500}}$$

$$ \Rightarrow $$$$\,\,\,$$ W = 9.81 gm
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is : (Molar mass of PbSO4 = 303 g mol$$-$$1)
A
22.8
B
15.2
C
7.6
D
11.4

Explanation

Pb(s) + SO$$_4^{ - 2}$$  $$ \to $$  PbSO4 + 2e$$-$$

$${{{n_{PbS{O_4}}}} \over 1} = {{{n_{e - }}} \over 2}$$

$$ \Rightarrow $$   $${{{n_{PbS{O_4}}}} \over 1} = {{0.05} \over 2}$$

$$ \therefore $$  Weight of PbSO4 $$=$$ $${{0.05} \over 2} \times 303 = 7.6g$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

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

Explanation

Higher the oxidation potential better will be reducing power.
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

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

Explanation

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

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