1

### 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 $=$ ${{0.05} \over 2} \times 303 = 7.6g$
2

### 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.
3

### 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
4

### JEE Main 2019 (Online) 11th January Morning Slot

For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below :

Mx+ (aq)/M(s) Au3+(aq)/Au(s) Ag+(aq)/Ag(s) Fe3+(aq)/Fe2+ (aq) Fe2+(aq)/Fe(s)
E0Mx+/M/(V) 1.40 0.80 0.77 $-$0.44

If $E_{z{n^{2 + }}/zn}^0$ = $-$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
A
Ag+/Ag
B
Fe3+/Fe2+
C
Au3+/Au
D
Fe2+/Fe

## Explanation

Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s)
---------------------------------------
Anode              Cathode

Eocell = Eocathode – Eoanode

(i) For Ag+/Ag :

Eocell = 0.80 – (– 0.76) = 1.56 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn2+ = 2

Valency factor of Ag+/Ag = 1

LCM of 1 and 2 = 2

$\therefore$ No of electrons transferred = 2

$\therefore$ Eocell per electron = ${{1.56} \over 2}$ = 0.78

(ii) For Fe3+/Fe2+ :

Eocell = 0.77 – (– 0.76) = 1.53 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn2+ = 2

Valency factor of Fe3+/Fe2+ = 1

LCM of 2 and 1 = 2

$\therefore$ No of electrons transferred = 2

$\therefore$ Eocell per electron = ${{1.53} \over 2}$ = 0.76

(iii) For Au3+/Au :

Eocell = 1.40 – (– 0.76) = 2.16 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn2+ = 2

Valency factor of Au3+/Au = 3

LCM of 2 and 3 = 6

$\therefore$ No of electrons transferred = 6

$\therefore$ Eocell per electron = ${{2.16} \over 6}$ = 0.36

(iv) For Fe2+/Fe :

Eocell = –0.44 – (– 0.76) = 0.32 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn2+ = 2

Valency factor of Fe2+/Fe = 2

LCM of 2 and 2 = 2

$\therefore$ No of electrons transferred = 2

$\therefore$ Eocell per electron = ${{0.32} \over 2}$ = 0.16

Eocell is maximum for EoAg+(aq)/Ag(s) .