1

### 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) .
2

### JEE Main 2019 (Online) 12th January Morning Slot

The standard electrode potential ${E^o }$ and its temperature coefficient $\left( {{{d{E^o }} \over {dT}}} \right)$ for a cell are 2V and $-$ 5 $\times$ 10$-$4 VK$-$1 at 300 K respectively.
The cell reaction is
Zn(s) + Cu2+ (aq) $\buildrel \, \over \longrightarrow$ Zn2+ (aq) + Cu(s)

The standard reaction enthalpy ($\Delta$rH${^o }$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
A
$-$ 412.8
B
$-$ 384.0
C
192.0
D
206.4

## Explanation

$\Delta$G = -nFEcell = -2$\times$96500$\times$2 = -386 kJ

$\Delta$S = nF$\left( {{{d{E^o }} \over {dT}}} \right)$ = 2$\times$96500$\times$ ($-$ 5 $\times$ 10$-$4) = -96.5 kJ

At 298 K
T$\Delta$S = 298 $\times$ (–96.5 J) = – 28.8 kJ

at constant T (=248 K) and pressure

$\Delta$G = $\Delta$H – T$\Delta$S

$\Rightarrow$ $\Delta$H = $\Delta$G + T$\Delta$S

= -386 - 28.8 = -412.8 kJ
3

### JEE Main 2019 (Online) 12th January Evening Slot

$\wedge _m^ \circ$ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is-

5 $\times$ 10–5 S cm–1, degree of dissociation of HA is -
A
0.50
B
0.125
C
0.25
D
0.75

## Explanation

$\wedge _m^ \circ$ (HA) = $\wedge _m^ \circ$ (HCl) + $\wedge _m^ \circ$ (NaA) $-$ $\wedge _m^ \circ$ (NaCl)

= 425.9 + 100.5 $-$ 126.4

= 400 S cm2 . mol$-$1

$\wedge _m^ c$ = ${{K \times 1000} \over M}$

= ${{5 \times {{10}^{ - 5}} \times 1000} \over {{{10}^{ - 3}}}}$

= 50 S cm2 mol$-$1

$\alpha$ = ${{\Lambda _m^c} \over {\Lambda _m^o}}$ = ${{50} \over {400}}$ = 0.125
4

### JEE Main 2019 (Online) 12th January Evening Slot

Given

(i)  C (graphite) + O2(g) $\to$ CO2(g); $\Delta$rH$^\Theta$ = x kJ mol$-$1

(ii)  C(graphite) + ${1 \over 2}$O2(g) $\to$ CO2(g); $\Delta$rH$^\Theta$ = y kJ mol$-$1

(iii)  CO(g) + ${1 \over 2}$ O2(g) $\to$ CO2(g); $\Delta$rH$^\Theta$ = z kJ mol$-$1

Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct ?
A
z = x + y
B
x = y + z
C
x = y – z
D
y = 2z – x

## Explanation

In reaction (i), the product is CO2

If we add reaction (ii) and (iii) we get,
C + O2  $\to$  CO2
here also product is CO2 from same reactant C and O2.

So according to hess law, we can say
x = y + z