1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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