1
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.
2
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
3
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) .
4
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

Questions Asked from Electrochemistry

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