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1

MCQ (Single Correct Answer)

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) + Cu^{2+} (aq) $$\buildrel \, \over
\longrightarrow $$ Zn^{2+} (aq) + Cu(s)

The standard reaction enthalpy ($$\Delta $$_{r}H$${^o }$$) at 300 K in kJ mol^{–1} is, [Use R = 8 JK^{–1} mol^{–1} and F = 96,000C mol^{–1}]

The cell reaction is

Zn(s) + Cu

The standard reaction enthalpy ($$\Delta $$

A

$$-$$ 412.8

B

$$-$$ 384.0

C

192.0

D

206.4

$$\Delta $$G = -nFE_{cell} = -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

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

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

2

MCQ (Single Correct Answer)

For the cell Zn(s) |Zn^{2+} (aq)| |M^{x+} (aq)| M(s), different half cells and their standard electrode potentials are given below :

If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of E^{o}_{cell} per electron transferred?

M^{x+} (aq)/M(s) |
Au^{3+}(aq)/Au(s) |
Ag^{+}(aq)/Ag(s) |
Fe^{3+}(aq)/Fe^{2+} (aq) |
Fe^{2+}(aq)/Fe(s) |
---|---|---|---|---|

E^{0}_{Mx+}_{/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 E

A

Ag^{+}/Ag

B

Fe^{3+}/Fe^{2+}

C

Au^{3+}/Au

D

Fe^{2+}/Fe

Zn(s) |Zn^{2+} (aq)| |M^{x+} (aq)| M(s)

---------------------------------------

Anode Cathode

E^{o}_{cell} = E^{o}_{cathode} – E^{o}_{anode}

(i) For Ag^{+}/Ag :

E^{o}_{cell} = 0.80 – (– 0.76) = 1.56 V

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

Valency factor of Zn(s) |Zn^{2+} = 2

Valency factor of Ag^{+}/Ag = 1

LCM of 1 and 2 = 2

$$ \therefore $$ No of electrons transferred = 2

$$ \therefore $$ E^{o}_{cell} per electron = $${{1.56} \over 2}$$ = 0.78

(ii) For Fe^{3+}/Fe^{2+} :

E^{o}_{cell} = 0.77 – (– 0.76) = 1.53 V

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

Valency factor of Zn(s) |Zn^{2+} = 2

Valency factor of Fe^{3+}/Fe^{2+} = 1

LCM of 2 and 1 = 2

$$ \therefore $$ No of electrons transferred = 2

$$ \therefore $$ E^{o}_{cell} per electron = $${{1.53} \over 2}$$ = 0.76

(iii) For Au^{3+}/Au :

E^{o}_{cell} = 1.40 – (– 0.76) = 2.16 V

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

Valency factor of Zn(s) |Zn^{2+} = 2

Valency factor of Au^{3+}/Au = 3

LCM of 2 and 3 = 6

$$ \therefore $$ No of electrons transferred = 6

$$ \therefore $$ E^{o}_{cell} per electron = $${{2.16} \over 6}$$ = 0.36

(iv) For Fe^{2+}/Fe :

E^{o}_{cell} = –0.44 – (– 0.76) = 0.32 V

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

Valency factor of Zn(s) |Zn^{2+} = 2

Valency factor of Fe^{2+}/Fe = 2

LCM of 2 and 2 = 2

$$ \therefore $$ No of electrons transferred = 2

$$ \therefore $$ E^{o}_{cell} per electron = $${{0.32} \over 2}$$ = 0.16

E^{o}_{cell} is maximum for E^{o}_{Ag+(aq)/Ag(s)
}.

---------------------------------------

Anode Cathode

E

(i) For Ag

E

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

Valency factor of Zn(s) |Zn

Valency factor of Ag

LCM of 1 and 2 = 2

$$ \therefore $$ No of electrons transferred = 2

$$ \therefore $$ E

(ii) For Fe

E

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

Valency factor of Zn(s) |Zn

Valency factor of Fe

LCM of 2 and 1 = 2

$$ \therefore $$ No of electrons transferred = 2

$$ \therefore $$ E

(iii) For Au

E

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

Valency factor of Zn(s) |Zn

Valency factor of Au

LCM of 2 and 3 = 6

$$ \therefore $$ No of electrons transferred = 6

$$ \therefore $$ E

(iv) For Fe

E

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

Valency factor of Zn(s) |Zn

Valency factor of Fe

LCM of 2 and 2 = 2

$$ \therefore $$ No of electrons transferred = 2

$$ \therefore $$ E

E

3

MCQ (Single Correct Answer)

In the cell

Pt$$\left| {\left( s \right)} \right|$$H_{2}(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 $$\left. {298} \right\}$$

Pt$$\left| {\left( s \right)} \right|$$H

the cell potential is 0.92 V when a 10

$$\left\{ {} \right.$$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

Anode : H_{2}(g) $$ \to $$ 2H^{+}(aq) + 2e^{-}

Cathode : AgCl(s) + e^{-} $$ \to $$ Ag(s) + Cl^{-}(aq)

---------------------------------------------------------------

H_{2}(g) + 2AgCl(s) $$ \to $$ 2Ag(s) + 2H^{+}(aq) + 2Cl^{-}(aq)

From Nernst equation we know,

E_{cell} = E^{0}_{cell} - $${{0.06} \over n}\log Q$$

Here,

E_{cell} = E^{0}_{cell} - $${{0.06} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}{{\left[ {C{l^ - }} \right]}^2}} \over {{P_{{H_2}}}}}$$

$$ \Rightarrow $$ 0.92 = E^{0}_{AgCl/AgCl -} - $$0.03\log {{{{\left[ {{{10}^{ - 6}}} \right]}^2}{{\left[ {{{10}^{ - 6}}} \right]}^2}} \over 1}$$

$$ \Rightarrow $$ E^{0}_{AgCl/AgCl} = 0.92 - 0.72 = 0.2 V

Cathode : AgCl(s) + e

---------------------------------------------------------------

H

From Nernst equation we know,

E

Here,

E

$$ \Rightarrow $$ 0.92 = E

$$ \Rightarrow $$ E

4

MCQ (Single Correct Answer)

Consider the following reduction processes :

Zn^{2+} + 2e^{–} $$ \to $$ Zn(s) ; E^{o} = – 0.76 V

Ca^{2+} + 2e^{–} $$ \to $$ Ca(s); E^{o} = –2.87 V

Mg^{2+} + 2e^{–} $$ \to $$ Mg(s) ; E^{o} = – 2.36 V

Ni^{2} + 2e^{–} $$ \to $$ Ni(s) ; E^{o} = – 0.25

The reducing power of the metals increases in the order :

Zn

Ca

Mg

Ni

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

Higher the oxidation potential better will be reducing power.

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