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Chemistry NEET Questions — Free MCQ Practice

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Calculate the quantity of electricity (i.e. charge) delivered by a Daniel cell initially containing 1L each of $1 M C u^{2 +}$ ion and $1 M Z n^{2 +},$ which is operated until its potential drops to 1V. (Given : $E_{Z n^{2 +} / Z n}^{0} = - 0.76 V; E_{C a^{2 +} / C a}^{0} = + 0.34 V)$

A.

$0.029 \times 10^{5} C .$

B.

$2.239 \times 10^{5} C$

C.

$1.029 \times 10^{5} C .$

D.

$3.123 \times 10^{5} C$

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Explanation

(C) At anode $Z n \to Z n^{2 +} + 2 e^{-}$

At cathode : $C u^{2 +} + 2 e^{-} \to C u$

Cell reaction : $Z n + C u^{2 +} \to Z n^{2 +} + C u$

As electricity is withdrawn from the cell, the concentration of $C u^{2 +}$ decreases while that of $Z n^{2 +}$ increases

$E_{c e l l} = {E^{°}}_{c e l l} - \frac{0.059}{n} \log \frac{[Z n^{2 +}]}{[C u^{2 +}]} 1 = 1.103 - \frac{0.059}{2} \log \frac{[Z n^{2 +}]}{[C u^{2 +}]} ∴ \frac{[Z n^{2 +}]}{[C u^{2 +}]} = 3041$

If x is the amount of $C u^{2 +}$ that is converted to Cu when the cell potential drops to 1V then

$\frac{[Z n^{2 +}]}{[C u^{2 +}]} = \frac{1 + x}{1 - x} = 3041; x = \frac{3040}{3042} = 0.9993$

To change $C u^{2 +} t o C u a n d Z n t o Z n^{2 +}, 2$ electrons are requried, hence quantity of charge drawn from the cell

$= 2 \times 0.9993 \times 96500 = 1.029 \times 10^{5} C .$

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Some half cell reaction & their standard potential are given which combination would result in a cell with the largest potential.

$(i) A + e^{-} \to A^{-} E ° = - 0.24$

$(i i) B^{-} + e^{-} \to B^{- 2} E ° = 1.25$

$(i i i) C^{-} + 2^{e -} \to C^{- 3} E^{-} = - 1.25$

$(i v) D + 2^{e -} \to D^{- 2} E ° = 0.06$

$(v) E + 4^{e -} \to E^{- 4} E ° = 0.38$

A.

i and ii

B.

ii and iii

C.

iv and v

D.

ii and v

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Explanation

(B) Since $B^{-} + e^{-} \to B^{- 2}; E ° = 1.25 . . . . (i) h a v i n g l a r g e s t r e d u c t i o n p o t e n t i a l a n d C^{- 3} \to C^{-} + 2 e^{-}; E ° = + 1.25 . . . . (i i i)$

having largest oxidation potential. So the cell having these two half cell reaction cell would result in maximum potential.

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A hydrogen electrode is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution. $(T a k e p_{H_{2_{}}} = 1 a t m) T = 298 K .$

A.

increase by 0.41 V

B.

increase by 59 mV

C.

decrease by 0.41 V

D.

decrease by 59 mV

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Explanation

(C). pH changes from 0 to 7.

$∴ [H^{+}]$ changes from 1 to $10^{- 7}$ M.

Accordingly $E_{r e d}$ decreases by 0.059 log $10^{- 7}$

i.e. 0.059 × (–7) = – 0.41 volt.

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The solubility product of silver iodide is $8.3 \times 10^{- 17}$ and the standard potential (reduction) of Ag, $A g^{+}$ electrode is + 0.800 volts at $25^{°} C .$ The standard potential of Ag, AgI/ $I^{-}$ electrode (reduction) from these data is-

A.

– 0.30 V

B.

+ 0.15 V

C.

+ 0.10 V

D.

– 0.15 V

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Explanation

(D). Solubility product reaction is

$A g I \to A g^{+} + I^{-}$

By calculating the EMF of this cell reaction from the given data and relating to $K_{s p}$ via the $∆ G °$ of the reaction, we can obtain $K_{s p}$ .

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How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-

A.

26.805 h

B.

53.61 h

C.

107.22 h

D.

None of these

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Explanation

(B) $2 O H^{-} \to H_{2} O + \frac{1}{2} O_{2} + 2 e^{-} (A n o d e) 2 F = 1 m o l H_{2} O d e c o m p o s e d; s o f o r 2 m o l o f H_{2} O, 4 F o f e l e c t r i c i t y r e q u i r e d . Q = i \times t (i = 2 a m p .) t = \frac{4 \times 96500}{2} = 193000 s e c = 53.61 h o u r s .$

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A certain current liberates 0.504 g of hydrogen in 2 hours. How many gram of copper can be liberated by the same current flowing for the same time in aqueous $C u S O_{4}$ solution :

A.

12.7

B.

16

C.

31.8

D.

63.5

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Explanation

(B) $E q . o f H_{2} = E q . o f C u ∴ \frac{0.154}{1} = \frac{w}{63.5 / 2} ∴ w_{C u} = 16 g$

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Calculate the maximum work that can be obtained from the Daniell cell given below -

$Z n (s) | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u (s) .$

Given that $E_{Z n^{2 +} / Z n}^{°} = - 0.76 V a n d E_{C u^{2 +} / C u}^{°} = + 0.34 V$

A.

– 212300 J

B.

– 202100 J

C.

– 513100 J

D.

– 232120 J

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Explanation

(A) Cell reaction is : $Z n (s) + C u^{2 +} (a q) \to C u (s) + Z n^{2 +} (a q)$

Here n=2

${E^{\circ}}_{c e l l} = {E^{\circ}}_{c a t h o d e} - {E^{\circ}}_{a n o d e}$

(On the basis of reduction potential)

= + 0.34 – (– 0.76) = 1.10 V

We know that : $W_{m a x} = ∆ G^{\circ} = - n F E^{\circ}$

$= - (2 m o l) \times (96500 C m o l) \times (1.10 V) = - 212300 J$

$o r W_{m a x} = - 212300 J .$

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The metal that cannot be produced on reduction of its oxide by aluminium is :

A.

K

B.

Mn

C.

Cr

D.

Fe

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Explanation

(A) ${E^{\circ}}_{O P} o f K > {E^{\circ}}_{O P} o f A l .$

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We have taken a saturated solution of $A g B r, K_{s p}$ of AgBr is $12 \times 10^{- 14} .$ If $10^{- 7}$ mole of $A g N O_{3}$ are added to 1 litre of this solution then the conductivity of this solution in terms of $10^{- 7}$ $S m^{- 1}$ units will be

$[λ °_{(A g^{+})} = 4 \times 10^{- 3} S m^{2} m o l^{- 1}; λ °_{(B r^{-})} = 6 \times 10^{- 3} S m^{2} m o l^{- 1}, λ °_{(N O_{3}^{-})} = 5 \times 10^{- 3} S m^{2} m o l^{- 1}]$

A.

39

B.

55

C.

15

D.

41

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Explanation

(A). The solubility of AgBr in presence of $10^{- 7}$ molar $A g N O_{3} i s 3 \times 10^{- 7} M .$

Therefore $[B r^{-}] = 3 \times 10^{- 4} m^{3}, [A g^{+}] = 4 \times 10^{- 4} m^{3} a n d [N {O_{3}}^{-}] = 10^{- 4} m^{3}$

Therefore $k_{t o t a l} = k_{B r^{-}} + k_{A g^{+}} + k_{N O_{3}^{-}} = 55 S m^{- 1}$

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The standard reduction potentials, $E^{\circ}$ , for the half reactions are as $Z n = Z n^{2 +} + 2 e; E ° = + 0.76 V$ and $F e = F e^{2 +} + 2 e; E ° = + 0.41$ V; the e.m.f. for the cell reaction, $F e^{2 +} + Z n = Z n^{2 +} + F e i s -$

A.

– 0.35 V

B.

+ 0.35 V

C.

+ 1.17 V

D.

– 1.17 V

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Explanation

(B). Since oxidation potential of Zn is higher than Fe so it will act as anode simultaneously Fe will act as cathode so EMF of cell will be Eº (ox.) of Zn + Eº (red) of Fe.

E = 0.76 + ( - 0.41) = + 0.35 V

Chemistry MCQs for NEET — Practice Questions with Answers

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