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Chemistry NEET Questions — Free MCQ Practice

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A hydrogen gas electrode is made by dipping platinum wire in a solution of HCI of pH = 10 and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be

A.

0.059 V

B.

0.59 V

C.

00.118 V

D.

1.18 V

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Explanation

(b) For hydrogen electrode, oxidation half reaction is

H₂ $\to$ 2H⁺ +2e^-

If Ph = 10

H⁺ = 1x10^-PH = 1x10^-10

From Nernst equation,

E_cell = E_cell - (0.0591/2) log[H⁺]²/P_H2

For hydrogen electrode, E_cell = 0

E_cell = - (0.0591/2) log[10^-10]²/1

= +(0.0591x2/2) log 1/10^-10

= 0.0591 log 10¹⁰

= 0.0591 x 10 = 0.591V

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The atomic mass of Al is 27. When a current of 5 faraday is passed through a solution of Al³⁺ ions, the mass of Al deposited is:

A.

27g

B.

36g

C.

45g

D.

9g

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Explanation

Concept used: Faraday's first law:

w = z x Q

$w = \frac{E}{96500} \times 5 \times 96500$ [1F=96500 C]

$w = \frac{27}{\frac{3}{96500}} \times 5 \times 96500 w = \frac{27 \times 5}{3}$

w= 45 g

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Cu⁺(aq) is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction

2Cu⁺(aq) $⇌$ Cu²⁺(aq) + Cu(s)

Choose the correct E0 for above reaction if E⁰_cu²⁺/cu = 0.34 V and E⁰_cu²⁺/cu+ = 0.15 V

A.

-0.38 V

B.

+4.9 V

C.

+0.38 V

D.

-0.19 V

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Explanation

(C) - $∆$ G⁰= nFE⁰

From given data,

(i) Cu(s) $\to$ Cu²⁺(aq) + 2e^-

- $∆$ G⁰₁= -2(-0.34)xF

(ii) Cu²⁺(aq) +e^- $\to$ Cu⁺(aq)

- $∆$ G⁰₂= -1(-0.15)xF

On addition

Cu(s) $\to$ Cu⁺(aq) + e^-

- $∆$ G⁰₃= -1xE⁰xF

- $∆$ G⁰₃ = $∆$ G⁰₁+- $∆$ G⁰₂

-n₃FE⁰ = - n₁FE⁰₁-n₂FE²

-E⁰ = -2(-0.34) - 1(0.15)

= (-2x-0.34) +(-1 x 0.15)

-E⁰ = +0.68 - 0.15 = 0.53

E⁰ = 0.53V

Reaction 2Cu⁺(aq) ^$â‡Œ$Cu²⁺(aq) + Cu(s)

So Cu⁺(aq) + e^- ^$â‡Œ$ Cu(s) E⁰ = 0.53V

Cu⁺(aq) ^$â‡Œ$ Cu²⁺(aq) +e^- E⁰ = -0.15V

2Cu⁺(aq) ^$â‡Œ$Cu²⁺(aq) + Cu(s) E⁰ = +0.38V

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On the basis of the information available from the reaction.

$\frac{4}{3} Al + O_{2} \overset{}{\to} \frac{2}{3} {Al}_{2} O_{3}, ∆ G = - 827 kJ {mol}^{- 1}$

of O₂, the minimum EMF required to carry out the electrolysis of Al₂O₃ is (F=96500 C mol^-1)

A.

2.14 V

B.

4.28 V

C.

6.42 V

D.

8.56 V

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Explanation

(a)

$\frac{4}{3} Al + O_{2} \overset{}{\to} \frac{2}{3} {Al}_{2} O_{3} ∆ G = - n E F - 827 \times 10^{3} J = - 4 \times E \times 96500 E = \frac{827 \times 10^{3}}{4 \times 96500} E = + 2.14 V$

n = 4 , as Al ⁰ to Al ⁺³ ,3 electron has been lost , so 4/3 aluminium will lose , 4/3x 3 = 4

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Which one of the following pairs of substances on reaction will not evolve H₂ gas?

A.

Iron and H₂SO₄ (aq)

B.

iron and steam

C.

Copper and HCl(aq)

D.

Sodium and ethyl alcohol

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Explanation

(c) Since copper is placed below hydrogen in the electrochemical series, thus copper does not give hydrogen with dilute acids. While all other pairs give hydrogen.

Fe + dil. H₂SO₄ → FeSO₄ + H₂ $↑$

3Fe + 4H₂O → Fe₃O₄ + 4H₂ $↑$

2Na + C₂H₅OH → 2C₂H₅ONa+H₂ $↑$

Cu + dil.-HCl → No reaction

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A certain current liberates 0.504 g of hydrogen in 2 hr. How many gram of copper can be liberated by the same current flowing for the same time in CuSO₄ solution?

A.

12.7

B.

16

C.

31.8

D.

63.5

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For the reductidon of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25⁰C. The value of standard Gibbs energy, - $∆$ G⁰will be (F = 96500 Cmol^-1)

A.

-89.0 KJ

B.

-89.0 J

C.

-44.5 KJ

D.

-98.0 KJ

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Explanation

(a) We know that, standard Gibbs energy,

$∆$ G⁰= -nFE⁰_cell

For the cell reaction,

2Ag⁺+ Cu $\to$ Cu²⁺+ 2Ag

$∆$ E⁰_cell= +0.46V

$∆$ G⁰= -nFE⁰_cell

n=2

$∆$ G⁰= -2x96500x0.46

= -88780J

= - 88.7 KJ

=-89.0 KJ

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A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively:

A.

H₂, O₂

B.

O₂, H₂

C.

O₂, Na

D.

O₂, SO₂

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Explanation

(a) At cathode: $2 H^{+} + 2 e \overset{}{\to} H_{2};$

At anode: $2 {OH}^{-} \overset{}{\to} H_{2} O + \frac{1}{2} O_{2} + 2 e$

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If a salt bridge is removed from the two half cells, the voltage:

A.

drops to zero

B.

does not change

C.

increases gradually

D.

increases rapidly

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Explanation

(a) The e.m.f. of cell decreases gradually and finally to zero due to liquid junction potential arised in cell after removal of salt bridge.

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Two electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be:

A.

3:1

B.

2:1

C.

1:1

D.

3:2

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Explanation

(d) Eq. of Fe²⁺ = Eq. of Fe³⁺

or m₁/(A/2) = m₂/(A/3)

or m_Fe²⁺/m_Fe³⁺ = 3/2

Chemistry MCQs for NEET — Practice Questions with Answers

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