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Electrochemistry NEET Questions — Free MCQ Practice

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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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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

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If the half-cell reaction A+e $\to$ A^- has a large negative reduction potential, it follows that:

A.

A is readily reduced

B.

A is readily oxidized

C.

A^- is readily reduced

D.

A^-is readily oxidized

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Explanation

(d) Large negative RP or more positive oxidation potential and thus, more is the tendency to get oxidized.

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Pick out the incorrect statement

A.

Equivalent conductance increases with dilution

B.

Molar conductance increases with dilution

C.

Specific conductance increases with dilution

D.

Specific resistance increases with dilution

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Explanation

Specific conductance Decreases with dilution since number of ions per unit volume decrease with dilution.

Electrochemistry MCQs for NEET — Chemistry Questions with Answers

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