Electrochemistry(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

ELECTROCHEMISTRY MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. Given standard electrode potentials

F e^{+ +} + 2 e^{-} \to F e; E^{\circ} = - 0.440 V

F e^{+ + +} + 3 e^{-} \to F e; E^{\circ} = - 0.036 V

The standard electrode potential

(E^{\circ})

for

F e^{+ + +} + e^{-} \to F e^{+ +}

- 0.476 V
- 0.404 V
+ 0.404 V
+ 0.772 V

Discuss Question

Answer: Option D. -> + 0.772 V
:
D

Δ G^{\circ} = - n E^{\circ} F

F e^{2 +} + 2 e^{-} \to F e

. . . . . . . (i)

Δ G^{\circ} = - 2 \times F \times (- 0.440 V) = 0.880 F

F e^{3 +} + 3 e^{-} \to F e

. . . . . . (i i)

Δ G^{\circ} = - 3 \times F \times (- 0.036) = 0.108 F

On substracting equation

(i)

from

(i i)

F e^{3 +} + e^{-} \to F e^{2 +}

Δ G^{\circ} = 0.108 F - 0.880 F = - 0.772 F

E^{\circ}

for the reaction

= - \frac{Δ G^{\circ}}{n F} = - \frac{(- 0.772 F)}{1 \times F} = + 0.772 V

Question 12. One of the most successful fuel cell uses the reaction of hydrogen with oxygen to form water. The cell was used for providing electrical power in the Apollo space program. The water vapors produced during the reaction were condensed and added to the drinking water supply for the astronauts. In the cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Which of the following can be used as catalysts in order to increase the rate of electrode reactions in this fuel cell?

Lithium
Palladium
Titanium
All of above

Discuss Question

Answer: Option B. -> Palladium
:
B
Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reactions.

Question 13. In a cell that utilizes the reaction

Z n_{(s)} + 2 H^{+} (a q) \to Z n^{2 +} (a q) + H_{2 (g)}

addition of

H_{2} S O_{4}

to cathode compartment, will

Increase the E and shift equilibrium to the right
Lower the E and shift equilibrium to the right
Lower the E and shift equilibrium to the left
Increase the E and shift equilibrium to the left

Discuss Question

Answer: Option A. -> Increase the E and shift equilibrium to the right
:
A

Z n_{(s)} + 2 H_{(a q)}^{+} ⇌ Z n_{(a q)}^{2 +} + H_{2 (g)}

E_{c e l l} = E_{c e l l}^{\circ} - \frac{.059}{2} l o g \frac{[Z n^{2 +}]}{[H^{+}]^{2}}

When

H_{2} S O_{4}

is added then

[H^{+}]

will increase therefore

E_{c e l l}

will also increases and equilibrium will shift towards right.

Question 14. Assertion : Copper reacts with hydrochloric acid and liberates hydrogen from the solution of dilute hydrochloric acid.
Reason : Hydrogen is present below copper in the electrochemical series.

If both assertion and reason are true and the reason is the correct explanation of the assertion.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.

Discuss Question

Answer: Option D. -> If the assertion and reason both are false.
:
D
Copper is present below hydrogen therefore hydrogen from HCl cannot be liberated by treating with copper. Hence assertion and reason both are false.

Question 15.

E^{\circ}

for the cell is

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

25^{\circ} C

, the equilibrium constant for the reaction

Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)

is of the order of

10−37
10−28
10+18
10+17

Discuss Question

Answer: Option A. -> 10−37
:
A

E_{c e l l}^{\circ} = \frac{0.059}{n} l o g K

l o g K = \frac{1.10 \times 2}{0.059} = 37.2881 \Rightarrow K = 10^{- 37}

Question 16. Which of the following methods cannot be employed in order to prevent corrosion?

Covering the surface by paint
Covering the surface by some chemical
Allowing the surface to interact with moisture
Providing the sacrificial electrode of another metal

Discuss Question

Answer: Option C. -> Allowing the surface to interact with moisture
:
C
Corrosion can be prevented by preventing the surface of of the metallic object to come in contact with the atmosphere. Options a, b and d shows methods of preventing the surface to come in contact with the atmosphere.

Question 17. Assertion: If

λ_{N a^{+}}^{\circ} + λ_{C l^{-}}^{\circ}

are molar limiting conductivity of the sodium and chloride ions respectively, then the limiting molar conducting for sodium chloride is given by the equation:

λ_{N a C l}^{\circ} = λ_{N a^{+}}^{\circ} + λ_{C l^{-}}^{\circ}

Reason:This is according to Kohlrausch law of independent migration of ions.

If both assertion and reason are true and the reason is the correct explanation of the assertion.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.

Discuss Question

Answer: Option A. -> If both assertion and reason are true and the reason is the correct explanation of the assertion.
:
A
According to Kohlrausch law, “Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte”.

Question 18. Calculate standard free energy change for the reaction

\frac{1}{2} C u (s) + \frac{1}{2} C l_{2} (g) ⇌ \frac{1}{2} C u^{2 +} + C l^{-}

taking place at in a cell whose standard e.m.f. is

1.02 v o l t s

- 98430 J
98430 J
96500 J
- 49215 J

Discuss Question

Answer: Option A. -> - 98430 J
:
A

Δ G = - n F E^{\circ}

\Rightarrow Δ G = - 1 \times 96500 \times 1.02 \Rightarrow Δ G = - 98430

Question 19. An aqueous solution containing one mole per litre of each

C u (N O_{3})_{2}, A g N O_{3}, H g_{2} (N O_{3})_{2}

and

M g (N O_{3})_{2}

, is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are

A g / A g^{+} = + 0.80, 2 H g / H g_{2}^{2 +} = + 0.79, C u / C u^{2 +} = + 0.34, M g / M g^{2 +} = - 2.37

with increasing voltage, the sequence of deposition of metals on the cathode will be

Ag,Hg,Cu,Mg
Mg,Cu,Hg,Ag
Ag,Hg,Cu
Cu,Hg,Ag

Discuss Question

Answer: Option C. -> Ag,Hg,Cu
:
C
A cationthat has thehighest reduction potential will be reduced first andso on.
However,

M g^{2 +}

in aqueous solution will not be reduced

(E_{M g^{2 +} / M g}^{\circ} < E_{_{H_{2} O} / \frac{1}{2} H_{2} + O H^{-}}) .

Instead water would be reduced in preference.

Question 20. The standard reduction potentials at 298 K for the following half reactions are given:
Which is the strongest reducing agent?

Z n^{2 +} (a q .) + 2 e ⇌ Z n (s); - 0.762

C r^{3 +} (a q) + 3 e ⇌ C r (s); - 0.740

2 H^{+} (a q) + 2 e ⇌ H_{2} (g); 0.00

F e^{3 +} (a q) + e ⇌ F e^{2 +} (a q); 0.770

Zn(s)
Cr(s)
H2(g)
Fe2+(aq)

Discuss Question

Answer: Option A. -> Zn(s)
:
A
More negative is the reduction potential, higher will be the reducing property, i.e. the power to give up electrons.