Electrochemistry Test

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Electrochemistry Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

for the cell reaction
$$Cu^{2+}(C_1aq)+Zn(s)=Zn^{2+}(C_2aq)+Cu(s)$$
of an electrochemical cell, the change in free energy at a given temperature is a function of

A
$$\ln (C_1)$$

B
$$\ln (C_2)$$

C
$$\ln(C_1+C_2)$$

D
$$\ln (C_2 /C_1)$$

Solution

Answer is option (D)
We know
$$ΔG=nFE$$
Again according to Nernst equation,
$$E=\dfrac{0.059}{n}$$ $$log \dfrac{[Zn^{2+} ]}{[Cu^{2+} ]}$$ $$ at $$25^o$$ C.

So $$ΔG$$ (free energy change) in an electrochemical cell at a given temperature is a function of
ln $$\dfrac{[Zn^{2+} ]}{ [Cu^{2+}]}$$ $$=ln\dfrac{C_2}{C_1}$$
Question 2
1 / -0

The main reason for not using a mercury electrolytic cell in NaOH manufacture is that

A
Hg is toxic

B
Hg is liquid

C
Hg has a high vapour pressure

D
Hg is a good conductor of electricity

Solution

The main reason for not using a mercury electrolytic cell in NaOH manufacture is because Hg is a good conductor of electricity.
Question 3
1 / -0

In which cell the free energy of a chemical reaction is directly converted into electricity

A
Lenclanche cell

B
Concentration cell

C
Fuel cell

D
Lead storage battery

Solution

In concentration cell the free energy of a chemical reaction is directly converted into electricity.
Correct answer is option (B)
Question 4
1 / -0

$$E^o_{Cell}$$ for some half-cell reactions are given below. On the basis of these mark the correct answers.

(a) $$H^- (aq) + e^- \rightarrow \frac{1}{2} H_2 (g); E^o_{Cell} = 1.23V$$

(b) $$2H_2O(l) \rightarrow O_2(g) + 4H^- (aq) + 4e^- ; E^o_{Cell} = 1.23V$$

(c) $$2SO^{2-}_4(aq) \rightarrow S_2O^{2-}_8(aq) + 2e^-; E^o_{Cell} = 1.96V$$

A
In dilute sulphuric acid solution, hydrogen will be reduced at cathode.

B
In concentrated sulphuric acid solution, water will be oxidised at anode.

C
In dilute sulphuric acid solution, water will be oxidised at anode.

D
In concentrated sulphuric acid solution, $$SO^{2-}_4$$ ion will be oxidized to tetrathionate ion at anode.

Solution

In the electrolysis of dil. H$$_2$$SO$$_4$$, above three reaction takes place. Oxidation half reaction occurs at anode, lower value of standard reduction potential will be preferred. At cathode hydrogen ion, will be converted into hydrogen.
Question 5
1 / -0

$$E_n = -\dfrac{313.6}{n^2}$$, if the value of $$E_i = -38.84$$ to which value 'n' corresponds?

A
2

B
4

C
1

D
3

Solution

Given :- $$E_n = \dfrac{-313.6}{n^2}$$
To calculate :- The value of n if the value of $$E_i = -38.84 kcal \ mol^{-1}$$
Sol :- $$E_i = -38.84$$

$$-38.84 = \dfrac{-313.6}{n^2}$$

$$n^2 = \dfrac{313.6}{38.84}$$

$$n^2 = 8$$

$$n = 2.82 \approx 3$$

So, the correct option is (D) 3
Question 6
1 / -0

Two electrolytic cells containing $$CuSO_{4}$$ and $$AgNO_{4}$$ respectively are connected in series and a current is passed through them unit 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately
[ Atomic weight of copper and silver are respectively 63.57 and 107.88]

A
$$1.7 mg$$

B
$$3.4 mg$$

C
$$5.1 mg$$

D
$$6.8 mg$$

Solution

$$ \dfrac{m_{1}}{m_{2}} = \dfrac{E_{1}}{E_{2}} $$ (By faraday law for same current and time)
Where $$ E_{1} $$ and $$ E_{2} $$ are the chemical equivalents and $$ m_{1} $$ and $$ m_{2} $$ are the masses of copper and silver respectively.
$$ E = \dfrac{Atomic\, weight}{Valency} E_{1} = \dfrac{63.57}{2} = 31 . 79 $$ and
$$ E_{2} = \dfrac{107.88}{1} =107 . 88 $$
$$ \therefore \dfrac{1mg}{m_{2}} = \dfrac{31.79}{107.88} \Rightarrow m_{2} = \dfrac{107.88}{31.79}mg = 3.4 mg $$
Question 7
1 / -0

Resistance of a voltameter is 2$$\Omega $$, it is connected in series to a battery of 10 V through a resistance of 3$$\Omega $$ . In a certain time mass deposited on cathode is 1 gm. Now the voltameter and the 3$$\Omega $$resistance are connected in parallel with the battery. Increase in the deposited mass on cathode in the same time will be

A
0

B
1.5 gm

C
2.5 gm

D
2 gm

Solution

Remember mass of the metal deposited on cathode depends on the current through the voltameter and not on the current supplied by the battery. Hence by using $$m =Zit$$, we can say $$\dfrac{m_{Parallel}}{m_{Series}}=\dfrac{i_{Parallel}}{i_{Series}}\Rightarrow m parallel = \dfrac{5}{2}\times 1=2.5 gm.$$
Question 8
1 / -0

If redox reaction takes place in cell then electromotive force (e.m.f) of cell will be:

A
positive

B
negative

C
zero

D
one

Solution
Question 9
1 / -0

Which of the following mixtures represents rusting of iron?

A
FeO and Fe(OH)$$_3$$

B
FeO and Fe(OH)$$_2$$

C
Fe$$_2$$O$$_3$$ and Fe(OH)$$_3$$

D
Fe$$_3$$O$$_4$$ and Fe(OH)$$_2$$

Solution
Question 10
1 / -0

The amount of charge required to liberate 9 gm of aluminium ( atomic weight = 27 and valency = 3 ) in the process of electrolysis is ( Faraday's number = 96500 coulombs/gm equivalent)

A
321660 coulombs

B
69500 coulombs

C
289500 coulombs

D
96500 coulombs

Solution

The atomic mass of $$Al=27 g/mol$$
$$9g $$ of $$Al=\dfrac{9}{27}=0.333 moles$$
1 mole $$Al= 3 $$ moles electrons.
$$0.333$$ moles $$Al=0.333\times 3=1 $$ mole electrons
1 mole electrons =1 faraday of electricity $$=96500 C.$$
Hence, option (D) is the correct answer.

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Electrochemistry Test - 74

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Question 1

$$\ln (C_1)$$

$$\ln (C_2)$$

$$\ln(C_1+C_2)$$

$$\ln (C_2 /C_1)$$

Solution

Question 2

Hg is toxic

Hg is liquid

Hg has a high vapour pressure

Hg is a good conductor of electricity

Solution

Question 3

Lenclanche cell

Concentration cell

Fuel cell

Lead storage battery

Solution

Question 4

In dilute sulphuric acid solution, hydrogen will be reduced at cathode.

In concentrated sulphuric acid solution, water will be oxidised at anode.

In dilute sulphuric acid solution, water will be oxidised at anode.

In concentrated sulphuric acid solution, $$SO^{2-}_4$$ ion will be oxidized to tetrathionate ion at anode.

Solution

Question 5

2

4

1

3

Solution

Question 6

$$1.7 mg$$

$$3.4 mg$$

$$5.1 mg$$

$$6.8 mg$$

Solution

Question 7

0

1.5 gm

2.5 gm

2 gm

Solution

Question 8

positive

negative

zero

one

Solution

Question 9

FeO and Fe(OH)$$_3$$

FeO and Fe(OH)$$_2$$

Fe$$_2$$O$$_3$$ and Fe(OH)$$_3$$

Fe$$_3$$O$$_4$$ and Fe(OH)$$_2$$

Solution

Question 10

321660 coulombs

69500 coulombs

289500 coulombs

96500 coulombs

Solution

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