Electrochemistry Test

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Electrochemistry Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

The standard electrode potentials $$(E_{M^+/M}^{o})$$ of four metals A, B, C and D are -1.2 V, 0.6 V, 0.85 V and -0.76 V, respectively. The increasing sequence of deposition of metals on applying potential is :

A
A, C, B, D

B
B, D, C, A

C
C, B, D, A

D
D, A, B, C

Solution

The standard electrode potentials $$(E_{M^+/M}^{o})$$ of four metals A, B, C and D are $$-1.2 V, 0.6 V, 0.85 V and -0.76 V$$ respectively. The increasing sequence of deposition of metals on applying potential is $$C, B, D, A$$. $$C$$ is having maximum standard reduction potential of $$0.85 V$$ and $$ A$$ is having minimum standard reduction potential of $$-1.2 V$$. The, increasing sequence of deposition of metals on applying potential is equal to the decreasing sequence of the standard reduction potentials.
Question 2
1 / -0

Given, $$\mathrm{E}_{\mathrm{F}\mathrm{e}^{3+}/\mathrm{F}\mathrm{e}}^{o}= -0.036\mathrm{V},\ \mathrm{E}_{\mathrm{F}\mathrm{e}^{2+}/\mathrm{F}\mathrm{e}}^{o}= -0.439 \mathrm{V}$$. The value of standard electrode potential for the change, $$\mathrm{F}\mathrm{e}^{3+}(\mathrm{a}\mathrm{q})+\mathrm{e}^{-}\rightarrow \mathrm{F}\mathrm{e}^{2+}(\mathrm{a}\mathrm{q})$$ will be:

A
$$-0.403\mathrm{V}$$

B
$$0.385 \mathrm{V}$$

C
$$0.770\mathrm{V}$$

D
$$-0.270\mathrm{V}$$

Solution

$$\mathrm{F}\mathrm{e}^{3+}+3\mathrm{e}\rightarrow \mathrm{F}\mathrm{e},\ \mathrm{E_1}^{\mathrm{o}}=-0.036\mathrm{V}$$

$$\mathrm{F}\mathrm{e}^{2+}+2\mathrm{e}\rightarrow \mathrm{F}\mathrm{e},\ \mathrm{E_2}^{\mathrm{o}}=-0.439\mathrm{V}$$

$$\mathrm{F}\mathrm{e}^{3+}(\mathrm{a}\mathrm{q})+\mathrm{e}^{-}\rightarrow \mathrm{F}\mathrm{e}^{2+}(\mathrm{a}\mathrm{q}) ;\ \mathrm{E_n}^{\mathrm{o}}=?$$

$$\Delta G$$ is an additive property:

$$\therefore\ \Delta G^o_n = \Delta G^o_1 -\ \Delta G^o_2$$

$$\Delta G^o =\ nFE^o$$

Now,

$$\ n_nFE^o_n=\ n_1FE^o-\ n_2FE^o_2$$

$$\therefore E^o_n =\dfrac{3\times (-0.036)-2\times (-0.439)}{1}= 0.770\ V$$
Question 3
1 / -0

In the electrolytic cell, flow of electrons is from:

A
cathode to anode in solution

B
cathode to anode through external supply

C
cathode to anode through internal supply

D
anode to cathode through internal supply
Solution
Hint: Electrons flow from the negative to a positive end.

Step 1: A cell in which Electrical energy is used to cause a non-spontaneous oxidation and reduction reaction is called an Electrolytic cell.
The electrolytic cell requires an energy input to proceed.

Step 2: Electrolytic cell contains three parts Cathode, Anode, and electrolyte.
Positive electrode is called anode.
Negative electrode is called Cathode.
An electrolyte is a substance in an aqueous or molten form that dissociates into ions and conducts electricity.
In a electrolytic cell, electrons are transferred from cathode to anode through internal supply.

Final answer: Therefore, the correct option is "C".
Question 4
1 / -0

When $$0.1$$ mol $$MnO_4^{2-}$$ is oxidised the quantity of electricity required to completely $$MnO_4^{2-}$$ to $$MnO_4{^-}$$ is:

A
$$96500\ C$$

B
$$2\times 96500\ C$$

C
$$9650\ C$$

D
$$96.50\ C$$

Solution

The oxidation reaction is as shown below.
$$MnO_4^{2-} + e^-\rightarrow MnO_4^-$$

Oxidation of one mole of $$MnO_4^{2-}$$ requires 1 mole of electrons.

Hence, the oxidation of 0.1 mole of electrons will require 0.1 mole of electrons.
This corresponds to $$ 96500 \times 0.1= 9650 C$$.
Question 5
1 / -0

1C electricity deposits:

A
10.8 g of Ag

B
electrochemical equivalent of Ag

C
half of electrochemical equivalent of Ag

D
96500 g of Ag

Solution

From Faraday's first law, $$w=Z\times Q$$

When $$Q=1\ C$$

$$Z = \dfrac {108}{1} = 108$$

$$w=Z$$

Where $$Z =$$ electrochemical equivalent of Ag.

Hence, option B is correct.
Question 6
1 / -0

In a galvanic cell electron flow will be from:

A
negative electrode to positive electrode

B
positive electrode to negative electrode

C
there will be no flow of electrons

D
cathode to anode in the extenal circuit

Solution

Electron flows opposite to the direction of current,
i.e. from low to high voltage.
$$\therefore$$ Electron flows from negative to the positive electrode.

Hence, the correct option is $$\text{A}$$
Question 7
1 / -0

An aqueous solution of the following concentration of Acetic acid is the best conductor.

A
10$$^{1}$$M

B
10$$^{-3}$$M

C
10$$^{-1}$$M

D
10$$^{2}$$M

Solution

for weak electrolytes
$$\lambda_m=\frac{K}{C}$$
$$\therefore \lambda_m \alpha \frac{1}{C}$$
$$\therefore$$ The one with lowest concentration has highest molar conductance.
Hence, option B is correct.
Question 8
1 / -0

In a galvanic cell, the reactions taking place in the anodic half cell and the cathodic half cell will be:

A
reduction

B
oxidation

C
oxidation and reduction respectively

D
reduction and oxidation respectively

Solution

In a galvanic cell in anode half cell, oxidation of Zn takes place and in cathodic half cell
$$Cu^{2t}$$ gets reduced.
Hence, option C is correct.
Question 9
1 / -0

Rusting of iron is catalysed by which of the following:

A
$$Fe$$

B
$$Zn$$

C
O$$_{2}$$

D
$$H^{+}$$

Solution

The rusting or iron occurs in acidic medium. Hence it is catalyzed by $$ { H }^{ + } $$ ions.
Hence, correct option is D.
Question 10
1 / -0

$$ Cu(S) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq)+2Ag(S) $$
In the given reaction, the reduction half cell reaction is:

A
$$Cu$$ $$^{2+}$$ $$+$$ $$2e$$$$^{-}$$ $$\rightarrow $$ $$Cu$$

B
$$Cu$$ $$-$$ $$2e$$ $$^{-}$$$$\rightarrow $$$$Cu$$$$^{2+}$$

C
$$Ag$$ $$^{+}$$ $$+$$ e$$^{-}$$$$\rightarrow $$ $$Ag$$

D
$$Ag$$ $$-$$ $$e$$ $$^{-}$$$$\rightarrow $$ $$Ag$$ $$^{+}$$

Solution

$$Cu$$ is getting oxidised and $$Ag^+$$ is getting reduced

$$\therefore$$ The reduction half-cell is

$$Ag^+ + e^-\rightarrow Ag$$

Hence, the correct option is $$\text{C}$$

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Electrochemistry Test - 12

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Electrochemistry Test - 12

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Question 1

A, C, B, D

B, D, C, A

C, B, D, A

D, A, B, C

Solution

Question 2

$$-0.403\mathrm{V}$$

$$0.385 \mathrm{V}$$

$$0.770\mathrm{V}$$

$$-0.270\mathrm{V}$$

Solution

Question 3

cathode to anode in solution

cathode to anode through external supply

cathode to anode through internal supply

anode to cathode through internal supply

Solution

Question 4

$$96500\ C$$

$$2\times 96500\ C$$

$$9650\ C$$

$$96.50\ C$$

Solution

Question 5

10.8 g of Ag

electrochemical equivalent of Ag

half of electrochemical equivalent of Ag

96500 g of Ag

Solution

Question 6

negative electrode to positive electrode

positive electrode to negative electrode

there will be no flow of electrons

cathode to anode in the extenal circuit

Solution

Question 7

10$$^{1}$$M

10$$^{-3}$$M

10$$^{-1}$$M

10$$^{2}$$M

Solution

Question 8

reduction

oxidation

oxidation and reduction respectively

reduction and oxidation respectively

Solution

Question 9

$$Fe$$

$$Zn$$

O$$_{2}$$

$$H^{+}$$

Solution

Question 10

$$Cu$$ $$^{2+}$$ $$+$$ $$2e$$$$^{-}$$ $$\rightarrow $$ $$Cu$$

$$Cu$$ $$-$$ $$2e$$ $$^{-}$$$$\rightarrow $$$$Cu$$$$^{2+}$$

$$Ag$$ $$^{+}$$ $$+$$ e$$^{-}$$$$\rightarrow $$ $$Ag$$

$$Ag$$ $$-$$ $$e$$ $$^{-}$$$$\rightarrow $$ $$Ag$$ $$^{+}$$

Solution

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