Electrochemistry Test

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Electrochemistry Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The figure shows a car bumper coated with chromium.

Why are car bumpers often coated with chromium?

A
(i) and (ii) only

B
(ii) and (iii) only

C
(iii) and (i) only

D
(i),(ii) and (iii) only

Solution

Car bumpers are coated with chromium because chromium forms an oxide layer and helps in the prevention of rust formation and corrosion.
Hence, option A is correct.
Question 2
1 / -0

1 faraday is defined as:

A
charge of one mole of electrons

B
smaller unit of charge

C
moles of electron involves in half reaction

D
none of the above

Solution

1 mol of electron $$\approx 96500 \ C$$ or 1 faraday.
Question 3
1 / -0

Which cell involves spontaneous oxidation and reduction?

A
Galvanic cell

B
Electrolytic cell

C
Fuel cell

D
Lead cell

Solution
Question 4
1 / -0

The given diagram shows an electrochemical cell in which the respective half cells contain aqueous 1.0 M solutions of the salts $$XCl_2$$ and $$YCl_3$$. Given that,

$$3X_{(s)} + 2Y^{3+}_{(aq)} \rightarrow 3X^{2+}_{(aq)} + 2Y_{(s)},\ E_{cell}> 0$$

Which of the following statement(s) is/are correct?

A
The electrode made from metal X has positive polarity.

B
Electrode Y is the anode.

C
The flow of electrons is from Y to X.

D
The reaction at electrode X is an oxidation reaction.

Solution

Oxidation occurs at the electrode X which is the anode and it has negative polarity.
Anode: $$X \rightarrow X^{+2}+2e^-$$
Cathode: $$Y^{3+}+3e^-\rightarrow Y$$
Here, the electron is flowing from X to Y.
Option D is correct.
Question 5
1 / -0

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is:
$$\mathrm{M}(\mathrm{s})|\mathrm{M}^{+}(\mathrm{a}\mathrm{q};0.05\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r})||\mathrm{M}^{+}(\mathrm{a}\mathrm{q})$$ , 1 $$\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r})|\mathrm{M}(\mathrm{s})$$
For the above electrolytic cell the magnitude of the cell potential $$|\mathrm{E}_{\mathrm{c}\mathrm{e}ll}|=70\mathrm{m}\mathrm{V}$$.

For the above cell:

A
$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}<0;\Delta \mathrm{G}>0$$

B
$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}>0;\Delta \mathrm{G}<0$$

C
$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}<0;\Delta \mathrm{G}^{\mathrm{o}}>0$$

D
$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}>0;\Delta \mathrm{G}^{\mathrm{o}}>0$$

Solution

$$\mathrm{M}(\mathrm{s})+\mathrm{M}{(\mathrm{a}\mathrm{q})1\mathrm{M}}^{+}\rightarrow
\mathrm{M}{(\mathrm{a}\mathrm{q})0.05\mathrm{M}}^{+}+\mathrm{M}(\mathrm{s})$$

According to Nernst equation,

$$\displaystyle\mathrm{E}_{\mathrm{c}\mathrm{e}ll}=0-\frac{2.303\mathrm{R}\mathrm{T}}{\mathrm{F}}\log\frac{\mathrm{M}_{05\mathrm{M}}^{+}}{\mathrm{M}_{1\mathrm{M}}^{+}}$$$$=0-\displaystyle \frac{2.303\mathrm{R}\mathrm{T}}{\mathrm{F}}\log(5\times 10^{-2})$$$$=+\mathrm{v}\mathrm{e}$$

Hence, $$|\mathrm{E}_{\mathrm{c}\mathrm{e}ll}|=\mathrm{E}_{\mathrm{c}\mathrm{e}ll}=0.70\mathrm{V}$$ and $$\Delta \mathrm{G}<0$$ for the feasibility of the reaction.

Option B is correct.
Question 6
1 / -0

What is $$E_{RP}^0$$ for the reaction : $$C{u^{2 + }} + 2e\xrightarrow{{}}Cu$$ in the half cell $$P{t_{{S^{2 - }}/CuS/Cu}}$$ if $$E_{C{u^{2 + }}/Cu}^0$$ is 0.34 V and $$K_{sp}$$ of $$CuS=10^{-35}$$ ?

A
0.34 V

B
-0.6925 V

C
+0.6925 V

D
-0.66 V
Question 7
1 / -0

In a voltaic cell, if iron and silver electrodes are connected with each other then current flows:

A
from iron to silver outside the cell

B
from silver to iron within the cell

C
from silver to iron outside the cell

D
current does not flow in this cell

Solution

In a voltaic cell, if iron and silver electrodes are connected with each other then current flows from silver to iron outside the cell.
Question 8
1 / -0

The electrode potentials for $$Cu^{2+}_{(aq)} + e^{-} \rightarrow Cu_{(aq)}^{+}$$ and $$Cu_{(aq)}^{+} + e^{-} \rightarrow Cu_{(s)}$$
Are $$+0.15 V$$ and $$+0.50 V$$ respectively.
The value of $$E_{Cu^{2+}/Cu}^{\circ}$$ will be

A
$$0.500 V$$

B
$$0.325 V$$

C
$$0.650 V$$

D
$$0.150 V$$

Solution

Hint: Gibb's Free energy ($$\Delta G^{\circ}$$) is an additive property.

Step 1 :
By question,
$$Cu^{2+}_{(aq)} + e^- \rightarrow Cu^+_{(aq)} \Rightarrow(1)$$ ; $$E^{\circ}_1 =+0.15\ V$$
$$Cu^+_{(aq)} + e^- \rightarrow Cu_{(s)} \Rightarrow(2)$$ ; $$E^{\circ}_2 = +50\ V$$
$$Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)} \Rightarrow(3)$$ ; $$E^{\circ}_3 =\ ?\ V$$

Formula of Gibb's free energy for Galvanic cell,
$$\Delta G^{\circ} = -nFE^{\circ}$$

Step 2:
Calculation of $$E^{\circ}_{Cu^{+2}/Cu}$$:
By adding equation $$(1)$$ and $$(2)$$, we get equation $$(3)$$ -

$$Cu^{2+}_{(aq)} + e^-\ \ \rightarrow Cu^+_{(aq)} $$
$$\underline{Cu^+_{(aq)} + e^-\ \ \rightarrow Cu_{(s)}}$$
$$Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$$

Now, $$\Delta G^{\circ}_3 =\Delta G^{\circ}_1 + \Delta G^{\circ}_2 $$

$$-nFE^{\circ}_3 = -nFE^{\circ}_1 + -nFE^{\circ}_2$$

$$-nFE^{\circ}_3 = -F(nE^{\circ}_1 + nE^{\circ}_2)$$

$$2\times E^{\circ}_3 = (1\times 0.15 + 1\times 0.50)$$

$$2\times E^{\circ}_3 = 0.65$$

$$E^{\circ}_3 = \dfrac{0.65}{2}$$

$$E^{\circ}_3 = 0.325\ V.$$

Final Step: Correct answer - (B) $$0.325\ V.$$
Question 9
1 / -0

In a daniel cell, if $$A(E^o = -0.76 \ C)$$ and $$B(E^o = -2.36 \ V)$$ half cells are taken then:

A
$$B$$ acts as anode

B
$$A$$ acts as anode

C
$$B$$ acts as cathode

D
cannot be predicted

Solution

Daniel cell converts chemical energy into electrical energy. Lower standard reduction potential is of anode and cathode has higher standard reduction potential. Emf of Daniel cell equals E$$_{red. (cathode)}$$ - E $$_{ red. (anode)}$$. Emf of cell has to be greater than zero for working of cell.
Question 10
1 / -0

At 300K, $$\Delta$$H for the reaction $$Zn(s) + 2AgCl \rightarrow ZnCl_2 + 2Ag$$, is $$- 218$$ kJ/mol while the emf of the cell was $$1.015V$$. $$(\dfrac{dE}{dT})_p$$ of the cell is :

A
$$-4.2\times 10^{-4}VK^{-1}$$

B
$$-3.81\times 10^{-4}VK^{-1}$$

C
$$0.11VK^{-1}$$

D
$$7.6\times 10^{-4}VK^{-1}$$

Solution

$$\Delta H = nF(\dfrac{dE}{dT})_pT-nFE_{cell}$$

$$-218\times 1000=2\times 96500\times 300(\dfrac{dE}{dT})_p-2\times 96500\times 1.015$$

$$(\dfrac{dE}{dT})_p=-3.81\times 10^{-4}VK^{-1}$$

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Electrochemistry Test - 55

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Electrochemistry Test - 55

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Question 1

(i) and (ii) only

(ii) and (iii) only

(iii) and (i) only

(i),(ii) and (iii) only

Solution

Question 2

charge of one mole of electrons

smaller unit of charge

moles of electron involves in half reaction

none of the above

Solution

Question 3

Galvanic cell

Electrolytic cell

Fuel cell

Lead cell

Solution

Question 4

The electrode made from metal X has positive polarity.

Electrode Y is the anode.

The flow of electrons is from Y to X.

The reaction at electrode X is an oxidation reaction.

Solution

Question 5

$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}<0;\Delta \mathrm{G}>0$$

$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}>0;\Delta \mathrm{G}<0$$

$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}<0;\Delta \mathrm{G}^{\mathrm{o}}>0$$

$$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}>0;\Delta \mathrm{G}^{\mathrm{o}}>0$$

Solution

Question 6

0.34 V

-0.6925 V

+0.6925 V

-0.66 V

Question 7

from iron to silver outside the cell

from silver to iron within the cell

from silver to iron outside the cell

current does not flow in this cell

Solution

Question 8

$$0.500 V$$

$$0.325 V$$

$$0.650 V$$

$$0.150 V$$

Solution

Question 9

$$B$$ acts as anode

$$A$$ acts as anode

$$B$$ acts as cathode

cannot be predicted

Solution

Question 10

$$-4.2\times 10^{-4}VK^{-1}$$

$$-3.81\times 10^{-4}VK^{-1}$$

$$0.11VK^{-1}$$

$$7.6\times 10^{-4}VK^{-1}$$

Solution

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